Two persons A and B are standing in front of a cliff in the same line 170 m apart as shown in the diagram.

Person B fires the gun and hears the echo in 3 s. Then the person A standing in front of the person B fires the gun.

(Speed of sound in air is 340 m s^-1)

(a) Calculate:

1. the distance of the person B from the cliff.
2. the minimum time in which B hears the gunshot fired by A.

(b) Fill in the blank.

The echo is softer (less loud) than the original sound due to the decrease in …………… [amplitude / frequency] of the wave.

Given,

• Distance between A and B = 170 m
• Time taken by the person B to hear the echo = 3 s
• Speed of sound in air = 340 ms^-1

(a)

1. Let, the distance of the person B from the cliff be d.

So,

$\text d = \dfrac {\text {Speed of sound in air} \times \text {Time taken}}{2} \\[1em] = \dfrac{340 \times 3}{2} \\[1em] = 170 \times 3 \\[1em] = 510\ \text m$

Hence, the distance of the person B from the cliff is 510 m.

2. Person B first hears the direct sound of the gunshot and only afterwards hears its echo from the cliff. Therefore, the minimum time in which B hears the gunshot is the time taken by the sound to travel directly from person A to person B, who is 170 m away.

Then, minimum time (t) is given by,

$\text t = \dfrac{\text {Distance between A and B}}{\text {Speed of sound in air}} \\[1em] = \dfrac{170}{340} \\[1em] = \dfrac{1}{2} \\[1em] = 0.5\ \text s$

Hence, the minimum time in which B hears the gunshot fired by A is 0.5 second.

(b) The echo is softer (less loud) than the original sound due to the decrease in amplitude of the wave.