Question

There are two walls a and b which are 180 m apart from each other. A boy standing at 60 m away from wall a claps his hands once. If the speed of sound in air is 330 m s-1, what is the time interval between the first and the second echo that he hears?

Answer 1

Explanation:

Time for 1st echo

=0.54 s

Time for 2nd echo

=0.18 s

Time interval

=0.36 s

Answer 2

Answer:

The time interval between the first and the second echo is 0.36s.

Explanation:

Given two walls a and b, the distance between the walls a and b is 180m.

Given the boy is standing at 60m away from wall a,

then the boy will be at $180-60=120m$ away from wall b.

Speed of sound in air, $v=330 m /s$

Echo is the repeated sound caused by the reflection of sound waves from a surface.

Let the first echo is due to the wall a, hence the time taken between the clap and the first echo from wall a be $t_a$.

Let the second echo is due to the wall b, hence the time taken between the clap and the second echo from wall b be $t_b$.

Velocity is defined as the distance travelled per unit time. i.e.,

$v=\dfrac{d}{t}$

Then the time given can be written as

$t=\dfrac{d}{v}$

Substituting the values,

For first echo, distance travelled is from boy to wall a and wall a to boy,

$t_a=\dfrac{60+60}{330}=\dfrac{120}{330}=0.36s$

For second echo, distance travelled is from boy to wall b and wall b to boy,

$t_b=\dfrac{120+120}{330}=\dfrac{240}{330}=0.72s$

The time interval between the two echoes is

$t=t_b-t_a$

 $=0.72-0.36=0.36s$

Therefore, the time interval between the first and the second echo is 0.36s.

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