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Physics

A person standing between two vertical cliffs and 640 m away from the nearest cliff produces a sound. He hears the first echo after 4 s and the second echo 3 s later. Calculate :

(a) The speed of sound in air and

(b) The distance between the cliffs

Sound

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Answer

Figure given below shows that two cliffs and the position of the person.

Figure given below shows that two cliffs and the position of the person. Concise Physics Solutions ICSE Class 10.

(a) First echo is heard from the nearest cliff.

Let d1, be the distance of the nearest cliff 1 from the person.

Total distance travelled by the sound in going and then coming back = 2d₁ = 2 x 640 m = 1280 m

Time taken (t) = 4 s

Speed of sound (v)=Tot. dist. sound travelsTime takenTime taken=2d1t=12804=320 ms−1\text {Speed of sound (v)} = \dfrac{\text {Tot. dist. sound travels} }{\text {Time taken}} \\[1em] {\text {Time taken}}=\dfrac{2d_1}{\text t} \\[1em] =\dfrac{1280}{4}=320\text{ ms}^{-1}

(b) The second echo is heard from the farther cliff 2. If d2 is the distance of the farther cliff 2 from the person, then total distance travelled by the sound in going and then coming back = 2d2.

Time taken (t) = 4 + 3 = 7 s

Now,

Speed of sound (v)=Tot. dist. sound travelsTime taken⇒v=2d2t⇒d2=vt2⇒d2=320×72⇒d2=1120 m\text {Speed of sound (v)} = \dfrac{\text {Tot. dist. sound travels} }{\text {Time taken}} \\[1em] \Rightarrow \text v = \dfrac{2d2}{\text t} \\[1em] \Rightarrow \text d2 = \dfrac{\text{vt}}{2} \\[1em] \Rightarrow \text d2 = \dfrac{320\times7}{2} \\[1em] \Rightarrow \text d2 = 1120 \text{ m}

Distance between the two cliffs = d1 + d2 = 640 m + 1120 m = 1760 m

Hence, speed of sound in air is 320 ms-1 and distance between the cliffs is 1760 m.

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