Oscillations — HOTS Questions

A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross-sectional area is 4.9 × 10^-7 m². If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad s^-1. If the Young's modulus of the material of the wire is n × 10⁹ N m^-2, find the value of n.

Answer

Given,

• Mass suspended, m = 0.1 kg
• Length of the wire, L = 1 m
• Area of cross-section, A = 4.9 × 10^-7 m²
• Angular frequency, ω = 140 rad s^-1
• Young's modulus, Y = n × 10⁹ N m^-2

The Young's modulus of the material of the wire is

$\text Y = \dfrac{\text F/\text A}{\text l/\text L} = \dfrac{\text{FL}}{\text{Al}} \quad \Rightarrow \quad \text F = \left(\dfrac{\text{YA}}{\text L}\right)\text l$

If the elongation of the wire on suspending the mass m is l, then the restoring force acting on the mass due to elasticity is

$\text F = \text k \text l$

Comparing the two equations, the force-constant of the wire is

$\text k = \dfrac{\text{YA}}{\text L}$

The angular frequency of the resulting S.H.M. is

$ω = \sqrt{\dfrac{\text k}{\text m}} = \sqrt{\dfrac{\text{YA}}{\text{mL}}}$

Substituting the values,

$140 = \left[\dfrac{\text n \times 10^9 \times 4.9 \times 10^{-7}}{0.1 \times 1}\right]^{\frac{1}{2}}$

$140 = \left[\dfrac{4.9 \times 10^2\text n}{0.1}\right]^{\frac{1}{2}} = \left[4900\text n\right]^{\frac{1}{2}} = 70\sqrt{\text n}$

$\sqrt{\text n} = \dfrac{140}{70} = 2 \quad \Rightarrow \quad \text n = 4$

Hence, the value of n is 4.

Based upon the following paragraph three multiple choice type questions are to be answered. Each question has four choices out of which only one is correct.

When a particle of mass m moves on the X-axis in a potential of the form U(x) = kx², it performs simple harmonic motion. The corresponding time period is proportional to $\sqrt{\dfrac{\text m}{\text k}}$, as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of x = 0 in a way different from kx² and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the X-axis. Its potential energy is U(x) = αx⁴ (α > 0) for | x | near the origin and becomes a constant equal to U₀ for | x | ≥ X₀ (see fig. below).

(i) If the total energy of the particle is E, it will perform periodic motion only if :

1. E < 0
2. E > 0
3. U₀ > E > 0
4. E > U₀

(ii) For periodic motion of small amplitude A, the time period T of this particle is proportional to :

1. $\text A\sqrt{\dfrac{\text m}{α}}$

2. $\dfrac{1}{\text A}\sqrt{\dfrac{\text m}{α}}$

3. $\text A\sqrt{\dfrac{α}{\text m}}$

4. $\dfrac{1}{\text A}\sqrt{\dfrac{α}{\text m}}$

(iii) The acceleration of this particle for | x | > X₀ is :

1. proportional to U₀

2. proportional to $\dfrac{\text U_0}{\text m \text X_0}$

3. proportional to $\sqrt{\dfrac{\text U_0}{\text m \text X_0}}$

4. zero

Answer

(i) U₀ > E > 0

The particle will perform periodic motion (and not escape to infinity) only if a restoring force continues to act upon it. This is possible only when the particle remains confined within the potential well, that is, when its total energy is less than the constant value U₀ attained by the potential energy outside the well,

$\text E \lt \text U_0$

Since the potential energy is positive everywhere, E must also be positive. Hence the condition is U₀ > E > 0.

(ii) $\dfrac{1}{\text A}\sqrt{\dfrac{\text m}{α}}$

The dimensions of α follow from U(x) = αx⁴,

$[α] = \dfrac{\text U(\text x)}{\text x^4} = \dfrac{\text{energy}}{(\text{length})^4} = \dfrac{[\text{ML}^2\text T^{-2}]}{[\text L^4]} = [\text{ML}^{-2}\text T^{-2}]$

Now checking the dimensions of the option $\dfrac{1}{\text A}\sqrt{\dfrac{\text m}{α}}$,

$\dfrac{1}{[\text L]}\sqrt{\dfrac{[\text M]}{[\text{ML}^{-2}\text T^{-2}]}} = \dfrac{1}{[\text L]}\sqrt{[\text L^2 \text T^2]} = \dfrac{[\text{LT}]}{[\text L]} = [\text T]$

This has the dimensions of time. Hence the time period is proportional to $\dfrac{1}{\text A}\sqrt{\dfrac{\text m}{α}}$.

(iii) Zero

For | x | > X₀, the potential energy is a constant, U = U₀. Therefore the force acting on the particle is

$\text F = -\dfrac{\text{dU}}{\text{dx}} = 0 \qquad [\because \text U = \text{constant}]$

Since no force acts on the particle in this region, its acceleration is zero.

A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant when it is at a distance $\dfrac{\text{2A}}{3}$ from equilibrium position. The new amplitude of the motion is :

1. 3A

2. $\text A\sqrt3$

3. $\dfrac{\text{7A}}{3}$

4. $\dfrac{\text A}{3}\sqrt{41}$

Answer

$\dfrac{\text{7A}}{3}$

Given,

• Original amplitude of the motion = A
• Displacement at which the speed is trebled, $\text x = \dfrac{2\text A}{3}$

The velocity of a particle in S.H.M. at displacement x is

$\text u = ω\sqrt{\text A^2 - \text x^2}$

At $\text x = \dfrac{2\text A}{3}$,

$\text u = ω\sqrt{\text A^2 - \left(\dfrac{2\text A}{3}\right)^2} = ω\sqrt{\text A^2 - \dfrac{4\text A^2}{9}} = ω\sqrt{\dfrac{5\text A^2}{9}} = \dfrac{\sqrt5ω\text A}{3}$

The speed at this position is now trebled, so the new speed is

$\text u_{new} = 3\text u = \sqrt5ω\text A$

The displacement is unchanged at this instant, so if A_new is the new amplitude,

$\text u_{new} = ω\sqrt{\text A_{new}^2 - \text x^2} = \sqrt5ω\text A$

Squaring both sides and cancelling ω²,

$\text A_{new}^2 - \text x^2 = 5\text A^2$

$\text A_{new}^2 = 5\text A^2 + \left(\dfrac{2\text A}{3}\right)^2 = 5\text A^2 + \dfrac{4\text A^2}{9} = \dfrac{45\text A^2 + 4\text A^2}{9} = \dfrac{49}{9}\text A^2$

$\text A_{new} = \dfrac{7}{3}\text A$

Hence, the new amplitude of the motion is $\dfrac{\text{7A}}{3}$.

A particle is executing S.H.M. along a straight line. Its velocities at distances x₁ and x₂ from the mean position are v₁ and v₂ respectively. Its time period is :

1. $\text{2π}\sqrt{\dfrac{\text x_2^2 - \text x_1^2}{\text v_1^2 - \text v_2^2}}$

2. $\text{2π}\sqrt{\dfrac{\text v_1^2 + \text v_2^2}{\text x_1^2 + \text x_2^2}}$

3. $\text{2π}\sqrt{\dfrac{\text v_1^2 - \text v_2^2}{\text x_1^2 - \text x_2^2}}$

4. $\text{2π}\sqrt{\dfrac{\text x_1^2 - \text x_2^2}{\text v_1^2 - \text v_2^2}}$

Answer

$\text{2π}\sqrt{\dfrac{\text x_2^2 - \text x_1^2}{\text v_1^2 - \text v_2^2}}$

The velocity of a particle in S.H.M. at a displacement x from the mean position is

$\text v = ω\sqrt{\text a^2 - \text x^2}$

where a is the amplitude. Squaring, at the two given displacements,

$\text v_1^2 = ω^2(\text a^2 - \text x_1^2) \qquad \dots(\text i)$

$\text v_2^2 = ω^2(\text a^2 - \text x_2^2) \qquad \dots(\text{ii})$

Subtracting equation (ii) from equation (i), the amplitude a is eliminated,

$\text v_1^2 - \text v_2^2 = ω^2(\text x_2^2 - \text x_1^2)$

$ω = \sqrt{\dfrac{\text v_1^2 - \text v_2^2}{\text x_2^2 - \text x_1^2}}$

But $ω = \dfrac{\text{2π}}{\text T}$, therefore

$\text T = \dfrac{\text{2π}}{ω} = \text{2π}\sqrt{\dfrac{\text x_2^2 - \text x_1^2}{\text v_1^2 - \text v_2^2}}$

A particle moves with simple harmonic motion in a straight line. In first τ s, after starting from rest it travels a distance a and in next τ s, it travels 2a, in same direction, then :

1. amplitude of motion is 3a
2. time period of oscillations is 8τ
3. amplitude of motion is 4a
4. time period of oscillations is 6τ

Answer

time period of oscillations is 6τ

Since the particle starts from rest, it starts from the extreme position. Hence its displacement is measured from the extreme position as

$\text x = \text A\cos ω\text t, \qquad \text{so that at } \text t = 0,\ \text x = \text A$

At t = τ : The particle has travelled a distance a, so x = A − a,

$\text A - \text a = \text A\cos ω\text{τ} \quad \Rightarrow \quad \text a = \text A(1 - \cos ω\text{τ}) \qquad \dots(\text i)$

At t = 2τ : The particle has travelled a further 2a, so the total distance travelled is 3a and x = A − 3a,

$\text A - 3\text a = \text A\cos 2ω\text{τ} \quad \Rightarrow \quad 3\text a = \text A(1 - \cos 2ω\text{τ}) \qquad \dots(\text{ii})$

Dividing equation (i) by equation (ii),

$\dfrac{1}{3} = \dfrac{1 - \cos ω\text{τ}}{1 - \cos 2ω\text{τ}} = \dfrac{1 - \cos ω\text{τ}}{2\sin^2 ω\text{τ}}$

Let x = cos ωτ. Then sin²ωτ = 1 − x², so

$\dfrac{1}{3} = \dfrac{1 - \text x}{2(1 - \text x^2)} = \dfrac{1 - \text x}{2(1 - \text x)(1 + \text x)} = \dfrac{1}{2(1 + \text x)}$

$2(1 + \text x) = 3 \quad \Rightarrow \quad \text x = \dfrac{1}{2} = \cos ω\text{τ}$

Therefore,

$ω\text{τ} = \dfrac{π}{3} \quad \Rightarrow \quad \dfrac{\text{2π}}{\text T}\text{τ} = \dfrac{π}{3}$

$\text T = 6\text{τ}$

Hence, the time period of oscillations is 6τ.

A small block is connected to one end of a massless spring of unstretched length 4.9 m. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by 0.2 m and released from rest at t = 0. It then executes simple harmonic motion with angular frequency $ω = \dfrac{π}{3}$ rad/s. Simultaneously at t = 0, a small pebble is projected with speed v from point P at an angle of 45° as shown in the figure. Point P is at a horizontal distance of 10 m from O. If the pebble hits the block at t = 1 s, the value of v is : (take g = 10 m/s²)

1. $\sqrt{50}$ m/s
2. $\sqrt{51}$ m/s
3. $\sqrt{52}$ m/s
4. $\sqrt{53}$ m/s

Answer

$\sqrt{50}$ m/s

Given,

• Unstretched length of the spring = 4.9 m
• Amplitude of oscillation, a = 0.2 m
• Angular frequency, $ω = \dfrac{π}{3}$ rad s^-1
• Horizontal distance OP = 10 m
• Angle of projection = 45°, time of flight = 1 s, g = 10 m s^-2

Position of the block at t = 1 s : The time period of the block is

$\text T = \dfrac{\text{2π}}{ω} = \dfrac{\text{2π} \times 3}{π} = 6\ \text s$

The block is released from the stretched (extreme) position, so its displacement from the mean position is

$\text x = \text a\cos ω\text t = 0.2\cos\left(\dfrac{\text{2π}}{\text T}\text t\right)$

At t = 1 s,

$\text x = 0.2\cos\left(\dfrac{\text{2π}}{6} \times 1\right) = 0.2\cos\dfrac{π}{3} = \dfrac{0.2}{2} = 0.1\ \text m$

Hence the position of the block measured from the wall at t = 1 s is

$4.9\ \text m + 0.1\ \text m = 5\ \text m$

Range of the pebble : Since P is at 10 m from O, the horizontal distance covered by the pebble is

$\text R = 10 - 5 = 5\ \text m$

The horizontal component of the velocity of the pebble remains constant, so

$\text v\cos 45° \times \text t = 5\ \text m$

$\text v \times \dfrac{1}{\sqrt2} \times 1 = 5 \quad \Rightarrow \quad \text v = 5\sqrt2 = \sqrt{50}\ \text{m s}^{-1}$

Hence, the value of v is $\sqrt{50}$ m/s.

A point mass is subjected to two simultaneous sinusoidal displacements in X-direction, x₁(t) = A sin ωt and $\text x_2(\text t) = \text A\sin\left(ω\text t + \dfrac{\text{2π}}{3}\right)$. Adding a third sinusoidal displacement x₃(t) = B sin (ωt + φ) brings the mass to a complete rest. The values of B and φ are :

1. $\sqrt2\text A, \dfrac{\text{3π}}{4}$
2. $\text A, \dfrac{\text{4π}}{3}$
3. $\sqrt3\text A, \dfrac{\text{5π}}{6}$
4. $\text A, \dfrac{π}{3}$

Answer

$\text A, \dfrac{\text{4π}}{3}$

Given,

• x₁(t) = A sin ωt
• $\text x_2(\text t) = \text A\sin\left(ω\text t + \dfrac{\text{2π}}{3}\right)$

The two displacements x₁ and x₂ have equal amplitudes A and a phase difference of $\dfrac{\text{2π}}{3}$. Their resultant amplitude is

$\text R = \sqrt{\text A^2 + \text A^2 + 2\text A \cdot \text A\cos\dfrac{\text{2π}}{3}}$

Since $\cos\dfrac{\text{2π}}{3} = -\dfrac{1}{2}$,

$\text R = \sqrt{2\text A^2 + 2\text A^2\left(-\dfrac{1}{2}\right)} = \sqrt{2\text A^2 - \text A^2} = \text A$

Since the two amplitudes are equal, this resultant bisects the angle between them, so its phase is $\dfrac{π}{3}$.

The third displacement x₃(t) = B sin (ωt + φ) will bring the mass to complete rest only when it is equal and opposite to this resultant. Hence,

$\text B = \text A \qquad \text{and} \qquad φ = π + \dfrac{π}{3} = \dfrac{\text{4π}}{3}$

Hence, B = A and $φ = \dfrac{\text{4π}}{3}$.

For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against its displacement d. Which one of the following represents these correctly?

(Graphs are schematic and not drawn to scale)

Answer

For a particle executing S.H.M. with amplitude a, the kinetic energy and the potential energy at a displacement d are

$\text K = \dfrac{1}{2}\text m ω^2(\text a^2 - \text d^2) \qquad \text{and} \qquad \text U = \dfrac{1}{2}\text m ω^2 \text d^2$

From these expressions,

• The kinetic energy is maximum at the mean position (d = 0) and becomes minimum (zero) at the extreme positions (d = ± a).
• The potential energy is minimum (zero) at the mean position and becomes maximum at the extreme positions.

Both graphs are parabolas, the KE curve opening downwards and the PE curve opening upwards, and their sum is the constant total energy. Only graph (b) shows the KE curve with its maximum at d = 0 and the PE curve with its minimum at d = 0.

A simple pendulum has a time period T₁. When its point of suspension is moved vertically upwards according as y = k t², where y is vertical distance covered and k = 1 m/s², its time period becomes T₂. Then, $\dfrac{\text T_1^2}{\text T_2^2}$ is (g = 10 m/s²) :

1. $\dfrac{5}{6}$

2. $\dfrac{11}{10}$

3. $\dfrac{6}{5}$

4. $\dfrac{5}{4}$

Answer

$\dfrac{6}{5}$

Given,

• Vertical distance covered by the point of suspension, y = kt²
• k = 1 m s^-2
• g = 10 m s^-2

The upward acceleration of the point of suspension is obtained by differentiating y twice with respect to time,

$\dfrac{\text{dy}}{\text{dt}} = \dfrac{\text d}{\text{dt}}(\text k \text t^2) = 2\text k \text t$

$\text a_y = \dfrac{\text d^2\text y}{\text{dt}^2} = \dfrac{\text d}{\text{dt}}(2\text k \text t) = 2\text k = 2 \times 1 = 2\ \text{m s}^{-2}$

When the support moves upward with acceleration a_y, the effective value of the acceleration due to gravity becomes (g + a_y). Hence the two time periods are

$\text T_1 = \text{2π}\sqrt{\dfrac{\text l}{\text g}} \qquad \text{and} \qquad \text T_2 = \text{2π}\sqrt{\dfrac{\text l}{\text g + \text a_y}}$

Squaring and dividing,

$\dfrac{\text T_1^2}{\text T_2^2} = \dfrac{\text g + \text a_y}{\text g} = \dfrac{(10 + 2)\ \text{m s}^{-2}}{10\ \text{m s}^{-2}} = \dfrac{12}{10} = \dfrac{6}{5}$

Hence, $\dfrac{\text T_1^2}{\text T_2^2} = \dfrac{6}{5}$.

Two particles are executing simple harmonic motion of the same amplitude A and frequency ω along the X-axis. Their mean position is separated by distance X₀ (X₀ > A). If the maximum separation between them is (X₀ + A), the phase difference between their motions is :

1. $\dfrac{π}{6}$

2. $\dfrac{π}{2}$

3. $\dfrac{π}{3}$

4. $\dfrac{π}{4}$

Answer

$\dfrac{π}{3}$

Given,

• Amplitude of each particle = A
• Angular frequency of each particle = ω
• Separation between the mean positions = X₀
• Maximum separation = X₀ + A

Let the displacements of the two particles be

$\text x_1 = \text A\sin ω\text t \qquad \text{and} \qquad \text x_2 = \text X_0 + \text A\sin(ω\text t + φ)$

The separation between them at any instant is

$\text x_2 - \text x_1 = \text X_0 + \text A\sin(ω\text t + φ) - \text A\sin ω\text t$

Using the identity $\sin \text C - \sin \text D = 2\cos\left(\dfrac{\text C + \text D}{2}\right)\sin\left(\dfrac{\text C - \text D}{2}\right)$,

$\text x_2 - \text x_1 = \text X_0 + 2\text A\sin\left(\dfrac{φ}{2}\right)\cos\left(ω\text t + \dfrac{φ}{2}\right)$

The maximum value of the cosine term is 1, so the maximum separation is

$(\text x_2 - \text x_1)_{max} = \text X_0 + 2\text A\sin\dfrac{φ}{2}$

But it is given that the maximum separation is X₀ + A. Therefore,

$\text X_0 + 2\text A\sin\dfrac{φ}{2} = \text X_0 + \text A$

$\sin\dfrac{φ}{2} = \dfrac{1}{2} \quad \Rightarrow \quad \dfrac{φ}{2} = \dfrac{π}{6} \quad \Rightarrow \quad φ = \dfrac{π}{3}$

Hence, the phase difference between their motions is $\dfrac{π}{3}$.

A piece of wire is bent in the space of a parabola y = k x² (Y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the X-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the Y-axis is :

1. $\dfrac{\text a}{\text g \text k}$

2. $\dfrac{\text a}{2\text g \text k}$

3. $\dfrac{2\text a}{\text g \text k}$

4. $\dfrac{\text a}{4\text g \text k}$

Answer

$\dfrac{\text a}{2\text g \text k}$

Given,

• Equation of the parabolic wire, y = kx²
• Horizontal acceleration of the wire = a

Let the bead settle at a point where the tangent to the parabola makes an angle θ with the horizontal. In the frame of the wire, the bead is in equilibrium under its weight mg, the normal reaction of the smooth wire, and the pseudo force ma acting horizontally opposite to the acceleration.

Resolving along and perpendicular to the wire, the condition of equilibrium gives

$\text{ma} = \text{mg}\tan θ \quad \Rightarrow \quad \text a = \text g\tan θ$

The slope of the parabola at the point (x, y) is

$\tan θ = \dfrac{\text{dy}}{\text{dx}} = \dfrac{\text d}{\text{dx}}(\text k \text x^2) = 2\text k \text x$

Substituting this value,

$\text a = \text g \times 2\text k \text x$

$\text x = \dfrac{\text a}{2\text g \text k}$

Hence, the new equilibrium position of the bead is at a distance $\dfrac{\text a}{2\text g \text k}$ from the Y-axis.

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and the cylinder have equal cross-sectional area A. When the piston is in equilibrium, the volume of the gas is V₀ and its pressure is P₀. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency :

1. $\dfrac{1}{\text{2π}}\dfrac{\text A γ \text P_0}{\text V_0 \text M}$

2. $\dfrac{1}{\text{2π}}\dfrac{\text V_0 \text M \text P_0}{\text A^2 γ}$

3. $\dfrac{1}{\text{2π}}\sqrt{\dfrac{\text A^2 γ \text P_0}{\text M \text V_0}}$

4. $\dfrac{1}{\text{2π}}\sqrt{\dfrac{\text M \text V_0}{\text A γ \text P_0}}$

Answer

$\dfrac{1}{\text{2π}}\sqrt{\dfrac{\text A^2 γ \text P_0}{\text M \text V_0}}$

Given,

• Mass of the piston = M
• Area of cross-section = A
• Equilibrium volume of the gas = V₀
• Equilibrium pressure of the gas = P₀

Since the system is completely isolated from its surroundings, the process is adiabatic. Hence, by Poisson's law,

$\text{PV}^γ = \text{a constant}$

Taking logarithms and differentiating,

$\dfrac{\text{dP}}{\text P} + γ\dfrac{\text{dV}}{\text V} = 0 \quad \Rightarrow \quad \text{dP} = -\dfrac{γ\text P\text{dV}}{\text V}$

If the piston is displaced through a small distance x, the change in volume is dV = Ax. The net restoring force on the piston is

$\text F_{net} = \text{dP} \times \text A = -\dfrac{γ\text P_0 (\text{Ax})}{\text V_0} \times \text A = -\left(\dfrac{γ\text P_0 \text A^2}{\text V_0}\right)\text x$

By Newton's second law of motion, F_net = Mα, so the acceleration of the piston is

$α = -\left(\dfrac{γ\text P_0 \text A^2}{\text M \text V_0}\right)\text x \qquad \dots(\text i)$

Since α ∝ − x, the motion of the piston is simple harmonic. Comparing equation (i) with α = − ω²x,

$ω^2 = \dfrac{γ\text P_0 \text A^2}{\text M \text V_0}$

The frequency of oscillation is

$\text n = \dfrac{ω}{\text{2π}} = \dfrac{1}{\text{2π}}\sqrt{\dfrac{\text A^2 γ \text P_0}{\text M \text V_0}}$

If x, v and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period T, then, which of the following does not change with time?

1. $\dfrac{\text a \text T}{\text v}$
2. a²T² + 4π²v²
3. $\dfrac{\text a \text T}{\text x}$
4. aT + 2πv

Answer

a²T² + 4π²v² and $\dfrac{\text a \text T}{\text x}$

For a particle executing S.H.M. with amplitude A,

$\text x = \text A\sin ω\text t, \qquad \text v = \text Aω\cos ω\text t, \qquad \text a = -\text Aω^2\sin ω\text t$

Checking a²T² + 4π²v² : Substituting the values and using $\text T = \dfrac{\text{2π}}{ω}$,

$\text a^2\text T^2 + 4π^2\text v^2 = \text A^2ω^4\sin^2 ω\text t \times \dfrac{4π^2}{ω^2} + 4π^2 \text A^2ω^2\cos^2 ω\text t$

$= 4π^2\text A^2ω^2\sin^2 ω\text t + 4π^2\text A^2ω^2\cos^2 ω\text t$

$= 4π^2\text A^2ω^2(\sin^2 ω\text t + \cos^2 ω\text t) = 4π^2\text A^2ω^2 = \text{a constant}$

Checking $\dfrac{\text a \text T}{\text x}$ :

$\dfrac{\text a\text T}{\text x} = \dfrac{-\text Aω^2\sin ω\text t \times \dfrac{\text{2π}}{ω}}{\text A\sin ω\text t} = -\text{2π}ω = \text{a constant}$

The remaining two expressions contain sin ωt and cos ωt separately and therefore change with time.

Hence, the quantities a²T² + 4π²v² and $\dfrac{\text a \text T}{\text x}$ do not change with time.

Note: The question is worded as though only one option is correct, but two of the given expressions, options 2 and 3, are both independent of time. The textbook answer key also marks both (b) and (c) as correct.

A particle of mass m is attached to one end of a massless spring of force constant k, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time t = 0 with an initial velocity v₀. When the speed of the particle is 0.5 v₀, it collides elastically with a rigid wall. After this collision :

1. the speed of the particle when it returns to its equilibrium position is v₀
2. the time at which the particle passes through the equilibrium position for the first time is $\text t = π\sqrt{\dfrac{\text m}{\text k}}$
3. the time at which the maximum compression of the spring occurs is $\text t = \dfrac{\text{4π}}{3}\sqrt{\dfrac{\text m}{\text k}}$
4. the time at which the particle passes through the equilibrium position for the second time is $\text t = \dfrac{\text{5π}}{3}\sqrt{\dfrac{\text m}{\text k}}$

Answer

Options 1 and 4

Since the particle starts from the equilibrium position, its displacement is

$\text x = \text a\sin ω\text t \quad \Rightarrow \quad \text v = \text aω\cos ω\text t = \text v_0\cos ω\text t$

Checking option 1 : The collision with the rigid wall is elastic, so no mechanical energy is lost. Hence, when the particle returns to its equilibrium position it again has the full speed v₀. Option 1 is correct.

Time of collision : At the instant of collision the speed is 0.5 v₀,

$\dfrac{\text v_0}{2} = \text v_0\cos ω\text t_1 \quad \Rightarrow \quad \cos ω\text t_1 = \dfrac{1}{2} \quad \Rightarrow \quad ω\text t_1 = \dfrac{π}{3} \quad \Rightarrow \quad \text t_1 = \dfrac{π}{3ω}$

Checking option 2 : After the elastic collision the particle retraces its path, so it returns to the equilibrium position after a further time t₁. Hence

$\text t_2 = 2\text t_1 = \dfrac{\text{2π}}{3ω} = \dfrac{\text{2π}}{3} \times \dfrac{1}{\text{2π}}\text T = \dfrac{\text{2π}}{3}\sqrt{\dfrac{\text m}{\text k}}$

using $ω = \sqrt{\dfrac{\text k}{\text m}}$. This is not $π\sqrt{\dfrac{\text m}{\text k}}$, so option 2 is incorrect.

Checking option 3 : After crossing the equilibrium position, the spring is compressed and the compression is maximum a quarter of a period later,

$\text t_3 = \text t_2 + \dfrac{\text T}{4} = \dfrac{\text{2π}}{3}\sqrt{\dfrac{\text m}{\text k}} + \dfrac{\text{2π}}{4}\sqrt{\dfrac{\text m}{\text k}} = \dfrac{\text{7π}}{6}\sqrt{\dfrac{\text m}{\text k}}$

This is not $\dfrac{\text{4π}}{3}\sqrt{\dfrac{\text m}{\text k}}$, so option 3 is incorrect.

Checking option 4 : The particle passes the equilibrium position for the second time a further quarter period later,

$\text t_4 = \text t_3 + \dfrac{\text T}{4} = \dfrac{\text{7π}}{6}\sqrt{\dfrac{\text m}{\text k}} + \dfrac{\text{2π}}{4}\sqrt{\dfrac{\text m}{\text k}} = \dfrac{\text{5π}}{3}\sqrt{\dfrac{\text m}{\text k}}$

Option 4 is correct.

Hence, options 1 and 4 are correct.

Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies ω₁ and ω₂ and have total energies E₁ and E₂, respectively. Then variations of their momenta p with positions x are shown in the figures. If $\dfrac{\text a}{\text b} = \text n^2$ and $\dfrac{\text a}{\text R} = \text n$, then the correct equations are :

1. E₁ω₁ = E₂ω₂
2. $\dfrac{ω_2}{ω_1} = \text n^2$
3. ω₁ω₂ = n²
4. $\dfrac{\text E_1}{ω_1} = \dfrac{\text E_2}{ω_2}$

Answer

Options 2 and 4

Given,

• $\dfrac{\text a}{\text b} = \text n^2$ and $\dfrac{\text a}{\text R} = \text n$
• Both oscillators have the same mass m

From the figures, the first oscillator gives an ellipse with semi-axes a (along x) and b (along p), while the second gives a circle of radius R.

First oscillator : The amplitude is a and the maximum momentum is b, so

$\text p_1 = \text m\text a ω_1 = \text b \qquad \dots(\text i)$

Second oscillator : The amplitude is R and the maximum momentum is also R, so

$\text p_2 = \text m \text R ω_2 = \text R \quad \Rightarrow \quad \text m ω_2 = 1 \qquad \dots(\text{ii})$

From equation (i), $\dfrac{\text b}{\text a} = \text m ω_1$. Dividing this by equation (ii),

$\dfrac{ω_1}{ω_2} = \dfrac{\text b}{\text a} = \dfrac{1}{\text n^2} \quad \Rightarrow \quad \dfrac{ω_2}{ω_1} = \text n^2$

Hence option 2 is correct.

Total energies : For a harmonic oscillator of mass m, amplitude A and angular frequency ω, the total energy is $\text E = \dfrac{1}{2}\text m ω^2 \text A^2$. Therefore

$\text E_1 = \dfrac{1}{2}\text m ω_1^2 \text a^2 \qquad \text{and} \qquad \text E_2 = \dfrac{1}{2}\text m ω_2^2 \text R^2$

Dividing,

$\dfrac{\text E_1}{\text E_2} = \dfrac{ω_1^2}{ω_2^2} \times \left(\dfrac{\text a}{\text R}\right)^2 = \dfrac{ω_1^2}{ω_2^2} \times \text n^2$

Substituting $\text n^2 = \dfrac{ω_2}{ω_1}$,

$\dfrac{\text E_1}{\text E_2} = \dfrac{ω_1^2}{ω_2^2} \times \dfrac{ω_2}{ω_1} = \dfrac{ω_1}{ω_2}$

$\dfrac{\text E_1}{ω_1} = \dfrac{\text E_2}{ω_2}$

Hence options 2 and 4 are correct.

(i) Time period of a particle in S.H.M. depends on the force-constant k and mass m of the particle : $\text T = \text{2π}\sqrt{\dfrac{\text m}{\text k}}$. A simple pendulum executes S.H.M. approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?

(ii) The motion of a simple pendulum is approximately simple harmonic for small angles of oscillation. For larger angles of oscillation, T is worked out to be greater than $\text{2π}\sqrt{\dfrac{\text l}{\text g}}$. Explain this result.

(iii) A man with a wrist watch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?

(iv) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is falling freely under gravity?

Answer

(i) In the case of a simple pendulum, the restoring force acting on the bob is

$\text F = -\left(\dfrac{\text{mg}}{\text l}\right)\text x = -\text k \text x$

so the force-constant is $\text k = \dfrac{\text{mg}}{\text l}$. Substituting this value in the relation $\text T = \text{2π}\sqrt{\dfrac{\text m}{\text k}}$,

$\text T = \text{2π}\sqrt{\dfrac{\text m}{\text{mg}/\text l}} = \text{2π}\sqrt{\dfrac{\text l}{\text g}}$

The mass m cancels out, because for a pendulum the force-constant itself is proportional to the mass. Hence the time period of a pendulum does not depend upon the mass of its bob.

(ii) For larger angles θ of oscillation, the restoring force on the bob of a simple pendulum is

$\text F = -\text{mg}\sin θ = -\text m \text g_e'$

where g_e′ (= g sin θ) is the effective value of g. For a small angle we take sin θ ≈ θ, so the effective value is g_e = gθ.

But sin θ < θ for every value of θ. Therefore

$\text g_e' \lt \text g_e$

that is, the effective value of g decreases for a larger value of θ. Since $\text T = \text{2π}\sqrt{\dfrac{\text l}{\text g}}$, a reduction in g implies an increase in T. Hence for larger angles the time period is greater than $\text{2π}\sqrt{\dfrac{\text l}{\text g}}$.

(iii) Yes, the wrist watch gives correct time during the free fall. The reason is that a wrist watch works on spring action; it has nothing to do with the acceleration due to gravity.

(iv) In free fall the effective value of the acceleration due to gravity becomes zero, that is, g = 0. Hence the frequency of oscillation of the simple pendulum,

$\text n = \dfrac{1}{\text{2π}}\sqrt{\dfrac{\text g}{\text l}}$

becomes zero. The pendulum does not oscillate at all.

Show that for a particle in linear S.H.M., the average kinetic energy over a period of oscillation is equal to the average potential energy over the same period.

Answer

Let a particle of mass m perform S.H.M. with time period T and angular frequency $ω\left(= \dfrac{\text{2π}}{\text T}\right)$. Its instantaneous displacement from the mean position is

$\text y = \text a\sin ω\text t$

The instantaneous kinetic energy and potential energy are

$\text K = \dfrac{1}{2}\text m ω^2(\text a^2 - \text y^2) = \dfrac{1}{2}\text m ω^2(\text a^2 - \text a^2\sin^2 ω\text t) = \dfrac{1}{2}\text m ω^2 \text a^2\cos^2 ω\text t$

$\text U = \dfrac{1}{2}\text m ω^2 \text y^2 = \dfrac{1}{2}\text m ω^2 \text a^2\sin^2 ω\text t$

Average kinetic energy over one oscillation :

$\text K_{av} = \dfrac{1}{\text T}\int_0^{\text T} \dfrac{1}{2}\text m ω^2 \text a^2\cos^2 ω\text t\ \text{dt}$

Using $\cos^2 ω\text t = \dfrac{1 + \cos 2ω\text t}{2}$,

$\text K_{av} = \dfrac{\text m ω^2 \text a^2}{2\text T}\int_0^{\text T}\dfrac{1 + \cos 2ω\text t}{2}\ \text{dt} = \dfrac{\text m ω^2 \text a^2}{4\text T}\left[\text t + \dfrac{\sin 2ω\text t}{2ω}\right]_0^{\text T}$

Since $\sin 2ω\text T = \sin \text{4π} = 0$, the second term vanishes at both the limits,

$\text K_{av} = \dfrac{\text m ω^2 \text a^2}{4\text T}[\text T] = \dfrac{1}{4}\text m ω^2 \text a^2 \qquad \dots(\text i)$

Average potential energy over one oscillation :

$\text U_{av} = \dfrac{1}{\text T}\int_0^{\text T}\dfrac{1}{2}\text m ω^2 \text a^2\sin^2 ω\text t\ \text{dt}$

Using $\sin^2 ω\text t = \dfrac{1 - \cos 2ω\text t}{2}$,

$\text U_{av} = \dfrac{\text m ω^2 \text a^2}{2\text T}\int_0^{\text T}\dfrac{1 - \cos 2ω\text t}{2}\ \text{dt} = \dfrac{\text m ω^2 \text a^2}{4\text T}\left[\text t - \dfrac{\sin 2ω\text t}{2ω}\right]_0^{\text T}$

$\text U_{av} = \dfrac{\text m ω^2 \text a^2}{4\text T}(\text T) = \dfrac{1}{4}\text m ω^2 \text a^2 \qquad \dots(\text{ii})$

From equations (i) and (ii),

$\text K_{av} = \text U_{av}$

Hence, for a particle in linear S.H.M. the average kinetic energy over a period of oscillation is equal to the average potential energy over the same period, each being one-half of the total energy.

A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillation is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant (torsional rigidity) of the wire.

Answer

Given,

• Mass of the disc, M = 10 kg
• Radius of the disc, R = 15 cm = 0.15 m
• Period of torsional oscillation, T = 1.5 s

The torsional oscillations of the disc are angular simple harmonic oscillations, whose time period is given by

$\text T = \text{2π}\sqrt{\dfrac{\text I}{\text c}}$

where I is the moment of inertia of the disc about the wire as axis and c is the torsional rigidity of the wire.

The moment of inertia of a circular disc about an axis passing through its centre and perpendicular to its plane is

$\text I = \dfrac{1}{2}\text{MR}^2 \\[1em] = \dfrac{1}{2} \times 10\ \text{kg} \times (0.15\ \text m)^2 \\[1em] = \dfrac{1}{2} \times 10 \times 0.0225 \\[1em] = 0.1125\ \text{kg m}^2 \\[1em]$

Squaring the expression for T and solving for c,

$\text c = \dfrac{4π^2 \text I}{\text T^2}$

Substituting the values,

$\text c = \dfrac{4 \times (3.14)^2 \times 0.1125\ \text{kg m}^2}{(1.5\ \text s)^2} \\[1em] = \dfrac{4 \times 9.8596 \times 0.1125}{2.25} \\[1em] = \dfrac{4.4368}{2.25}$

$\text c = 1.97\ \text{N m}$

Hence, the torsional spring constant of the wire is 1.97 N m.

A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (i) 5 cm, (ii) 3 cm, (iii) 0 cm.

Answer

Given,

• Amplitude, a = 5 cm = 0.05 m
• Time period, T = 0.2 s

The angular frequency of the motion is

$ω = \dfrac{\text{2π}}{\text T} = \dfrac{\text{2π}}{0.2\ \text s} = 10π\ \text s^{-1}$

For a body in S.H.M. at displacement y, the acceleration and the velocity are

$α = -ω^2 \text y \qquad \text{and} \qquad \text u = ω\sqrt{\text a^2 - \text y^2}$

(i) When y = 5 cm = 0.05 m :

$α = -(10π\ \text s^{-1})^2(0.05\ \text m) = -5π^2\ \text{m s}^{-2}$

$\text u = (10π\ \text s^{-1})\sqrt{(0.05\ \text m)^2 - (0.05\ \text m)^2} = 0$

(ii) When y = 3 cm = 0.03 m :

$α = -(10π\ \text s^{-1})^2(0.03\ \text m) = -3π^2\ \text{m s}^{-2}$

$\text u = (10π\ \text s^{-1})\sqrt{(0.05\ \text m)^2 - (0.03\ \text m)^2} = (10π\ \text s^{-1})\sqrt{0.0025 - 0.0009}$

$\text u = (10π\ \text s^{-1})(0.04\ \text m) = 0.4π\ \text{m s}^{-1}$

(iii) When y = 0 :

$α = -ω^2(0) = 0$

$\text u = ω\sqrt{\text a^2 - 0} = ω\text a = (10π\ \text s^{-1})(0.05\ \text m) = 0.5π\ \text{m s}^{-1}$

Hence, at y = 5 cm the acceleration is − 5π² m s^-2 and the velocity is zero; at y = 3 cm the acceleration is − 3π² m s^-2 and the velocity is 0.4π m s^-1; and at y = 0 the acceleration is zero and the velocity is 0.5π m s^-1.

Note: The velocity at the mean position is printed in the textbook as 0.05π m s^-1. Substituting ω = 10π s^-1 and a = 0.05 m in u = ωa gives 0.5π m s^-1, so the printed value carries a decimal-place slip.

A mass attached to a spring is free to oscillate, with angular velocity ω in a horizontal plane without friction. It is pulled to a distance x₀ and pushed towards the centre with a velocity v₀ at time t = 0. Determine the amplitude of the resulting oscillation in terms of the parameters ω, x₀ and v₀.

Answer

Given,

• Initial displacement at t = 0 = x₀
• Initial velocity at t = 0 = v₀
• Angular velocity of the oscillation = ω

Let the general equation for the displacement of the mass be

$\text x = \text a\cos(ω\text t + θ)$

where a is the amplitude and θ the initial phase. Differentiating with respect to time, the velocity of the particle is

$\text v = \dfrac{\text{dx}}{\text{dt}} = -\text aω\sin(ω\text t + θ)$

At t = 0, x = x₀ and v = v₀. Therefore,

$\text x_0 = \text a\cos θ \quad \Rightarrow \quad \dfrac{\text x_0}{\text a} = \cos θ \qquad \dots(\text i)$

$\text v_0 = -\text aω\sin θ \quad \Rightarrow \quad \dfrac{\text v_0}{\text aω} = -\sin θ \qquad \dots(\text{ii})$

Squaring and adding equations (i) and (ii), and using sin²θ + cos²θ = 1,

$\left(\dfrac{\text x_0}{\text a}\right)^2 + \left(\dfrac{\text v_0}{\text aω}\right)^2 = 1$

$\text a^2 = \text x_0^2 + \left(\dfrac{\text v_0}{ω}\right)^2$

$\text a = \sqrt{\text x_0^2 + \left(\dfrac{\text v_0}{ω}\right)^2}$

Hence, the amplitude of the resulting oscillation is $\text a = \sqrt{\text x_0^2 + \left(\dfrac{\text v_0}{ω}\right)^2}$.

A particle of mass m placed on smooth horizontal surface is attached to three identical springs A, B and C each of force constant k as shown in the figure. If the particle of mass m is pushed slightly against the spring A and released, find the time period of oscillation.

Answer

Given,

• Mass of the particle = m
• Force-constant of each of the three springs = k
• The springs B and C are inclined at 45° on either side of the line of the spring A

Let the particle at O be pushed through a small distance y in the direction of the spring A. Then the spring A is compressed by y, while the springs B and C are each stretched by

$\text y' = \text y\cos 45°$

The restoring forces due to B and C act along their own lengths; their components along the line OA are F_B cos 45° and F_C cos 45°. Hence the total restoring force acting on m along AO is

$\text F = -(\text F_A + \text F_B\cos 45° + \text F_C\cos 45°)$

$= -(\text{ky} + \text{ky}'\cos 45° + \text{ky}'\cos 45°) = -(\text{ky} + 2\text{ky}'\cos 45°)$

Substituting y′ = y cos 45°,

$\text F = -\bigg\lbrace\text{ky} + 2\text k(\text y\cos 45°)\cos 45°\bigg\rbrace = -\bigg\lbrace\text{ky} + 2\text{ky}\left(\dfrac{1}{\sqrt2}\right)\left(\dfrac{1}{\sqrt2}\right)\bigg\rbrace$

$\text F = -(\text{ky} + \text{ky}) = -2\text{ky}$

By Newton's second law of motion, F = mα, so

$\text m α = -2\text{ky} \quad \Rightarrow \quad α = -\left(\dfrac{2\text k}{\text m}\right)\text y$

Since α ∝ − y, the motion of the particle is simple harmonic. Comparing with α = − ω²y,

$ω^2 = \dfrac{2\text k}{\text m}$

The time period of oscillation is

$\text T = \dfrac{\text{2π}}{ω} = \text{2π}\sqrt{\dfrac{\text m}{2\text k}}$

Hence, the time period of oscillation of the particle is $\text T = \text{2π}\sqrt{\dfrac{\text m}{2\text k}}$.