Waves — Practice & Self Evaluation

The speed of sound in air depends on:

1. the amplitude of the sound
2. the frequency of the sound
3. the temperature of the air
4. the wavelength of the sound.

Answer

the temperature of the air

Reason — The speed of sound in a gas is given by $\text v = \sqrt{\dfrac{\gamma \text{RT}}{\text M}}$, so that $\text v \propto \sqrt{\text T}$. As the temperature of the air rises, the molecules move faster and the speed of sound increases. The amplitude, the frequency and the wavelength of the sound are not properties of the medium, so they do not decide the speed of sound in air.

The speed of a mechanical wave is determined by:

1. the medium through which it travels
2. its frequency
3. its wavelength
4. its amplitude.

Answer

the medium through which it travels

Reason — The speed of a mechanical wave depends only upon the properties of the medium through which it travels, that is, upon its elasticity and its inertia (density). For example, the speed of a transverse wave in a stretched string is $\text v = \sqrt{\dfrac{\text T}{\text m}}$ and that of sound in a gas is $\text v = \sqrt{\dfrac{\gamma \text P}{\text d}}$. Neither expression contains the frequency, the wavelength or the amplitude of the wave.

When two waves of slightly different frequencies are sounded together, the phenomenon of beats occurs. The beat frequency is equal to :

1. the sum of the two frequencies
2. the product of the two frequencies
3. the average of the two frequencies.
4. the difference between the two frequencies

Answer

the difference between the two frequencies

Reason — When two waves of nearly equal frequencies n₁ and n₂ are sounded together, the intensity of the resultant sound rises and falls (n₁ − n₂) times in one second. Hence the number of beats per second, that is, the beat frequency, is equal to the difference in the frequencies of the two sound-sources.

In stationary waves, the points of maximum amplitude are called:

1. nodes
2. antinodes
3. crests
4. troughs.

Answer

antinodes

Reason — In a stationary wave, the points at which the displacement of the medium particles is always maximum are called antinodes, and the points at which the displacement remains permanently zero are called nodes. At an antinode constructive interference takes place, so the amplitude there is maximum.

The wave equation for a progressive wave is y = a sin (ωt − kx). The term ω represents:

1. wave number
2. angular frequency
3. amplitude
4. wavelength.

Answer

angular frequency

Reason — In the equation of a progressive wave y = a sin (ωt − kx), the quantity ω is the angular frequency of the oscillating particles, and is related to the frequency n and the time-period T by

$\omega = 2\pi \text n = \dfrac{2\pi}{\text T}$

Here a is the amplitude and $\text k = \dfrac{2\pi}{\lambda}$ is the propagation constant, or wave number.

In a closed organ pipe, the fundamental frequency is determined by:

1. the length of the pipe
2. the diameter of the pipe
3. the frequency of the source
4. the temperature of the air.

Answer

the length of the pipe

Reason — In a closed organ pipe a node is formed at the closed end and an antinode at the open end, so that $\text l = \dfrac{\lambda}{4}$. Hence the fundamental frequency is

$\text n = \dfrac{\text v}{4\text l}$

which is decided by the length l of the pipe. The frequency of the source only selects which mode is excited; it does not decide the natural frequency of the pipe.

The phase difference between two points in a wave separated by one wavelength is:

1. 0
2. π/2 rad
3. π rad
4. 2π rad.

Answer

2π rad

Reason — The phase difference between two points whose path difference is Δx is

$\Delta \phi = \dfrac{2\pi}{\lambda} \times \Delta \text x$

Putting Δx = λ,

$\Delta \phi = \dfrac{2\pi}{\lambda} \times \lambda = 2\pi\ \text{radian}$

Thus, two particles separated by one wavelength are in the same phase of oscillation.

The speed of a wave on a string is given by $\sqrt{\dfrac{\text T}{\text m}}$. In this equation, m represents:

1. mass of the string
2. tension in the string
3. length of the string
4. mass per unit length of the string.

Answer

mass per unit length of the string

Reason — In the formula $\text v = \sqrt{\dfrac{\text T}{\text m}}$, T is the tension in the string and m is the mass per unit length, that is, the linear mass density of the string. It is not the mass of the whole string. If r be the radius of the string and d the density of its material, then m = πr²d.

The phenomenon of beats occurs when two waves:

1. are in phase
2. have slightly different frequencies
3. travel in opposite directions
4. have the same frequency but different amplitudes.

Answer

have slightly different frequencies

Reason — Beats are produced by the superposition of two waves of nearly equal frequencies travelling in the same direction. If the frequencies were exactly equal, the phase difference between the waves at any point would remain steady and the intensity of sound at that point would remain constant, so no beats would be heard.

The distance between two consecutive nodes in a stationary wave is:

1. equal to one wavelength
2. half a wavelength
3. quarter of a wavelength
4. one-third of a wavelength.

Answer

half a wavelength

Reason — In a stationary wave the nodes are situated at x = 0, $\dfrac{\lambda}{2}$, $\dfrac{2\lambda}{2}$, $\dfrac{3\lambda}{2}$, ... Hence the distance between two consecutive nodes, and also between two consecutive antinodes, is $\dfrac{\lambda}{2}$, while the distance between a node and its neighbouring antinode is $\dfrac{\lambda}{4}$.

In an open organ pipe, the fundamental frequency is:

1. directly proportional to the length of the pipe
2. inversely proportional to the length of the pipe
3. directly proportional to the square of the length of the pipe
4. independent of the length of the pipe.

Answer

inversely proportional to the length of the pipe

Reason — In an open organ pipe an antinode is formed at each end, so that $\text l = \dfrac{\lambda}{2}$. Hence the fundamental frequency is

$\text n = \dfrac{\text v}{2\text l} \qquad \Rightarrow \qquad \text n \propto \dfrac{1}{\text l}$

that is, the fundamental frequency is inversely proportional to the length of the pipe.

In a stationary wave, the points where the displacement is always zero are called:

1. crests
2. troughs
3. nodes
4. antinodes.

Answer

nodes

Reason — In a stationary wave certain points of the bounded medium, situated at equal distances, always remain in the position of rest, that is, their displacement remains zero. These points are called nodes, and destructive interference takes place there.

In a wave, the speed of propagation is affected by:

1. the frequency of the wave
2. the amplitude of the wave
3. the properties of the medium
4. the wavelength of the wave.

Answer

the properties of the medium

Reason — The speed of propagation of a wave is decided by the elasticity and the density of the medium alone. For example, $\text v = \sqrt{\dfrac{\text E}{\text d}}$ for a longitudinal wave. Changing the frequency of the source changes the wavelength in such a way that the product nλ, that is, the speed, remains the same.

The wavelength of a sound wave in air is determined by:

1. its amplitude
2. its frequency and the speed of sound in air
3. the temperature of the air
4. the density of the air.

Answer

its frequency and the speed of sound in air

Reason — The wavelength of a sound wave in air is the ratio of the speed of sound in air to its frequency,

$\lambda = \dfrac{\text v}{\text n}$

so it is decided by the frequency of the wave together with the speed of sound in air.

Note: The book's answer key prints option 3, the temperature of the air, but its own explanation states that the wavelength is the ratio of the speed of sound to its frequency, which is option 2. The temperature affects the wavelength only indirectly, through its effect on the speed of sound.

In a closed organ pipe, the first overtone corresponds to:

1. the fundamental frequency
2. the second harmonic
3. the third harmonic
4. the first harmonic.

Answer

the third harmonic

Reason — A closed organ pipe produces only odd harmonics, whose frequencies are in the ratio n₁ : n₂ : n₃ ... = 1 : 3 : 5 ... The tone next above the fundamental is therefore three times the fundamental frequency. This tone is called the first overtone or the third harmonic of the closed pipe.

A sound wave with a frequency of 300 Hz travels through a medium with a speed of 330 m/s. The wavelength of the sound wave is:

1. 1.1 m
2. 0.9 m
3. 1.2 m
4. 0.5 m.

Answer

0.9 m

Reason — Given, n = 300 Hz and v = 330 m/s. From v = nλ,

$\lambda = \dfrac{\text v}{\text n} = \dfrac{330\ \text{m/s}}{300\ \text{Hz}} = 1.1\ \text m$

Note: The book's answer key prints option (b), 0.9 m, but its own explanation carries out exactly this calculation and obtains 1.1 m. Hence the correct answer should be 1.1 m.

In an open organ pipe, the number of nodes for the second harmonic is:

1. One
2. Two
3. Three
4. Four.

Answer

Three

Reason — In an open organ pipe an antinode is formed at each of the two ends. In the second harmonic the length of the pipe is l = λ, so there are antinodes at x = 0, $\dfrac{\text l}{2}$ and l, that is, three antinodes, and nodes at $\dfrac{\text l}{4}$ and $\dfrac{3\text l}{4}$, that is, two nodes.

Note: The book's answer key prints option (c), Three, and its explanation states that the second harmonic of an open pipe has two antinodes and three nodes. In fact the counts are the other way round — three antinodes and two nodes — so the number of nodes is two and the correct answer should be Two.

In a stationary wave, the distance between two consecutive antinodes is:

1. half the wavelength
2. equal to the wavelength
3. twice the wavelength
4. a quarter of the wavelength.

Answer

half the wavelength

Reason — In a stationary wave the antinodes are situated at x = $\dfrac{\lambda}{4}$, $\dfrac{3\lambda}{4}$, $\dfrac{5\lambda}{4}$, ... Hence the distance between two consecutive antinodes is $\dfrac{\lambda}{2}$, which is the same as the distance between two consecutive nodes.

Two waves having slightly different frequencies and travelling in the same direction produce a phenomenon called:

1. resonance
2. diffraction
3. beats
4. doppler effect.

Answer

beats

Reason — When two sound waves of nearly equal frequencies travelling in the same direction superpose, the intensity of the resultant sound rises and falls alternately with time. This periodic rise and fall in the intensity of sound is called the phenomenon of beats.

The beat frequency between two sound waves of frequencies 1000 Hz and 1005 Hz is:

1. 1000 Hz
2. 1005 Hz
3. 5 Hz
4. 500 Hz.

Answer

5 Hz

Reason — The beat frequency is the difference between the frequencies of the two sound waves,

$\text n_1 - \text n_2 = 1005\ \text{Hz} - 1000\ \text{Hz} = 5\ \text{Hz}$

That is, five beats are heard in one second.

The phenomenon in which waves bend around obstacles or spread as they pass through openings is known as:

1. reflection
2. refraction
3. diffraction
4. interference.

Answer

diffraction

Reason — The bending of waves around obstacles, or their spreading out on passing through narrow openings, is called diffraction. It is the confirmative evidence of the wave nature of a disturbance.

In an organ pipe open at both ends, the wavelength of the fundamental mode is:

1. equal to the length of the pipe
2. twice the length of the pipe
3. four times the length of the pipe
4. half the length of the pipe.

Answer

twice the length of the pipe

Reason — In an organ pipe open at both ends there is an antinode at each end and a node in the middle. Hence for the fundamental mode,

$\text l = \dfrac{\lambda}{2} \qquad \Rightarrow \qquad \lambda = 2\text l$

that is, the wavelength of the fundamental mode is twice the length of the pipe.

The speed of sound in air increases when:

1. the temperature of the air decreases
2. the temperature of the air increases
3. the pressure of the air decreases
4. the density of the air increases.

Answer

the temperature of the air increases

Reason — The speed of sound in air is directly proportional to the square-root of its absolute temperature, $\text v \propto \sqrt{\text T}$. Numerically,

$\text v_\text t = (332 + 0.61\text t)\ \text{m s}^{-1}$

so the speed increases by about 0.61 m s^-1 for each °C rise in temperature. A change in pressure alone has no effect, since $\dfrac{\text P}{\text d}$ then remains constant.

The principle of superposition states that when two waves overlap:

1. their amplitudes add up
2. their frequencies add up
3. their wavelengths change
4. their velocities add up.

Answer

their amplitudes add up

Reason — According to the principle of superposition, when two or more waves overlap at a point, the resultant displacement at that point is the algebraic sum of the displacements produced by the individual waves,

$\text y = \text y_1 + \text y_2$

Each wave, however, retains its own frequency, wavelength and velocity even after the superposition.

In a string fixed at both ends, the fundamental mode of vibration corresponds to:

1. one antinode and two nodes
2. two antinodes and three nodes
3. one node and one antinode
4. two nodes and one antinode.

Answer

one antinode and two nodes

Reason — A string fixed at both ends must have a node at each of the two fixed ends. In the fundamental mode the string vibrates as a whole in one segment, with $\text l = \dfrac{\lambda}{2}$, so a single antinode is formed at the middle. Hence there are two nodes and one antinode.

In a sonometer experiment, the frequency of the vibrating string is proportional to:

1. the square of the length
2. the square root of the tension
3. the inverse of the mass per unit length
4. the inverse of the tension.

Answer

the square root of the tension

Reason — The fundamental frequency of a vibrating sonometer wire is

$\text n = \dfrac{1}{2\text l}\sqrt{\dfrac{\text T}{\text m}}$

For a given length and a given mass per unit length, $\text n \propto \sqrt{\text T}$. This is the law of tension of the transverse vibrations of a stretched string. The frequency is inversely proportional to the square-root of m, not to m itself.

The wave speed in a string will increase if:

1. the mass per unit length increases
2. the tension decreases
3. the tension increases
4. the amplitude increases.

Answer

the tension increases

Reason — The speed of a transverse wave in a stretched string is

$\text v = \sqrt{\dfrac{\text T}{\text m}}$

so the speed increases with increasing tension and decreases with increasing mass per unit length. The amplitude of the wave has no effect on its speed.

In a progressive wave, the velocity of the wave is related to its frequency and wavelength by:

1. v = nλ
2. v = n/λ
3. v = n²λ
4. v = λ/n.

Answer

v = nλ

Reason — In one time-period T the wave travels a distance equal to one wavelength λ. Hence the distance travelled in one second is

$\text v = \dfrac{\lambda}{\text T} = \text n \lambda$

that is, speed = frequency × wavelength.

In a sonometer experiment, the frequency of the vibrating string increases if :

1. the tension is decreased
2. the length of the string is increased
3. the tension is increased
4. the mass per unit length is increased.

Answer

the tension is increased

Reason — The frequency of a vibrating sonometer wire is

$\text n = \dfrac{1}{2\text l}\sqrt{\dfrac{\text T}{\text m}}$

Hence $\text n \propto \sqrt{\text T}$ and $\text n \propto \dfrac{1}{\text l}$ and $\text n \propto \dfrac{1}{\sqrt{\text m}}$. The frequency therefore increases only when the tension is increased; increasing the length or the mass per unit length lowers the frequency.

The number of harmonics present in a vibrating string depends on :

1. the length of the string
2. the tension in the string
3. the nature of the boundary conditions (whether the ends are fixed or free)
4. the amplitude of the wave.

Answer

the nature of the boundary conditions (whether the ends are fixed or free)

Reason — The boundary conditions decide which stationary waves can be set up in the medium. A string fixed at both ends must have a node at each end and so gives both even and odd harmonics in the ratio 1 : 2 : 3 : 4 ..., whereas a string fixed at one end and free at the other gives only the odd harmonics. The length and the tension decide the values of the frequencies, not which harmonics are permitted.

In an open organ pipe, the fundamental mode of vibration is characterized by:

1. one node and one antinode
2. two nodes
3. two antinodes
4. no nodes.

Answer

two antinodes

Reason — In an open organ pipe the air particles at both the ends have the greatest freedom to vibrate, so an antinode is formed at each end. In the fundamental mode there is one node in the middle of the pipe. Hence the fundamental mode is characterised by two antinodes, one at each open end.

When two waves meet and interfere constructively, the amplitude of the resulting wave:

1. increases
2. decreases
3. stays the same
4. becomes zero.

Answer

increases

Reason — In constructive interference the two waves meet in the same phase, that is, φ = 0. The resultant amplitude is then

$\text a_{max} = \text a_1 + \text a_2$

which is larger than the amplitude of either wave. Hence the amplitude of the resulting wave increases.

Two waves, described by y₁ = a sin (kx − ωt) and y₂ = a sin (kx + ωt), travel in opposite directions. The resultant wave:

1. is a travelling wave
2. is a stationary wave
3. has double the amplitude of the original waves
4. has zero displacement at all points.

Answer

is a stationary wave

Reason — The two waves have the same amplitude and the same frequency and travel in opposite directions. By the principle of superposition,

$\text y = \text y_1 + \text y_2 = \text a \sin (\text k\text x - \omega \text t) + \text a \sin (\text k\text x + \omega \text t) \\[1em] = 2\text a \sin \text k\text x \cos \omega \text t$

Here x and t occur in separate harmonic functions, so the resultant is a stationary wave. Its amplitude, 2a sin kx, varies from point to point and is not double the original amplitude everywhere.

If the length of an air column in a closed organ pipe is increased by a factor of 3, the fundamental frequency of the pipe will:

1. remain unchanged
2. increase by a factor of 3
3. decrease by a factor of 3
4. decrease by a factor of $\sqrt{3}$.

Answer

decrease by a factor of 3

Reason — The fundamental frequency of a closed organ pipe is

$\text n = \dfrac{\text v}{4\text l} \qquad \Rightarrow \qquad \text n \propto \dfrac{1}{\text l}$

If the length of the air column is made 3 l, the fundamental frequency becomes $\dfrac{\text v}{4(3\text l)} = \dfrac{\text n}{3}$, that is, it decreases by a factor of 3.

For two waves undergoing interference, if the phase difference between the waves is 2π, the resultant intensity will be:

1. maximum
2. zero
3. half of the original intensity
4. equal to the sum of the individual intensities.

Answer

maximum

Reason — A phase difference of 2π corresponds to a path difference of one full wavelength, so the two waves arrive at the point in the same phase. Constructive interference takes place, the resultant amplitude is a₁ + a₂, and since I ∝ a² the resultant intensity is maximum.

In progressive waves, the particles of the medium:

1. Move with the wave
2. Do not move
3. Oscillate about their mean positions
4. Move randomly.

Answer

Oscillate about their mean positions

Reason — In a progressive wave the energy is transferred from one point to another, but the medium itself does not travel along with the wave. Each particle of the medium merely oscillates about its own mean position, and only the phase of oscillation changes from particle to particle.

In a string of length l with fixed ends, the third harmonic has:

1. one node and two antinodes
2. three nodes and two antinodes
3. four nodes and three antinodes
4. two nodes and three antinodes.

Answer

four nodes and three antinodes

Reason — In a string of length l fixed at both ends, the pth mode of vibration has (p + 1) nodes and p antinodes. For the third harmonic p = 3, so the string vibrates in three segments with

$\text{nodes at}\ \text x = 0,\ \dfrac{\text l}{3},\ \dfrac{2\text l}{3},\ \text l$

that is, four nodes, and three antinodes between them.

The phase difference between two points separated by a distance equal to $\dfrac{\lambda}{4}$ in a progressive wave is:

1. $\dfrac{\pi}{2}$
2. π radian
3. 2π radian
4. $\dfrac{\pi}{3}$ radian.

Answer

$\dfrac{\pi}{2}$

Reason — The phase difference corresponding to a path difference Δx is

$\Delta \phi = \dfrac{2\pi}{\lambda} \times \Delta \text x$

Putting $\Delta \text x = \dfrac{\lambda}{4}$,

$\Delta \phi = \dfrac{2\pi}{\lambda} \times \dfrac{\lambda}{4} = \dfrac{\pi}{2}\ \text{radian}$

In a standing wave pattern on a string, if the distance between a node and the adjacent antinode is 0.25 m, the wavelength of the wave is:

1. 0.5 m
2. 1 m
3. 2 m
4. 4 m.

Answer

1 m

Reason — The distance between a node and the adjacent antinode in a stationary wave is $\dfrac{\lambda}{4}$. Given that this distance is 0.25 m,

$\dfrac{\lambda}{4} = 0.25\ \text m \qquad \Rightarrow \qquad \lambda = 4 \times 0.25$

$\lambda = 1\ \text m$

In a progressive wave, if the amplitude at a certain point becomes zero, it indicates that:

1. the wave has reached a node
2. the wave has reached an antinode
3. the wave is undergoing destructive interference
4. the wave is undergoing constructive interference.

Answer

the wave is undergoing destructive interference

Reason — A progressive wave advances in the medium with a definite velocity and all its particles vibrate with the same amplitude, so it has no permanent points of zero amplitude. If the amplitude at a certain point becomes zero, it means that another wave has arrived there in the opposite phase and the two have cancelled each other. Hence the wave is undergoing destructive interference.

The wave velocity in a vibrating string depends on:

1. tension and mass per unit length
2. frequency and amplitude
3. length of the string
4. number of nodes formed.

Answer

tension and mass per unit length

Reason — The speed of a transverse wave in a vibrating string is

$\text v = \sqrt{\dfrac{\text T}{\text m}}$

so it depends only on the tension T in the string and the mass per unit length m of the string. It does not depend upon the frequency, the amplitude, the length of the string or the number of nodes formed.

In the phenomenon of beats, the amplitude of the resultant wave:

1. varies periodically
2. remains constant
3. becomes zero
4. increases linearly with time.

Answer

varies periodically

Reason — In the phenomenon of beats the resultant displacement is

$\text y = \left[2\text a \cos \pi (\text n_1 - \text n_2)\text t\right]\sin \pi (\text n_1 + \text n_2)\text t$

The quantity within the brackets is the amplitude, and it depends upon t. Hence the amplitude of the resultant wave varies periodically with time between 2a and zero, giving alternate constructive and destructive interference.

In a string fixed at one end and free at the other, the fundamental frequency n is given by:

1. $\dfrac{\text v}{4\text l}$

2. $\dfrac{\text v}{2\text l}$

3. $\dfrac{\text v}{\text l}$

4. v × 2l.

Answer

$\dfrac{\text v}{4\text l}$

Reason — A string fixed at one end and free at the other has a node at the fixed end and an antinode at the free end. Hence in the fundamental mode

$\text l = \dfrac{\lambda}{4} \qquad \Rightarrow \qquad \lambda = 4\text l$

so the fundamental frequency is

$\text n = \dfrac{\text v}{\lambda} = \dfrac{\text v}{4\text l}$

In a sonometer experiment, if the tension in the wire is quadrupled, keeping other factors fixed, the frequency of the vibrating wire will:

1. Remain unchanged
2. Double
3. Increase by a factor of $\sqrt{2}$
4. Increase by a factor of 4.

Answer

Double

Reason — The frequency of a vibrating sonometer wire is

$\text n = \dfrac{1}{2\text l}\sqrt{\dfrac{\text T}{\text m}} \qquad \Rightarrow \qquad \text n \propto \sqrt{\text T}$

If the tension is made 4 T, the new frequency is

$\text n' = \dfrac{1}{2\text l}\sqrt{\dfrac{4\text T}{\text m}} = 2\left(\dfrac{1}{2\text l}\sqrt{\dfrac{\text T}{\text m}}\right) = 2\text n$

that is, the frequency is doubled.

If two tuning forks of frequencies 256 Hz and 260 Hz are sounded together, the beat frequency observed will be:

1. 516 Hz
2. 258 Hz
3. 4 Hz
4. 2 Hz.

Answer

4 Hz

Reason — The number of beats heard per second is equal to the difference in the frequencies of the two tuning forks,

$\text n_1 - \text n_2 = 260\ \text{Hz} - 256\ \text{Hz} = 4\ \text{Hz}$

A progressive wave is described by the equation y = 0.05 sin (4πx − 200 πt), where x is in meters and t in seconds. The wavelength of the wave is:

1. 0.5 m
2. 1 m
3. 2 m
4. 4 m.

Answer

0.5 m

Reason — Comparing y = 0.05 sin (4πx − 200 πt) with the standard equation y = a sin (kx − ωt),

$\text k = \dfrac{2\pi}{\lambda} = 4\pi$

$\lambda = \dfrac{2\pi}{4\pi} = 0.5\ \text m$

In a sonometer, if the length of the vibrating string is halved, keeping other factors constant, the fundamental frequency will:

1. remain unchanged
2. double
3. increase by a factor of $\sqrt{2}$
4. become zero.

Answer

double

Reason — The fundamental frequency of a sonometer wire is

$\text n = \dfrac{1}{2\text l}\sqrt{\dfrac{\text T}{\text m}} \qquad \Rightarrow \qquad \text n \propto \dfrac{1}{\text l}$

This is the law of length. If the vibrating length is halved, the new frequency is

$\text n' = \dfrac{1}{2\left(\dfrac{\text l}{2}\right)}\sqrt{\dfrac{\text T}{\text m}} = 2\text n$

that is, the fundamental frequency is doubled.

If the amplitude of a wave is doubled, the energy carried by the wave becomes:

1. doubled
2. quadrupled
3. halved
4. unchanged.

Answer

quadrupled

Reason — The energy carried by a wave is directly proportional to the square of its amplitude,

$\text E \propto \text a^2$

If the amplitude is doubled, the energy becomes (2)² = 4 times its original value, that is, it is quadrupled.

Two waves are given by y₁ = a sin (kx − ωt) and y₂ = a cos (kx − ωt). The resultant wave will have an amplitude of :

1. a
2. 2a
3. $\text a\sqrt{2}$
4. a/2.

Answer

$\text a\sqrt{2}$

Reason — The second wave may be written as

$\text y_2 = \text a \cos (\text k\text x - \omega \text t) = \text a \sin \left(\text k\text x - \omega \text t + \dfrac{\pi}{2}\right)$

so the phase difference between the two waves is $\phi = \dfrac{\pi}{2}$, that is, 90°. The resultant amplitude is

$\text A = \sqrt{\text a^2 + \text a^2 + 2\text a \times \text a \cos 90^\circ} = \sqrt{2\text a^2}$

$\text A = \text a\sqrt{2}$

Assertion (A): The speed of sound is greater in solids than in gases.

Reason (R): In solids, particles are closely packed, allowing for faster transmission of energy.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The speed of a longitudinal wave in a medium is $\text v = \sqrt{\dfrac{\text E}{\text d}}$, where E is the modulus of elasticity. Solids possess a very large modulus of elasticity, so the speed of sound is minimum in gases, more in liquids and maximum in solids.

Reason (R) is also correct: In a solid the particles are closely packed and are held in place by strong elastic forces, so a disturbance is passed on from one particle to the next very rapidly.

Because the particles are closely spaced and strongly bound, the energy is transferred more quickly between them, which is exactly why the speed of sound is greater in solids than in gases. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): The speed of a wave on a string increases with increasing tension.

Reason (R): The wave speed on a string is directly proportional to the square root of tension in the string.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: On increasing the tension in a stretched string, the transverse wave travels along it with a greater speed.

Reason (R) is also correct: The speed of a transverse wave in a stretched string is

$\text v = \sqrt{\dfrac{\text T}{\text m}} \qquad \Rightarrow \qquad \text v \propto \sqrt{\text T}$

for a string of given mass per unit length. Since the speed varies directly as the square-root of the tension, an increase in T necessarily increases v. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): The speed of sound in air increases with an increase in temperature.

Reason (R): At higher temperatures, air molecules move faster, increasing the rate of energy transfer.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: From $\text v = \sqrt{\dfrac{\gamma \text{RT}}{\text M}}$ it follows that $\text v \propto \sqrt{\text T}$, so the speed of sound in air increases with a rise in temperature. Numerically, v_t = (332 + 0.61 t) m s^-1.

Reason (R) is also correct: At higher temperatures the molecules of air move faster, so a disturbance is handed on from one molecule to the next more rapidly and the rate of energy transfer increases.

The faster molecular motion at a higher temperature is the physical cause of the greater speed of sound, so the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): A progressive wave transfers energy from one point to another.

Reason (R): In a progressive wave, the displacement of particles is perpendicular to the direction of wave propagation.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: A progressive wave advances in the medium with a definite velocity and transmits energy from one point to another, although the particles of the medium merely oscillate about their mean positions.

Reason (R) is also correct, but only for one kind of progressive wave: In a transverse progressive wave the displacement of the particles is perpendicular to the direction of wave propagation.

A progressive wave may be transverse or longitudinal, and a longitudinal progressive wave, in which the particles oscillate along the direction of propagation, also transfers energy. Hence the perpendicular displacement is not the reason for the transfer of energy. The Reason therefore does not explain the Assertion.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

Assertion (A): In stationary waves, no energy is transferred across the medium.

Reason (R): In stationary waves, nodes and antinodes are formed due to the superposition of two waves travelling in opposite directions.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Stationary waves do not advance in the medium, and so they do not transmit energy across it. The energy remains confined within the segments between the nodes.

Reason (R) is also correct: Stationary waves are formed by the superposition of two identical waves travelling in opposite directions with the same speed, and this superposition produces permanently fixed nodes and antinodes.

Because the two component waves carry equal energy in opposite directions, and because the nodes remain permanently at rest, there is no net transfer of energy across the medium. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): A wave with a higher frequency will have a shorter wavelength if the speed remains constant.

Reason (R): The speed of a wave is directly proportional to its frequency.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: From v = nλ, at a constant speed $\lambda \propto \dfrac{1}{\text n}$. Hence a wave of higher frequency has a shorter wavelength.

Reason (R) is incorrect: The speed of a wave is the product of its frequency and its wavelength, v = nλ, and it is decided by the properties of the medium alone. It is not directly proportional to the frequency; if the frequency of the source is changed, the wavelength changes in such a way that the speed remains unaltered.

Therefore, assertion is true but reason is false.

Assertion (A): A string with a higher linear mass density will have a slower wave speed for the same tension.

Reason (R): Wave speed on a string is inversely proportional to the square root of the linear mass density.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For a given tension, a string of larger linear mass density carries the transverse wave with a smaller speed.

Reason (R) is also correct: The speed of a transverse wave in a stretched string is

$\text v = \sqrt{\dfrac{\text T}{\text m}} \qquad \Rightarrow \qquad \text v \propto \dfrac{1}{\sqrt{\text m}}$

Since the speed varies inversely as the square-root of the linear mass density, a larger value of m gives a smaller value of v. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): Superposition of two waves can result in either constructive or destructive interference.

Reason (R): The resultant displacement at any point is the sum of the displacements of the individual waves.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When two waves superpose in the same phase the resultant amplitude is a₁ + a₂, which is constructive interference, and when they superpose in opposite phases the resultant amplitude is a₁ ~ a₂, which is destructive interference.

Reason (R) is also correct: By the principle of superposition, the resultant displacement at any point is the algebraic sum of the displacements produced by the individual waves,

$\text y = \text y_1 + \text y_2$

Whether this algebraic sum is large or small depends upon the phase difference between the waves, which is precisely why both constructive and destructive interference are possible. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): A sound wave is a longitudinal wave.

Reason (R): In a longitudinal wave, the particles of the medium oscillate parallel to the direction of wave propagation.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Sound waves in air propagate in the form of compressions and rarefactions, and are therefore longitudinal waves.

Reason (R) is also correct: In a longitudinal wave the particles of the medium oscillate parallel to the direction of wave propagation.

In a sound wave the air molecules move back and forth along the direction in which the sound advances, producing the compressions and rarefactions. This is exactly the defining property of a longitudinal wave, so the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): Beats are produced when two sound waves of slightly different frequencies interfere.

Reason (R): The beat frequency is the difference between the two frequencies of the interfering waves.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When two sound waves of nearly equal frequencies are sounded together, the intensity of the resultant sound rises and falls alternately, producing beats.

Reason (R) is also correct: The number of beats heard per second is equal to the difference in the frequencies of the two interfering waves, that is, beat frequency = n₁ ~ n₂.

If the frequencies were exactly equal, the difference would be zero and the phase difference at any point would remain steady, so no beats would be heard. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): In organ pipes, the fundamental frequency is higher for shorter pipes.

Reason (R): The wavelength of the fundamental frequency is inversely proportional to the length of the pipe.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For an organ pipe the fundamental frequency is $\text n = \dfrac{\text v}{2\text l}$ for an open pipe and $\text n = \dfrac{\text v}{4\text l}$ for a closed pipe, so in either case $\text n \propto \dfrac{1}{\text l}$. Hence a shorter pipe gives a higher fundamental frequency.

Reason (R) is also correct: For a given type of pipe the wavelength of the fundamental mode is a fixed multiple of the length of the pipe, so a shorter pipe supports a shorter wavelength and hence a higher frequency.

The Reason therefore explains the relationship between the length of the pipe and its fundamental frequency.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): A closed organ pipe produces a fundamental frequency that is half the wavelength of the pipe's length.

Reason (R): The closed end of the organ pipe acts as a node, and the open end acts as an antinode.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For a closed organ pipe, in the fundamental mode

$\text l = \dfrac{\lambda}{4}$

that is, the length of the pipe is one-fourth of the wavelength of the fundamental note.

Reason (R) is also correct: The closed end of the pipe acts as a node, because the air particles there have no freedom to vibrate, and the open end acts as an antinode, because the air particles there have the greatest freedom to vibrate.

These very boundary conditions fix the relation between the length of the pipe and the wavelength of the fundamental tone, so the Reason explains the Assertion.

Note: The Assertion as printed reads "a fundamental frequency that is half the wavelength of the pipe's length", which compares a frequency with a length and is therefore not dimensionally meaningful. It has been read here in the sense intended by the book's own explanation, namely the relation between the length of the pipe and the wavelength of the fundamental tone.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): In a vibrating string, the frequency of the wave increases if the tension is increased.

Reason (R): The wave speed on a string is directly proportional to the square root of the tension.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The frequency of a vibrating string is

$\text n = \dfrac{1}{2\text l}\sqrt{\dfrac{\text T}{\text m}}$

so the frequency increases when the tension is increased.

Reason (R) is also correct: The speed of a wave on a string is $\text v = \sqrt{\dfrac{\text T}{\text m}}$, that is, it is directly proportional to the square-root of the tension.

Since $\text n = \dfrac{\text v}{2\text l}$, and the length of the string is unchanged, an increase in the wave speed produced by the greater tension directly raises the frequency. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): The amplitude of a stationary wave is maximum at the nodes.

Reason (R): Nodes are points where destructive interference occurs, resulting in no displacement.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: The nodes are the points of a stationary wave at which the displacement of the particles remains permanently zero. The amplitude is maximum at the antinodes, not at the nodes.

Reason (R) is correct: At a node the two component waves always arrive in opposite phases, so destructive interference takes place there and the resultant displacement is always zero.

Therefore, assertion is false but reason is true.

Assertion (A): In a progressive wave, particles of the medium transfer energy without transferring matter.

Reason (R): Progressive waves involve the oscillation of particles around their equilibrium positions.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A progressive wave carries energy from one point of the medium to another, but the medium itself is not carried forward along with the wave.

Reason (R) is also correct: In a progressive wave each particle of the medium merely oscillates about its own equilibrium position, only the phase of the oscillation changing from particle to particle.

Since every particle returns to its equilibrium position after each oscillation, there is no permanent displacement of matter, and yet the disturbance, and with it the energy, is handed on from particle to particle. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): In a closed organ pipe, only odd harmonics are produced.

Reason (R): The closed end of the pipe acts as a node, and the open - end acts as an antinode, allowing only odd harmonics.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In a closed organ pipe the frequencies of the fundamental tone and the overtones are in the ratio n₁ : n₂ : n₃ ... = 1 : 3 : 5 ..., that is, only the odd harmonics are produced.

Reason (R) is also correct: The closed end of the pipe acts as a node and the open end as an antinode, so that

$\lambda = \dfrac{4\text l}{2\text m - 1}, \qquad \text m = 1, 2, 3, \dots$

These boundary conditions permit only those wavelengths for which the length of the pipe is an odd multiple of $\dfrac{\lambda}{4}$, and therefore permit only the odd harmonics. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): The principle of superposition applies only to mechanical waves.

Reason (R): Superposition occurs when two or more waves overlap and the resultant displacement is the sum of individual displacements.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: The principle of superposition applies to all types of waves, including light waves, water waves and sound waves, provided the medium behaves linearly. It is not restricted to mechanical waves.

Reason (R) is correct: Superposition occurs when two or more waves overlap at a point, and the resultant displacement there is the algebraic sum of the displacements produced by the individual waves.

Therefore, assertion is false but reason is true.

Assertion (A): In a sound wave, compression corresponds to the maximum displacement of air molecules.

Reason (R): In a longitudinal wave, compressions are regions of high pressure and density.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) as printed is not correct: In a longitudinal sound wave a compression is a region of maximum pressure and maximum density, and it is a displacement node — the particles there have zero displacement. The maximum displacement of the air molecules occurs midway between a compression and a rarefaction.

Reason (R) is correct: In a longitudinal wave the compressions are indeed the regions of high pressure and high density.

Note: The book's answer key marks option (b), and its own printed explanation states that maximum displacement occurs between compressions and rarefactions — which shows the Assertion to be false. On that reading the correct choice would be If assertion is false but reason is true.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

Assertion (A): In S.H.M., the velocity of a particle is maximum at the equilibrium position.

Reason (R): The acceleration is zero at the equilibrium position in S.H.M.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In S.H.M. the velocity of the particle is $\text v = \omega\sqrt{\text a^2 - \text y^2}$, which is maximum, equal to ωa, at the equilibrium position where y = 0.

Reason (R) is also correct: The acceleration in S.H.M. is a_acc. = − ω²y, which is zero at the equilibrium position where y = 0.

Since the restoring force, and hence the retardation, vanishes at the equilibrium position, the particle has been accelerated over the whole of its journey up to that point and there attains its greatest speed. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): Beats can only occur with sound waves.

Reason (R): Beats are the result of interference between two waves of slightly different frequencies.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: Beats can be produced with any type of wave whose frequencies differ slightly and which can superpose. They are not restricted to sound waves alone.

Reason (R) is correct: Beats are indeed the result of the interference of two waves of slightly different frequencies, the intensity of the resultant rising and falling alternately with time.

Therefore, assertion is false but reason is true.

Assertion (A): The frequency of a wave depends only on the source of the wave and not on the medium.

Reason (R): The speed of a wave depends on the properties of the medium through which it travels.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: The frequency of a wave is decided by the source producing it and remains unchanged when the wave passes from one medium into another. Only the speed and the wavelength change.

Reason (R) is also correct: The speed of a wave is determined by the elasticity and the density of the medium through which it travels.

The two statements are about different quantities. The fact that the speed depends upon the medium does not by itself explain why the frequency is fixed by the source; frequency is a characteristic of the wave-emitting source. Hence the Reason does not explain the Assertion.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

Assertion (A): The intensity of a sound wave decreases as it moves away from the source.

Reason (R): The energy of the wave spreads over a larger area as the distance from the source increases.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The intensity of a sound wave goes on decreasing as the wave travels away from the source.

Reason (R) is also correct: The energy sent out by the source spreads over a spherical surface whose area increases as the wave advances, so

$\text I = \dfrac{\text P}{4\pi \text r^2}$

where P is the power of the source and r the distance from it. Since the same energy is distributed over a larger and larger area, the intensity falls off as $\dfrac{1}{\text r^2}$. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): A standing wave is formed when two waves of the same frequency and amplitude travel in the same direction.

Reason (R): Standing waves are a result of constructive and destructive interference.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: A standing wave is formed when two identical waves travel in a bounded medium in opposite directions and superpose. Two waves travelling in the same direction can only produce interference or beats, never a standing wave.

Reason (R) is correct: Standing waves are indeed the result of constructive interference at the antinodes and destructive interference at the nodes.

Therefore, assertion is false but reason is true.

Assertion (A): The first harmonic of a vibrating string is also called the fundamental frequency.

Reason (R): The first harmonic corresponds to the longest wavelength that can form on the string.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The lowest frequency of vibration of a stretched string is called its fundamental frequency, and the corresponding tone is the first harmonic.

Reason (R) is also correct: In the first mode the string vibrates as a whole in a single segment, so that

$\text l = \dfrac{\lambda_1}{2} \qquad \Rightarrow \qquad \lambda_1 = 2\text l$

which is the longest wavelength that the string can support.

Since $\text n = \dfrac{\text v}{\lambda}$, the longest wavelength corresponds to the lowest frequency. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): The speed of sound in helium is higher than in air.

Reason (R): The speed of sound in a gas depends on its molecular weight, and helium has a lower molecular weight than air.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: At the same temperature the speed of sound in helium is greater than that in air.

Reason (R) is also correct: The speed of sound in a gas is

$\text v = \sqrt{\dfrac{\gamma \text{RT}}{\text M}} \qquad \Rightarrow \qquad \text v \propto \dfrac{1}{\sqrt{\text M}}$

that is, it is inversely proportional to the square-root of the molecular mass of the gas.

Helium has a molecular mass of 4, which is much less than the average molecular mass of air, so the speed of sound in helium comes out to be greater. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): The fundamental frequency of an open organ pipe is lower than that of a closed organ pipe of the same length.

Reason (R): The fundamental frequency of a closed pipe is half that of an open pipe of the same length.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: For pipes of the same length, the fundamental frequency of an open pipe is $\dfrac{\text v}{2\text l}$ and that of a closed pipe is $\dfrac{\text v}{4\text l}$. Hence the fundamental frequency of an open pipe is twice, that is, higher than, that of a closed pipe of the same length, not lower.

Reason (R) is correct: The fundamental frequency of a closed pipe is indeed half that of an open pipe of the same length,

$\dfrac{\text n_{closed}}{\text n_{open}} = \dfrac{\text v/4\text l}{\text v/2\text l} = \dfrac{1}{2}$

Therefore, assertion is false but reason is true.

Assertion (A): The amplitude of a sound wave is directly proportional to its speed.

Reason (R): The amplitude of a wave determines its energy, but not its speed.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: The speed of a sound wave is decided by the elasticity and the density of the medium alone, and is completely independent of the amplitude of the wave.

Reason (R) is correct: The amplitude of a wave determines the energy it carries, since E ∝ a², but it has no effect on the speed with which the wave travels.

Therefore, assertion is false but reason is true.

Assertion (A): The frequency of beats increases as the frequency difference between two interfering sound waves decreases.

Reason (R): Beat frequency is the absolute difference between the frequencies of the two interfering waves.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: The beat frequency is n₁ ~ n₂. Hence the frequency of the beats decreases, and not increases, as the frequency difference between the two interfering waves decreases.

Reason (R) is correct: The beat frequency is indeed the absolute difference between the frequencies of the two interfering waves.

Therefore, assertion is false but reason is true.

Assertion (A): In a stationary wave, the points of zero displacement are called antinodes.

Reason (R): In a stationary wave, energy is transferred between nodes and antinodes.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: In a stationary wave the points of zero displacement are called nodes. The antinodes are the points of maximum displacement.

Reason (R) is correct as regards the confinement of energy: In a stationary wave the energy remains confined within the segments and merely changes back and forth between the kinetic and the potential forms; there is no transfer of energy across the medium from one segment to another.

Therefore, assertion is false but reason is true.

Assertion (A): The wavelength of a sound wave decreases as it moves from air to water.

Reason (R): The speed of sound in water is higher than in air.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is incorrect as printed: The frequency of a sound wave is fixed by the source and does not change on passing from air into water. Since v = nλ and the speed of sound in water is greater than that in air, the wavelength increases on going from air to water, and does not decrease.

Reason (R) is correct: The speed of sound in water, about 1450 m s^-1, is indeed much greater than that in air, about 332 m s^-1, because water has a far greater bulk modulus.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

Note: The book's answer key marks option (b), and its printed explanation states that the wavelength increases rather than decreases, which makes the Assertion false. On that reading the correct choice would be If assertion is false but reason is true.

Assertion (A): The speed of a wave on a string is independent of the frequency of the wave.

Reason (R): The speed of a wave on a string depends on the tension and the linear mass density of the string.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The speed of a wave on a string does not change when the frequency of the source vibrating it is changed; only the wavelength changes so that the product nλ stays the same.

Reason (R) is also correct: The speed of a wave on a string is

$\text v = \sqrt{\dfrac{\text T}{\text m}}$

which contains only the tension in the string and its linear mass density.

Since the expression for the speed involves no term of frequency, the speed is necessarily independent of the frequency. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): In S.H.M., the potential energy is maximum at the equilibrium position.

Reason (R): The potential energy in S.H.M. is proportional to the square of the displacement.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: In S.H.M. the potential energy is $\text U = \dfrac{1}{2}\text m\omega^2 \text y^2$, which is zero at the equilibrium position where y = 0, and maximum at the extreme positions where y = ± a.

Reason (R) is correct: The potential energy in S.H.M. is indeed proportional to the square of the displacement from the mean position.

Therefore, assertion is false but reason is true.

Assertion (A): The superposition of two waves always results in constructive interference.

Reason (R): Constructive interference occurs when the waves are in phase.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: The superposition of two waves may result in either constructive or destructive interference, depending upon the phase difference between them. The resultant amplitude lies anywhere between a₁ + a₂ and a₁ ~ a₂.

Reason (R) is correct: Constructive interference does occur when the two waves meet in the same phase, that is, when φ = 0, and then a_max = a₁ + a₂.

Therefore, assertion is false but reason is true.

Assertion (A): In a closed pipe, the fundamental frequency is higher than the second harmonic.

Reason (R): The second harmonic does not exist in a closed pipe.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In a closed organ pipe the frequencies present are n₁, 3n₁, 5n₁, ..., so the tone next above the fundamental is the third harmonic. Since the second harmonic 2n₁ is missing altogether, the fundamental is the lowest frequency the pipe can produce and no note of frequency 2n₁ exists below or above it.

Reason (R) is also correct: The second harmonic does not exist in a closed pipe, because the boundary conditions — a node at the closed end and an antinode at the open end — permit only those modes for which the length is an odd multiple of $\dfrac{\lambda}{4}$.

The absence of the second harmonic is thus the very reason for the statement made in the Assertion, so the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): The speed of mechanical waves in a medium depends on the frequency of the waves.

Reason (R): The wave speed in a medium is determined by the properties of the medium, not by the frequency.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: The speed of a mechanical wave in a medium does not depend upon the frequency of the wave. Sound waves of different frequencies travel through air with the same speed, although their wavelengths in air are different.

Reason (R) is correct: The wave speed in a medium is determined by the properties of the medium, that is, by its elasticity and its density, and not by the frequency.

Therefore, assertion is false but reason is true.

Assertion (A): The velocity of a particle in S.H.M. is zero at the equilibrium position.

Reason (R): In S.H.M., velocity is maximum at the extreme positions.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: In S.H.M. the velocity of the particle is maximum, equal to ωa, at the equilibrium position, and it is zero at the extreme positions.

Reason (R) is correct in the sense that it identifies where the velocity is not maximum, but as printed it too is wrong: the velocity in S.H.M. is zero, and not maximum, at the extreme positions.

Therefore, assertion is false but reason is true.

Note: The book's answer key marks option (d). Strictly, both the Assertion and the Reason as printed are incorrect statements about S.H.M., since the velocity is maximum at the equilibrium position and zero at the extreme positions.

Assertion (A): In an open organ pipe, all harmonics are possible.

Reason (R): Both ends of an open organ pipe act as antinodes.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In an open organ pipe the frequencies of the fundamental tone and the overtones are in the ratio n₁ : n₂ : n₃ ... = 1 : 2 : 3 ..., that is, both the even and the odd harmonics are produced.

Reason (R) is also correct: Both the ends of an open organ pipe are open to the atmosphere, so the air particles there have the greatest freedom to vibrate and an antinode is formed at each end.

These boundary conditions give $\lambda = \dfrac{2\text l}{\text m}$ with m = 1, 2, 3, ..., which permits every integral harmonic. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): A node is a point on a stationary wave where the displacement is maximum.

Reason (R): At a node, destructive interference results in zero displacement.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: A node is a point of a stationary wave at which the displacement is always zero, not maximum. The points of maximum displacement are the antinodes.

Reason (R) is correct: At a node the two component waves always arrive in opposite phases, so destructive interference takes place and the resultant displacement there is zero.

Therefore, assertion is false but reason is true.

Assertion (A): The first harmonic of a string is also called the second overtone.

Reason (R): Overtones are numbered starting from the second harmonic.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: The first harmonic of a string is the fundamental tone itself. The second overtone is the third harmonic, whose frequency is three times the fundamental frequency.

Reason (R) is correct: The overtones are numbered from the second harmonic onwards — the second harmonic is the first overtone, the third harmonic is the second overtone, and so on.

Therefore, assertion is false but reason is true.

Assertion (A): The fundamental frequency of a vibrating string decreases as the length of the string increases.

Reason (R): The frequency is inversely proportional to the length of the vibrating string.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: As the vibrating length of a string is increased, the fundamental frequency of the note emitted by it falls.

Reason (R) is also correct: The fundamental frequency of a stretched string is

$\text n = \dfrac{1}{2\text l}\sqrt{\dfrac{\text T}{\text m}} \qquad \Rightarrow \qquad \text n \propto \dfrac{1}{\text l}$

for a given tension and a given mass per unit length. This is the law of length, and it shows directly that lengthening the string lowers the frequency. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Can you hear on earth the sound of an extremely violent explosion on moon ?

Answer

No, the sound of an extremely violent explosion on the moon cannot be heard on the earth.

Sound is a mechanical wave and it requires a material medium for its propagation. There is no atmosphere on the moon, and the space between the moon and the earth is a vacuum. Since a vacuum has no particles to carry the wave's energy, sound cannot travel through it.

The transverse waves cannot be produced in gases, why?

Answer

Transverse waves cannot be produced in gases because gases do not have rigidity.

The propagation of a transverse wave requires the medium to sustain shearing stress, and only a medium possessing a modulus of rigidity can do so. A gas has no definite shape and yields at once to shearing stress, so a transverse wave cannot be set up in it. Only longitudinal waves, for which the medium is alternately compressed and rarefied, can travel in a gas.

Sound is produced simultaneously at one end of two strings of the same length, one of rubber and the other of steel. In which string will the sound reach the other end earlier and why ?

Answer

The sound will reach the other end earlier in the steel string.

The speed of a longitudinal wave in a solid rod is

$\text v = \sqrt{\dfrac{\text Y}{\text d}}$

where Y is the Young's modulus of the material and d its density. The value of $\dfrac{\text Y}{\text d}$ is much larger for steel than for rubber, so the speed of sound in steel is much greater. Since both the strings are of the same length, the sound takes less time in the steel string.

What will be the speed of sound in a perfectly rigid rod?

Answer

The speed of sound in a perfectly rigid rod would be infinite.

The speed of a longitudinal wave in a rod is $\text v = \sqrt{\dfrac{\text Y}{\text d}}$. For a perfectly rigid rod the Young's modulus Y is infinite, since such a rod suffers no strain whatever the stress applied. Hence the speed of sound in it comes out to be infinite.

In which substances among gas, liquid and solid, the velocity of sound is maximum and minimum ?

Answer

The velocity of sound is maximum in solids and minimum in gases.

The speed of a longitudinal wave in a medium is $\text v = \sqrt{\dfrac{\text E}{\text d}}$, where E is the modulus of elasticity of the medium. Solids possess a very large modulus of elasticity, liquids a smaller one and gases the smallest. Hence the speed of sound is minimum in gases, more in liquids and maximum in solids.

A man produces sound by striking the railway track. Another man listens two sounds by putting his ear on the railway track at a distance of 1.0 km from the first man. Give reason. Which sound is heard first and why ?

Answer

The sound travelling through the railway track is heard first.

The sound produced by striking the track travels to the second man by two separate paths — through the steel rail and through the air. The speed of sound in steel is far greater than that in air, since the ratio $\dfrac{\text Y}{\text d}$ is much larger for steel. Hence the sound reaching through the rail arrives earlier, and the sound reaching through the air is heard a little later. This is why two sounds are heard at a small interval of time.

The velocity of sound is 330 m/s in air at constant temperature and pressure. Explain giving reason how velocity of sound changes on halving the pressure and keeping temperature constant.

Answer

There will be no change in the velocity of sound; it will remain 330 m/s.

The speed of sound in a gas is $\text v = \sqrt{\dfrac{\gamma \text P}{\text d}}$. From the gas equation, at a constant temperature

$\dfrac{\text P}{\text d} = \dfrac{\text{RT}}{\text M} = \text{constant}$

Hence when the pressure is halved, the density is also halved in the same proportion, so the ratio $\dfrac{\text P}{\text d}$ does not change. Therefore there is no effect of the pressure-change on the speed of sound.

At normal temperature and pressure, the speed of sound in air is 332 m s^-1. What will be the speed of sound on doubling the pressure ?

Answer

The speed of sound will remain 332 m s^-1.

On doubling the pressure at a constant temperature, the density of the air is also doubled, so that the ratio $\dfrac{\text P}{\text d}$ remains constant. Since $\text v = \sqrt{\dfrac{\gamma \text P}{\text d}}$, the speed of sound is unaltered.

What is the effect of temperature on the speed of sound in air ?

Answer

As the temperature of the air increases, the speed of sound in air increases.

The speed of sound in a gas is

$\text v = \sqrt{\dfrac{\gamma \text{RT}}{\text M}} \qquad \Rightarrow \qquad \text v \propto \sqrt{\text T}$

that is, the speed of sound in a gas is directly proportional to the square-root of its absolute temperature.

What is the increase in the velocity of sound in air when the temperature of air rises by 1°C ?

Answer

The velocity of sound in air increases by 0.61 m s^-1 for each 1°C rise in temperature.

Taking the speed of sound in air at 0°C to be 332 m s^-1, the speed at t°C is

$\text v_\text t = (332 + 0.61\text t)\ \text{m s}^{-1}$

so that a rise of 1°C raises the speed by 0.61 m s^-1.

Explain the effect of humidity on the velocity of sound in air.

Answer

The speed of sound in air increases with increase in humidity.

The density of water-vapour is less than that of air, so the density of moist air, that is, air mixed with water-vapour, is less than that of dry air. Assuming the value of γ for moist air to be the same as that for dry air, it follows from

$\text v = \sqrt{\dfrac{\gamma \text P}{\text d}} \qquad \Rightarrow \qquad \text v \propto \dfrac{1}{\sqrt{\text d}}$

that the speed of sound in moist air is slightly greater than in dry air. This is why the sirens of mills and the whistles of trains are heard up to longer distances in the rainy season.

Write the relation between root-mean-square speed v_rms of the molecules of a gas and the speed of sound v in the same gas.

Answer

The speed of sound in a gas and the root-mean-square speed of its molecules are related by

$\dfrac{\text v}{\text v_{rms}} = \sqrt{\dfrac{\gamma}{3}} \qquad \text{or} \qquad \text v_{rms} = \sqrt{\dfrac{3}{\gamma}}\ \text v$

where γ (> 1) is the ratio of the two molar specific heats of the gas. Since the value of $\sqrt{\dfrac{\gamma}{3}}$ is less than 1, the speed of sound in a gas is smaller than the root-mean-square speed of the molecules of that gas.

Write the equation of a plane progressive wave and explain the symbols used.

Answer

The equation of a plane progressive wave travelling along the positive direction of the X-axis is

$\text y = \text a \sin 2\pi\left(\dfrac{\text t}{\text T} - \dfrac{\text x}{\lambda}\right)$

The symbols used have the following meanings :

• y — the displacement of the particle situated at a distance x from the origin, at any instant t
• a — the amplitude of the wave, that is, the maximum displacement of the particle from its rest position
• T — the time-period of the oscillation of the particle
• λ — the wavelength of the wave
• x — the distance of the particle from the origin, measured along the direction of propagation
• t — the instant of time at which the displacement is measured

If the wave travels along the − X direction, the minus sign inside the bracket is replaced by a plus sign.

In the wave equation y = a sin (ω t − k x), what are ω and k ? What is represented by ω/k?

Answer

In the wave equation y = a sin (ω t − k x),

• ω is the angular frequency of the oscillating particle, related to the frequency n and the time-period T by $\omega = 2\pi \text n = \dfrac{2\pi}{\text T}$
• k is the propagation constant, or wave number, given by $\text k = \dfrac{2\pi}{\lambda}$, which defines how rapidly the phase of the wave changes with distance

The ratio $\dfrac{\omega}{\text k}$ represents the speed of the wave,

$\dfrac{\omega}{\text k} = \dfrac{2\pi \text n}{2\pi/\lambda} = \text n \lambda = \text v$

A plane progressive wave has an amplitude A metre, velocity v metre/second and frequency 'n' hertz. Write down the equation of the wave.

Answer

Given, the amplitude is A metre, the velocity is v metre/second and the frequency is n hertz.

The equation of a plane progressive wave travelling along the positive direction of the X-axis is

$\text y = \text A \sin 2\pi \text n\left(\text t - \dfrac{\text x}{\text v}\right)$

where y is the displacement, at any instant t, of the particle situated at a distance x from the origin.

In the equation of a progressive wave y = a sin (ω t − k x), y and a are displacement and amplitude respectively. ω is angular frequency, t is time, k is propagation constant and x is the distance on X-axis. What is the phase and the velocity of the wave at time t ?

Answer

In the equation y = a sin (ω t − k x), the argument of the sine gives the phase of the particle. Hence

Phase at time t :

$\phi = (\omega \text t - \text k \text x)$

Velocity of the wave : The phase of a given point of the wave remains constant as the wave advances, so

$\text v = \dfrac{\omega}{\text k}$

since $\omega = 2\pi \text n$ and $\text k = \dfrac{2\pi}{\lambda}$, which gives $\dfrac{\omega}{\text k} = \text n \lambda = \text v$.

Two waves meet at a point in opposite phases. What may be the possible phase difference between them?

Answer

Two waves meet in opposite phases when the phase difference between them is an odd multiple of π, that is,

$\Delta \phi = \pi,\ 3\pi,\ 5\pi,\ \dots$

or, in general, Δφ = (2m − 1)π, where m = 1, 2, 3, ... At such places destructive interference takes place and the resultant amplitude is a₁ ~ a₂.

Write the relation between phase difference (Δφ) and path difference (Δx) between two points.

Answer

The relation between the phase difference (Δφ) and the path difference (Δx) between two points is

$\Delta \phi = \dfrac{2\pi}{\lambda} \times \Delta \text x$

where λ is the wavelength of the wave. Thus a path difference of one full wavelength corresponds to a phase difference of 2π, which means that the two particles are in the same phase of oscillation.

What are beats?

Answer

Beats : When two sound waves of nearly equal frequencies are produced simultaneously, the intensity of the resultant sound produced by their superposition increases and decreases alternately with time. This rise and fall in the intensity of sound is called the phenomenon of beats.

One rise and one fall of the intensity together form one beat, and the number of times the intensity of sound rises and falls in one second is called the beat frequency. It is equal to the difference in the frequencies of the two sound-sources, that is, n₁ ~ n₂.

As in sound, can beats be observed by two light-sources? Explain.

Answer

No, beats cannot be observed with two light-sources.

To observe beats, the phase difference between the two sources must change regularly with time. In light-sources this change takes place at random, because a light-source consists of innumerable atoms and each atom emits wave independently of the others. Hence the phase difference between two light-sources does not vary in a regular manner, and no beats are observed.

To hear the beats clearly, what should be the maximum difference in the frequencies of the two sound-sources?

Answer

The maximum difference in the frequencies of the two sound-sources should be 10.

The sensitivity of our ear is $\dfrac{1}{10}$ second, that is, the effect of any sound remains on the ear for $\dfrac{1}{10}$ second. Hence the ear can clearly hear a maximum of 10 beats per second. If the difference in the frequencies exceeds 10, the successive beats overlap on the ear and cannot be distinguished separately.

Write the distances between (i) a node and the nearest antinode, and (ii) two successive antinodes in a stationary wave, in terms of the wavelength.

Answer

In a stationary wave,

(i) The distance between a node and the nearest antinode is

$\dfrac{\lambda}{4}$

(ii) The distance between two successive antinodes is

$\dfrac{\lambda}{2}$

which is also the distance between two successive nodes.

What is the phase difference between a node and its nearest antinode in a stationary wave?

Answer

The phase difference between a node and its nearest antinode in a stationary wave is $\dfrac{\pi}{2}$ radian.

The distance between a node and its nearest antinode is $\dfrac{\lambda}{4}$. Hence, using $\Delta \phi = \dfrac{2\pi}{\lambda} \times \Delta \text x$,

$\Delta \phi = \dfrac{2\pi}{\lambda} \times \dfrac{\lambda}{4} = \dfrac{\pi}{2}\ \text{radian}$

What is the phase difference between the particles of medium on the two sides of a node? What for the particles between two nodes?

Answer

Particles on the two sides of a node : The phase difference is π. At any instant, the phase of vibration of the points on one side of a node is opposite to the phase of vibration of the points on the other side.

Particles lying between two nodes : The phase difference is zero. All the points between two successive nodes vibrate in the same phase — they reach their positions of maximum displacement simultaneously and pass simultaneously through their mean positions.

A light wave is reflected from a mirror and the incident and the reflected waves superpose to form stationary wave, but the nodes and antinodes are not seen. Why?

Answer

The nodes and antinodes are not seen because they are far too close together to be resolved.

The wavelength of light is of the order of 10^-7 metre. The distance between two consecutive nodes, or between two consecutive antinodes, is $\dfrac{\lambda}{2}$, which is therefore also of the order of 10^-7 metre. It is not possible to resolve such a small distance by the eye or by an ordinary optical instrument, and so the nodes and antinodes are not seen.

An organ pipe emits a fundamental note of frequency 128 hertz. When blown forcefully, its first overtone of 384 hertz is emitted. Is the pipe closed or open?

Answer

The pipe is closed at one end.

Given, the fundamental frequency n₁ = 128 hertz and the frequency of the first overtone = 384 hertz. The ratio of the two is

$\dfrac{384}{128} = 3$

A closed organ pipe produces only odd harmonics, so its first overtone is the third harmonic, that is, 3 n₁. An open pipe would have given the first overtone as the second harmonic, 2 n₁ = 256 hertz. Hence the pipe is closed.

Two cylindrical open organ pipes A and B have the same length, but the diameter of A is double that of B. Which pipe has higher frequency?

Answer

The pipe B, which has the smaller diameter, has the higher frequency.

Applying the end correction, the fundamental frequency of an open organ pipe of radius r is

$\text n = \dfrac{\text v}{2(\text l + 1.2\text r)}$

As r increases, the denominator increases and so the frequency n decreases. The diameter of A is double that of B, so A has the larger radius and therefore the lower frequency. Hence pipe B has the higher fundamental frequency.

Write the frequencies of the second and the third harmonics of a stretched string in terms of fundamental frequency.

Answer

Let n be the fundamental frequency of a stretched string. Since a stretched string gives both the even and the odd harmonics, in the ratio n₁ : n₂ : n₃ ... = 1 : 2 : 3 ...,

Frequency of the second harmonic = 2n

Frequency of the third harmonic = 3n

The second harmonic is also called the first overtone and the third harmonic the second overtone.

Write the formula for the speed of transverse waves in a stretched string, explaining the symbols used.

Answer

The speed of a transverse wave in a stretched string is

$\text v = \sqrt{\dfrac{\text T}{\text m}}$

where

• T — the tension in the string
• m — the mass per unit length of the string, that is, its linear mass density, and not the mass of the whole string

If r be the radius of the string and d the density of its material, then m = πr²d, so that the speed may also be written as

$\text v = \sqrt{\dfrac{\text T}{\pi \text r^2 \text d}} = \sqrt{\dfrac{\text{stress}}{\text{density}}}$

Write down the formula for the speed (v) of sound in a medium in terms of the modulus of elasticity (E) and the density (d) of the medium.

Answer

The speed of sound in a medium, in terms of the modulus of elasticity E and the density d of the medium, is

$\text v = \sqrt{\dfrac{\text E}{\text d}}$

Here E is the appropriate modulus of elasticity of the medium — Young's modulus for a long rod, and the bulk modulus for a liquid or a gas.

Write the formula for the speed of longitudinal waves in a metallic rod in terms of the density and Young's modulus of the metal.

Answer

The speed of longitudinal waves in a metallic rod is

$\text v = \sqrt{\dfrac{\text Y}{\text d}}$

where Y is the Young's modulus of the metal and d is its density.

Write Newton's formula for the speed of longitudinal waves in a gaseous medium.

Answer

Newton's formula : Newton assumed that when longitudinal waves travel in a gas, the temperature of the gas remains constant, so that the elasticity concerned is the isothermal elasticity, whose value is equal to the pressure P of the gas. Hence

$\text v = \sqrt{\dfrac{\text P}{\text d}}$

where P is the pressure of the gas and d is its density.

Write the Laplace's formula for the speed of sound in a gas.

Answer

Laplace's formula : Laplace showed that the compressions and rarefactions occur so rapidly that the exchange of heat cannot take place, so the process is adiabatic and the elasticity concerned is the adiabatic elasticity, whose value is γP. Hence

$\text v = \sqrt{\dfrac{\gamma \text P}{\text d}}$

where $\gamma = \dfrac{\text C_\text p}{\text C_\text v}$ is the ratio of the specific heat of the gas at constant pressure to that at constant volume, P is the pressure of the gas and d is its density.

Write down the formula for the speed of transverse wave on a stretched string.

Answer

The speed of a transverse wave on a stretched string is

$\text v = \sqrt{\dfrac{\text T}{\text m}}$

where T is the tension in the string and m is the mass per unit length of the string.

Draw a graph between the pressure P of a gas and the speed v of sound travelling through the gas.

Answer

The graph will be a straight line parallel to the P-axis.

At a constant temperature the ratio $\dfrac{\text P}{\text d}$ is a constant, since a change in P is accompanied by a proportional change in d. Hence from $\text v = \sqrt{\dfrac{\gamma \text P}{\text d}}$ the speed v does not change with P, and the graph between P and v is a straight line parallel to the pressure axis.

What is the ratio of the speed of sound in a polyatomic gas (γ = 4/3) and the rms speed of the molecules in that gas?

Answer

Given, γ = 4/3 for the polyatomic gas.

The ratio of the speed of sound in a gas to the root-mean-square speed of its molecules is

$\dfrac{\text v}{\text v_{rms}} = \sqrt{\dfrac{\gamma}{3}}$

Substituting the value of γ,

$\dfrac{\text v}{\text v_{rms}} = \sqrt{\dfrac{4/3}{3}} = \sqrt{\dfrac{4}{9}}$

$\dfrac{\text v}{\text v_{rms}} = \dfrac{2}{3}$

Hence, the required ratio is 2 : 3.

Find the ratio of the speed of sound in hydrogen gas and root-mean-square speed of hydrogen molecules.

Answer

Hydrogen is a diatomic gas, so $\gamma = \dfrac{7}{5}$.

The ratio of the speed of sound in a gas to the root-mean-square speed of its molecules is

$\dfrac{\text v}{\text v_{rms}} = \sqrt{\dfrac{\gamma}{3}}$

Substituting the value of γ,

$\dfrac{\text v}{\text v_{rms}} = \sqrt{\dfrac{7/5}{3}} = \sqrt{\dfrac{7}{15}}$

Hence, the required ratio is $\sqrt{\dfrac{7}{15}}$.

The fundamental frequency of an open organ pipe is 512 Hz. What will be the fundamental frequency if its one end is closed?

Answer

Given, the fundamental frequency of the open organ pipe, n_open = 512 Hz.

For an open pipe of length l,

$\text n_{open} = \dfrac{\text v}{2\text l}$

and for a pipe of the same length closed at one end,

$\text n_{closed} = \dfrac{\text v}{4\text l} = \dfrac{1}{2}\text n_{open}$

Substituting the value,

$\text n_{closed} = \dfrac{512}{2}$

$\text n_{closed} = 256\ \text{hertz}$

Hence, the fundamental frequency becomes 256 hertz when one end is closed.

The frequency of a closed-end organ pipe is 50 Hz. What will be its resonating frequencies if the other end of the pipe is opened?

Answer

Given, the frequency of the closed-end organ pipe, n_closed = 50 Hz.

For a closed pipe of length l,

$\text n_{closed} = \dfrac{\text v}{4\text l} = 50\ \text{Hz}$

When the other end is also opened, the pipe becomes an open pipe of the same length, and its fundamental frequency is

$\text n_{open} = \dfrac{\text v}{2\text l} = 2 \times 50 = 100\ \text{Hz}$

An open pipe produces both the even and the odd harmonics, in the ratio 1 : 2 : 3 : 4 ... Hence the resonating frequencies are

$100\ \text{Hz},\ 200\ \text{Hz},\ 300\ \text{Hz},\ \dots$

Hence, the resonating frequencies of the opened pipe are 100 Hz, 200 Hz, 300 Hz, and so on.

The frequency of the fundamental note produced by a pipe closed at one end is 150 per second. What will be the frequency of the fundamental note from another pipe of the same type but of half its length?

Answer

Given, the frequency of the fundamental note of the closed pipe, n = 150 per second, and the length of the second pipe is half that of the first.

The fundamental frequency of a pipe closed at one end is

$\text n = \dfrac{\text v}{4\text l} \qquad \Rightarrow \qquad \text n \propto \dfrac{1}{\text l}$

For the second pipe of length $\dfrac{\text l}{2}$,

$\text n' = \dfrac{\text v}{4\left(\dfrac{\text l}{2}\right)} = 2\left(\dfrac{\text v}{4\text l}\right) = 2\text n$

$\text n' = 2 \times 150 = 300\ \text{per second}$

Hence, the frequency of the fundamental note of the second pipe is 300 per second.

The frequency of the fundamental note of a closed organ pipe and that of an open organ pipe are the same. What is the ratio between their lengths?

Answer

Let l₁ be the length of the closed organ pipe and l₂ that of the open organ pipe.

The fundamental frequencies are

$\text n_1 = \dfrac{\text v}{4\text l_1} \qquad \text{and} \qquad \text n_2 = \dfrac{\text v}{2\text l_2}$

It is given that the two fundamental frequencies are the same, so n₁ = n₂,

$\dfrac{\text v}{4\text l_1} = \dfrac{\text v}{2\text l_2}$

$\dfrac{\text l_1}{\text l_2} = \dfrac{2}{4} = \dfrac{1}{2}$

Hence, the ratio of the length of the closed pipe to that of the open pipe is 1 : 2.

In the given diagrams is shown a stretched string vibrating between two points P and Q . Write the ratio of the frequencies of the string in the two cases. The tension in the string remains unchanged.

Answer

In the two diagrams the stretched string vibrates between the fixed points P and Q, so a node is formed at each of these ends.

First case : The string vibrates in one segment, that is, in the fundamental mode. If l be the length PQ, then

$\text l = \dfrac{\lambda_1}{2} \qquad \Rightarrow \qquad \text n_1 = \dfrac{\text v}{2\text l}$

Second case : The string vibrates in three segments, with two additional nodes between P and Q. Hence

$\text l = \dfrac{3\lambda_2}{2} \qquad \Rightarrow \qquad \text n_2 = \dfrac{3\text v}{2\text l} = 3\text n_1$

The tension in the string remains unchanged, so the speed v is the same in both the cases.

$\dfrac{\text n_1}{\text n_2} = \dfrac{1}{3}$

Hence, the ratio of the frequencies of the string in the two cases is 1 : 3.

In a plane progressive wave, the maximum particle velocity is twice the wave velocity. Find the ratio of the wavelength and the amplitude of the wave.

Answer

Given, the maximum particle velocity is twice the wave velocity.

For a plane progressive wave the maximum particle velocity is

$\text u_{max} = \text a \omega = \text a (2\pi \text n)$

and the wave velocity is

$\text v = \text n \lambda$

It is given that u_max = 2v, so

$\text a (2\pi \text n) = 2\text n \lambda$

$\dfrac{\lambda}{\text a} = \pi$

Hence, the ratio of the wavelength to the amplitude of the wave is π : 1.

Sound waves of frequency 660 Hz fall normally on a perfectly reflecting wall. Determine the shortest distance from the wall at which the air particles have a maximum amplitude of vibration. The speed of sound in air is 330 m/s.

Answer

Given,

• Frequency of the sound waves, n = 660 Hz
• Speed of sound in air, v = 330 m/s

The wave falls normally on a perfectly reflecting wall, so the incident and the reflected waves superpose and form a stationary wave. The wall is a rigid boundary, so a node is formed at the wall.

The wavelength of the wave is

$\lambda = \dfrac{\text v}{\text n} = \dfrac{330\ \text{m/s}}{660\ \text{Hz}} = 0.5\ \text m$

The air particles have a maximum amplitude of vibration at an antinode, and the distance of the nearest antinode from the node at the wall is $\dfrac{\lambda}{4}$. Hence the shortest distance is

$\dfrac{\lambda}{4} = \dfrac{0.5}{4}$

$= 0.125\ \text m$

Hence, the shortest distance from the wall at which the air particles have a maximum amplitude of vibration is 0.125 m.

Write the equation of a stationary wave. Explain the meaning of symbols used.

Answer

The equation of a stationary wave is

$\text y = 2\text a \cos \dfrac{2\pi \text x}{\lambda} \sin \dfrac{2\pi \text t}{\text T}$

The symbols used have the following meanings :

• y — the resultant displacement, at the instant t, of the particle situated at a distance x from the origin
• a — the amplitude of each of the two component progressive waves, so that 2a is the maximum amplitude at an antinode
• λ — the wavelength of the component waves
• T — the time-period of the component waves
• x — the distance of the particle from the origin
• t — the instant of time

In this wave the positions x = 0, $\dfrac{\lambda}{2}$, $\dfrac{2\lambda}{2}$, ... are the antinodes, and the positions x = $\dfrac{\lambda}{4}$, $\dfrac{3\lambda}{4}$, $\dfrac{5\lambda}{4}$, ... are the nodes.

The fundamental frequency of pipe closed at one end is 200 hertz. What will be the fundamental frequency of another similar pipe of the same length, open at both ends?

Answer

Given, the fundamental frequency of the pipe closed at one end, n_closed = 200 hertz.

For a pipe of length l closed at one end,

$\text n_{closed} = \dfrac{\text v}{4\text l} = 200\ \text{hertz}$

For a similar pipe of the same length open at both ends,

$\text n_{open} = \dfrac{\text v}{2\text l} = 2\left(\dfrac{\text v}{4\text l}\right) = 2 \times 200$

$\text n_{open} = 400\ \text{hertz}$

Hence, the fundamental frequency of the pipe open at both ends is 400 hertz.

In the figure, two vibrating air columns are shown. Find the ratio of frequencies in the two cases shown.

Answer

In the figure the two vibrating air columns are those of pipes closed at one end, since each has a node at the closed end and an antinode at the open end.

First case : The figure shows the first overtone, that is, the third harmonic, of the closed pipe. Its frequency is

$\text n_1 = \dfrac{3\text v}{4\text l}$

Second case : The figure shows the second overtone, that is, the fifth harmonic, of the closed pipe. Its frequency is

$\text n_2 = \dfrac{5\text v}{4\text l}$

Therefore,

$\dfrac{\text n_1}{\text n_2} = \dfrac{3\text v/4\text l}{5\text v/4\text l} = \dfrac{3}{5}$

Hence, the ratio of the frequencies in the two cases is 3 : 5.

Two strings of equal lengths are stretched by equal forces. The strings are made of the same material. If their diameters are in the ratio 3 : 2, what will be the ratio of their fundamental frequencies?

Answer

Given, the two strings are of equal length, are stretched by equal forces, are made of the same material, and their diameters are in the ratio 3 : 2, that is, r₁ : r₂ = 3 : 2.

The fundamental frequency of a stretched string is

$\text n = \dfrac{1}{2\text l}\sqrt{\dfrac{\text T}{\text m}} = \dfrac{1}{2\text l}\sqrt{\dfrac{\text T}{\pi \text r^2 \text d}} \qquad \Rightarrow \qquad \text n \propto \dfrac{1}{\text r}$

since l, T and d are the same for both the strings. Therefore,

$\dfrac{\text n_1}{\text n_2} = \dfrac{\text r_2}{\text r_1} = \dfrac{2}{3}$

Hence, the ratio of their fundamental frequencies is 2 : 3.

The frequency of the third harmonic of a stretched string is 255 hertz. Determine its fundamental frequency.

Answer

Given, the frequency of the third harmonic of the stretched string, n₃ = 255 hertz.

A stretched string gives both the even and the odd harmonics, whose frequencies are in the ratio n₁ : n₂ : n₃ ... = 1 : 2 : 3 ... Hence the third harmonic is three times the fundamental frequency,

$\text n_3 = 3\text n_1$

$\text n_1 = \dfrac{\text n_3}{3} = \dfrac{255}{3}$

$\text n_1 = 85\ \text{hertz}$

Hence, the fundamental frequency of the string is 85 hertz.

The speed of sound does not depend upon its frequency. Can you verify this statement from some experience of your daily life ?

Answer

Yes. If sounds are produced simultaneously by different musical instruments in an orchestra, they are all heard by the ear at the same time.

The notes produced by the various instruments have widely different frequencies. If the speed of sound depended upon the frequency, then the notes of different frequencies would reach the listener at different instants of time and the music would be heard jumbled. Since we hear them together and in the proper sequence, the speed of sound in air must be the same for all frequencies. This is why we can enjoy an orchestra even when seated far away.

How will the frequency of a note emitted by an organ pipe be affected by rise in temperature?

Answer

The frequency of the note emitted by an organ pipe increases with a rise in temperature.

The frequency of the note emitted by an organ pipe is

$\text n \propto \dfrac{\text v}{\text l}$

With a rise in temperature the speed of sound in air increases, since $\text v \propto \sqrt{\text T}$. The length of the pipe is practically unchanged, so the frequency of the note also increases.

How will the fundamental frequency of a closed organ pipe be affected if instead of air it is filled with a gas heavier than air?

Answer

The fundamental frequency of the closed organ pipe will decrease.

The fundamental frequency of a closed organ pipe is

$\text n = \dfrac{\text v}{4\text l}$

where v is the speed of sound in the gas filling the pipe. Since $\text v \propto \dfrac{1}{\sqrt{\text d}}$, a gas heavier than air has a greater density and so the speed of sound in it is smaller. The length of the pipe being unchanged, the fundamental frequency therefore decreases.

A tuning fork produces resonance in a closed pipe, but the same tuning fork does not produce resonance in an open pipe of the same length. Why?

Answer

This happens because of the end correction.

Applying the end correction e, the fundamental frequencies of a closed pipe and of an open pipe of the same length l are

$\text n_{closed} = \dfrac{\text v}{4(\text l + \text e)} \qquad \text{and} \qquad \text n_{open} = \dfrac{\text v}{2(\text l + 2\text e)}$

Hence the fundamental frequency of the open pipe is not exactly twice that of the closed pipe of the same length, but slightly less than twice. Since the harmonics of the open pipe therefore do not coincide exactly with the frequency of the fork that resonates with the closed pipe, the same fork does not produce resonance in the open pipe.

If the tension of a stretched string is made four times its initial value, then the final velocity of the wave will be how many times of its initial velocity?

Answer

The final velocity of the wave will be twice its initial velocity.

The speed of a transverse wave in a stretched string is

$\text v = \sqrt{\dfrac{\text T}{\text m}} \qquad \Rightarrow \qquad \text v \propto \sqrt{\text T}$

If the tension is made 4 T, then

$\text v' = \sqrt{\dfrac{4\text T}{\text m}} = 2\sqrt{\dfrac{\text T}{\text m}} = 2\text v$

If the radius of a stretched wire is reduced to half, then what will be the new wave speed compared to its initial value?

Answer

The new wave speed will be double its initial value.

The speed of a transverse wave in a stretched wire of radius r is

$\text v = \sqrt{\dfrac{\text T}{\pi \text r^2 \text d}} \qquad \Rightarrow \qquad \text v \propto \dfrac{1}{\text r}$

for a given tension and a given material. If the radius is reduced to $\dfrac{\text r}{2}$, then

$\text v' = 2\text v$

A sonometer wire of length l is in unison with a tuning fork of frequency n. If the length of the wire is reduced to one-half then with what frequency will it be in unison?

Answer

The wire will be in unison with a fork of frequency 2 n.

The fundamental frequency of a sonometer wire is

$\text n = \dfrac{1}{2\text l}\sqrt{\dfrac{\text T}{\text m}} \qquad \Rightarrow \qquad \text n \propto \dfrac{1}{\text l}$

for a given tension and a given wire. If the length is reduced to $\dfrac{\text l}{2}$, the frequency becomes

$\text n' = 2\text n$

Write (i) λ/4 in forms of phase difference and time difference and (ii) π/3 in forms of time difference and wavelength.

Answer

(i) A path difference of $\dfrac{\lambda}{4}$ corresponds to

$\text{phase difference} = \dfrac{2\pi}{\lambda} \times \dfrac{\lambda}{4} = \dfrac{\pi}{2}$

$\text{time difference} = \dfrac{\text T}{\lambda} \times \dfrac{\lambda}{4} = \dfrac{\text T}{4}$

(ii) A phase difference of $\dfrac{\pi}{3}$ corresponds to

$\text{time difference} = \dfrac{\text T}{2\pi} \times \dfrac{\pi}{3} = \dfrac{\text T}{6}$

$\text{path difference} = \dfrac{\lambda}{2\pi} \times \dfrac{\pi}{3} = \dfrac{\lambda}{6}$

Equation of motion of a progressive wave is $\text y = 0.09 \sin 8\pi\left[\text t - \dfrac{\text x}{20}\right]$. The amplitude of reflected wave remains 2/3 of that of progressive wave when the wave is reflected back from a rigid wall. Determine the equation of the reflected wave.

Answer

Given, the equation of the progressive wave is

$\text y = 0.09 \sin 8\pi\left[\text t - \dfrac{\text x}{20}\right]$

and the amplitude of the reflected wave is $\dfrac{2}{3}$ of that of the incident wave.

Amplitude of the reflected wave :

$\text a' = \dfrac{2}{3} \times 0.09 = 0.06$

Direction of travel : The incident wave travels along the positive direction of the X-axis, so the reflected wave travels along the negative direction. Hence x is replaced by − x, that is, the minus sign inside the bracket becomes a plus sign.

Phase change : The wave is reflected back from a rigid wall, so it suffers a phase change of π. Therefore

$\text y = 0.06 \sin \bigg\lbrace 8\pi\left[\text t + \dfrac{\text x}{20}\right] + \pi \bigg\rbrace$

Using sin (θ + π) = − sin θ,

$\text y = - 0.06 \sin 8\pi\left(\text t + \dfrac{\text x}{20}\right)$

Hence, the equation of the reflected wave is $\text y = - 0.06 \sin 8\pi\left(\text t + \dfrac{\text x}{20}\right)$.

Why did Laplace introduce a correction in Newton's formula for velocity of sound ?

Answer

Newton assumed that when longitudinal waves travel in a gas, the temperature of the gas remains constant. On this assumption the elasticity concerned is the isothermal elasticity, whose value is equal to the pressure P of the gas, and the speed of sound is

$\text v = \sqrt{\dfrac{\text P}{\text d}}$

Taking P = 1.01 × 10⁵ N m^-2 and d = 1.29 kg m^-3 for air at 0°C, this gives v = 280 m s^-1, whereas the experimental value is nearly 331 m s^-1. The discrepancy is far too large.

Laplace pointed out that Newton's assumption is wrong. When sound waves travel in a gaseous medium, the states of compression and rarefaction occur alternately at every point. At a compression some heat is developed and at a rarefaction some heat is lost. The compressions and rarefactions occur so rapidly that the heat produced during compression cannot go out into the surroundings and the heat lost during rarefaction cannot come in from them. Besides this, gases are bad conductors of heat, so no exchange of heat takes place at all. Hence the temperature at a point rises during compression and falls during rarefaction, and the process is adiabatic, not isothermal.

Therefore the elasticity concerned is the adiabatic elasticity, whose value is γP, and the corrected formula is

$\text v = \sqrt{\dfrac{\gamma \text P}{\text d}}$

With γ = 1.41 for air, this gives $\text v = \sqrt{1.41} \times 280 = 332$ m s^-1, which agrees closely with the experimental value.

The speed of sound in moist air is greater than in dry air, why ? Will the speed of sound in moist hydrogen be greater than in dry hydrogen ?

Answer

The speed of sound in moist air is greater than in dry air.

The density of water-vapour is less than that of air, so the density of air mixed with water-vapour, that is, moist air, is less than that of dry air. Since

$\text v = \sqrt{\dfrac{\gamma \text P}{\text d}} \qquad \Rightarrow \qquad \text v \propto \dfrac{1}{\sqrt{\text d}}$

at a given pressure, the smaller density of moist air gives a greater speed of sound.

No, the speed of sound in moist hydrogen will be less than in dry hydrogen.

The density of water-vapour is more than that of hydrogen, which is the lightest of all gases. Hence the density of moist hydrogen is greater than that of dry hydrogen, and by the same relation $\text v \propto \dfrac{1}{\sqrt{\text d}}$ the speed of sound in moist hydrogen is less than in dry hydrogen.

Sound waves propagate in air with a speed of 300 m s^-1. Will the speed of these waves in hydrogen at the same temperature less or more ? In carbon-dioxide ?

Answer

In hydrogen the speed will be more, and in carbon-dioxide it will be less than 300 m s^-1.

At a given temperature the speed of sound in a gas is

$\text v = \sqrt{\dfrac{\gamma \text{RT}}{\text M}} \qquad \Rightarrow \qquad \text v \propto \dfrac{1}{\sqrt{\text M}}$

that is, the speed of sound in a gas is inversely proportional to the square-root of its molecular-mass.

The molecular mass of hydrogen, 2, is much less than the average molecular mass of air, about 29, so the speed of sound in hydrogen is much more than in air.

The molecular mass of carbon-dioxide, 44, is more than that of air, so the speed of sound in carbon-dioxide is less than in air.

What is meant by stationary waves? Write equation.

Answer

Stationary waves : When two identical transverse, or longitudinal, progressive waves travel in a bounded medium with the same speed but in opposite directions, then by their superposition a new type of wave is produced which appears stationary in the medium. This wave is called a stationary (or standing) wave.

The characteristic of the stationary wave is that some particles of the medium remain permanently at rest, while some other particles undergo maximum displacement compared to others. The former are called the nodes and the latter the antinodes.

Equation : If two waves of amplitude a travelling in opposite directions superpose, the equation of the resulting stationary wave is

$\text y = 2\text a \cos \left(\dfrac{2\pi \text x}{\lambda}\right)\sin \left(\dfrac{2\pi \text t}{\text T}\right)$

where λ is the wavelength and T the time-period of the component waves.

Write the conditions for the formation of a stationary wave.

Answer

The conditions for the formation of a stationary wave are :

(i) The medium should not be unlimited; it should have a boundary. That is, the medium must be a bounded medium, so that the wave propagating in it is reflected at the boundary.

(ii) The two superposing waves must travel in opposite directions along the same line.

(iii) The two waves must be of the same kind, that is, both transverse or both longitudinal.

(iv) The two waves must have the same frequency (or wavelength) and the same speed.

(v) The amplitudes of the two waves should be equal, or nearly equal, so that the nodes are well defined.

A vessel is placed below a water-tap. We can estimate the height of the water level reached in the vessel from a distance simply by listening the sound. How?

Answer

The air column above the water level in the vessel behaves as an air column closed at one end, the water surface being the closed end. The frequency of the note emitted by such an air column is

$\text n = \dfrac{\text v}{4\text l} \qquad \Rightarrow \qquad \text n \propto \dfrac{1}{\text l}$

that is, the frequency is inversely proportional to the length of the air column.

As the water level in the vessel rises, the length l of the air column above it goes on decreasing, so the frequency of the emitted note goes on increasing, that is, the note becomes more and more shrill.

Hence, simply by listening to the sound and noting how shrill it has become, we can estimate from a distance the height up to which the vessel has been filled with water.

An organ pipe is in resonance with a tuning fork. What change will have to be done in its length l, if (i) the temperature rises, (ii) hydrogen is filled in the pipe in place of air, (iii) pressure becomes higher?

Answer

The organ pipe is in resonance with the tuning fork, so its frequency $\text n \propto \dfrac{\text v}{\text l}$ must be kept unchanged.

(i) Temperature rises : With a rise in temperature the speed of sound in air increases, since $\text v \propto \sqrt{\text T}$. To keep the frequency of the pipe unchanged, the length l of the pipe must therefore be increased.

(ii) Hydrogen is filled in place of air : Hydrogen is far lighter than air, and since $\text v \propto \dfrac{1}{\sqrt{\text d}}$, the speed of sound is more in hydrogen than in air. Hence again, to keep the frequency unchanged, the length l must be increased.

(iii) Pressure becomes higher : There is no effect of a pressure rise on the speed of sound, because at a constant temperature the ratio $\dfrac{\text P}{\text d}$ remains constant. Hence no change is required in l.

What is meant by 'harmonics'? Explain giving examples.

Answer

Harmonics : If the frequencies of the fundamental tone and the overtones produced by a source of sound are in a harmonic series, then these tones are called harmonics. The fundamental tone is the first harmonic, and the tone whose frequency is m times the fundamental frequency is the mth harmonic.

Examples :

(i) Closed organ pipe : The frequencies of the fundamental tone and the overtones are in the ratio

$\text n_1 : \text n_2 : \text n_3 \dots = 1 : 3 : 5 : 7 \dots$

that is, a closed pipe produces only the odd harmonics.

(ii) Open organ pipe : The frequencies of the fundamental tone and the overtones are in the ratio

$\text n_1 : \text n_2 : \text n_3 : \text n_4 \dots = 1 : 2 : 3 : 4 \dots$

that is, an open pipe produces both the even and the odd harmonics. A stretched string also gives both types of harmonics in this same ratio.

Mention the differences between fundamental frequency, harmonics and overtones.

Answer

What is resonance? Explain.

Answer

Resonance : When an external periodic force is applied over a body and the frequency of the force is different from the natural frequency of the body, then the body executes forced vibrations with the frequency of the applied force but with a small amplitude. If, however, the frequency of the external force is equal to the natural frequency of the body, or to its integral multiple, then the amplitude of the forced vibrations of the body becomes quite large. This phenomenon is called resonance. Thus, resonance is a particular case of forced vibrations.

Explanation : When the frequency of the external force is equal to the natural frequency of the body, then at each step the force is in phase with the oscillating body. Hence the successive impulses given by the periodic force to the body are added up and increase the amplitude of oscillation continuously. But with increasing amplitude the air resistance and the internal friction also increase, so that the loss of energy from the body also increases. Finally a stage is reached when the energy supplied by the external force becomes equal to the energy lost by the body, and the amplitude becomes steady.

Example : If a tuning fork of frequency 256 Hz is vibrated over the mouth of a resonance tube and the length of the air column is gradually adjusted, then at one particular length the air column emits a loud sound. This happens when the natural frequency of the air column becomes equal to the frequency of the fork.

Show that in air the velocity of sound waves increases by 0.61 m s^-1 for each °C rise in temperature of the air.

Answer

The speed of sound in a gas is directly proportional to the square-root of its absolute temperature,

$\text v \propto \sqrt{\text T}$

If v₀ be the speed of sound at 0°C and v_t the speed at t°C, then

$\text v_0 = \sqrt{\dfrac{\gamma \text R}{\text M}} \times \sqrt{0 + 273} \qquad \dots(\text i)$

$\text v_\text t = \sqrt{\dfrac{\gamma \text R}{\text M}} \times \sqrt{\text t + 273} \qquad \dots(\text{ii})$

Dividing equation (ii) by equation (i),

$\dfrac{\text v_\text t}{\text v_0} = \sqrt{\dfrac{\text t + 273}{273}} = \sqrt{1 + \dfrac{\text t}{273}}$

$\text v_\text t = \text v_0\left(1 + \dfrac{\text t}{273}\right)^{1/2}$

Since t is small compared with 273, the binomial theorem may be applied and the terms beyond the first power of $\dfrac{\text t}{273}$ neglected,

$\text v_\text t = \text v_0\left(1 + \dfrac{1}{2} \times \dfrac{\text t}{273}\right) = \text v_0\left(1 + \dfrac{\text t}{546}\right)$

Taking the speed of sound in air at 0°C as v₀ = 332 m s^-1,

$\text v_\text t = 332\left(1 + \dfrac{\text t}{546}\right) = 332 + \dfrac{332\text t}{546}$

$\text v_\text t = (332 + 0.61\text t)\ \text{m s}^{-1}$

Hence, the velocity of sound in air increases by 0.61 m s^-1 for each °C rise in the temperature of the air.

Draw a graph between the absolute temperature (T) of a gas and the square of the speed of sound (v²) in the gas.

Answer

The graph will be a straight line passing through the origin.

The speed of sound in a gas is

$\text v = \sqrt{\dfrac{\gamma \text{RT}}{\text M}} \qquad \Rightarrow \qquad \text v^2 = \left(\dfrac{\gamma \text R}{\text M}\right)\text T$

For a given gas the quantity $\dfrac{\gamma \text R}{\text M}$ is a constant, so

$\text v^2 \propto \text T$

Since v² is directly proportional to T, and v² = 0 when T = 0, the graph between the absolute temperature T and the square of the speed of sound v² is a straight line passing through the origin.

Express the equation $\text y = \text a \sin 2\pi\left(\dfrac{\text t}{\text T} - \dfrac{\text x}{\lambda}\right)$ in two other forms.

Answer

Given, the equation of the progressive wave is

$\text y = \text a \sin 2\pi\left(\dfrac{\text t}{\text T} - \dfrac{\text x}{\lambda}\right)$

First other form : In wave motion $\dfrac{1}{\text T} = \dfrac{\text v}{\lambda}$. Substituting this,

$\text y = \text a \sin 2\pi\left(\dfrac{\text v}{\lambda}\text t - \dfrac{\text x}{\lambda}\right)$

$\text y = \text a \sin \dfrac{2\pi}{\lambda}(\text{vt} - \text x)$

Second other form : In wave motion $\text n = \dfrac{\text v}{\lambda}$, so that $\dfrac{2\pi}{\lambda} = \dfrac{2\pi \text n}{\text v}$. Substituting this in the first form,