Echo

Echo

When we shout or clap near a suitable reflecting surface such as a tall building or a mountain, we will hear the same sound again a little later. This sound which we hear is called an echo. The sensation of sound persists in our brain for about 0.1s.

Hence, to hear a distinct echo the time interval between the original sound and the reflected sound must be at least 0.1s. Let us consider the speed of sound to be 340 m s^–1 at 25° C. The sound must go to the obstacle and return to the ear of the listener on reflection after 0.1 s. Thus, the total distance covered by the sound from the point of generation to the reflecting surface and back should be at least 340 m s^–1 × 0.1 s = 34 m.

Thus, for hearing distinct echoes, the minimum distance of the obstacle from the source of sound must be half of this distance i.e. 17 m. This distance will change with the temperature of air. Echoes may be heard more than once due to successive or multiple reflections. The roaring of thunder is due to the successive reflections of the sound from a number of reflecting surfaces, such as the clouds at different heights and the land.

Example 3

A person claps his hands near a cliff and hears the echo after 5 s. What is the distance of the cliff from the person if the speed of the sound is taken as 330 m s^–1?

Solution:

Speed of sound, v = 330 m s^–1

Time taken to hear the echo, t = 5 s

Distance travelled by the sound, d = v × t

= 330 × 5

= 1650 m (or) 1.65 km

In 5s sound travels twice the distance between the cliff and person. Hence the distance between the cliff and the person

= 1650 /2 = 825 m.

Example 4

A man fires a gun and hears its echo after 5 s. The man then moves 310 m towards the hill and fires his gun again. If he hears the echo after 3 s, calculate the speed of sound.

Solution:

Distance (d) = velocity (v) × time (t)

Distance travelled by sound when  gun fires first time, 2d = v × 5 (1)

Distance travelled by sound when gun  fires second time, 2d – 620 = v × 3 (2)

Rewriting equation (2) as,

2d = (v × 3) + 620      (3)

Equating (1) and (3), 5v = 3v + 620

2v = 620

Velocity of sound, v = 310 m s^–1