Question

A man stands in between two vertical cliffs × and Y such that he is at a distance of 660 m from X. When he explodes cracker, he hears the first echo after 4 sec and second echo 6 second later. Calculate the speed of sound in air and the distance of cliff Y from the man.

Answer 1

Answer:

Question



A man stands in between two cliffs x and y, such that he is at a distance of 66m from x. When he blows a whistle he hears first echo after 0.4s and second echo after 1.2s. Calculate: (i) speed of sound (ii) distance of cliff y from man.

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Solution



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From the given data we have,

dx=66m

t1=0.4s

t2=1.2s

v=?

dy=?

When he blows the whistle, the sound generated travels everywhere, including in the direction towards the cliff. 

The sound wave travelling towards the cliff, hits it, gets reflected from it, moves backwards and reaches his ear.

This is how he hears the echo.

This shows that the sound wave travels twice the distance between him and the cliff. Hence, while calculating, we'll have to double the distances.

Case 1:- when the 1st echo is due to cliff x and 2nd due to y

        v=t12dx=0.42×66=330m/s

    2dy=v×t2

  ∴dy=2v×t2=2330×1.2=198metres

Case 2:- when the 1st echo is due to cliff y and 2nd due to x

        v=t22dx=1.22×66

Answer 2

Given : Two vertical cliffs × and Y such that he is at a distance of 660 m                        from x                  

       The first echo after 4 sec and second echo 6 second later

To Find : Speed of sound in air and the distance of cliff Y from the man.

Solution :

From the given data we have,

$\[{d_x}\]$ =660m

$\[{t_1}\]$=4 sec

$\[{t_2}\]$ = 6 sec

When he blows the whistle, the sound generated travels everywhere, including in the direction towards the cliff.  

The sound wave travelling towards the cliff, hits it, gets reflected from it, moves backwards and reaches his ear.

This is how he hears the echo.

This shows that the sound wave travels twice the distance between him and the cliff. Hence, while calculating, we'll have to double the distances.

Case 1:- when the 1st echo is due to cliff x and 2nd due to y

     $\[v = \frac{{2{d_x}}}{{{t_1}}} = \frac{{2 \times 660}}{4} = 330m/s\]$

from given equation,

$\[2{d_y} = v \times {t_2}\]$

$\[{d_y} = \frac{{330 \times 6}}{2} = 990metre\]$.

Case 2:- when the 1st echo is due to cliff y and 2nd due to x

$\[v = \frac{{2{d_x}}}{{{t_2}}} = \frac{{2 \times 660}}{6} = 220m/s\]$      

$\[2{d_y} = v \times {t_1}\]$

$\[{d_y} = 440metre\]$

We know that the speed of sound in air is approx. 330m/s

Thus, Case 1 is correct.

Hence, Speed of sound in air = 330 m/s and distance of cliff Y from the man is 990 meter.