(a) Explanation Given info: The time after which the echo was heard is 1.5 s . The speed of the sound is 343 m / s . The depth of the well is, d = v × t (1) Here, v is the speed of the sound in air. t is time taken by the sound of splash to come out of the well. Form equation of motion. d = 1 2 g T 2 (2) Here, g is the acceleration due to gravity. T is time taken by the stone to hit the water in the well. Compare equation (1) and (2). v × t = 1 2 g T 2 (3) The total time after the echo is heard, t + T = 1.5 s . (4) Rearrange the above equation. t = 1.5 s − T Substitute 1.5 s − T for t in equation (3). v ( 1.5 s − T ) = 1 2 g T 2 1.5 v − v T = 1 2 g T 2 T 2 + 2 v T g − 3 v g = 0 Substitute 343 m / s for v and 9.8 m / s 2 for g in the above equation. T 2 + 2 ( 343 m / s ) T 9 (b) To determine The change in the total time to hear echo if the depth is doubled.

Physics, Books a la Carte Edition (5th Edition)

5th Edition

ISBN: 9780134020853

Author: James S. Walker

Publisher: PEARSON

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Wave Motion

A wave motion is defined as the propagation of disturbances or transfer of energy and momentum from the state of rest or equilibrium to another point in a medium without any actual transfer of matter between them. In a simple explanation, it is the propa…

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Chapter 14, Problem 32PCE

(a)

To determine

The depth of the well.

(b)

To determine

The change in the total time to hear echo if the depth is doubled.

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Chapter 14, Problem 31PCE

Chapter 14, Problem 33PCE

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The depth of the well.

Solution Summary: The author explains that the depth of the well is 10.5m.

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The depth of the well.

The change in the total time to hear echo if the depth is doubled.

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Chapter 14 Solutions