A man P faces a vertical wall, 660 m away from
him. A second man Q stands 200 m behind P. When
P fires a gun, Q hears the shot and an echo at a time
interval of 'Z' seconds between the two sounds. Find
Z (the speed of sound is 330 m/s).
Answers
Answer:
ANSWER
From the given data we have,
d
x
=66m
t
1
=0.4s
t
2
=1.2s
v=?
d
y
=?
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=
t
1
2d
x
=
0.4
2×66
=330m/s
2d
y
=v×t
2
∴d
y
=
2
v×t
2
=
2
330×1.2
=198metres
Case 2:- when the 1st echo is due to cliff y and 2nd due to x
v=
t
2
2d
x
=
1.2
2×66
=110m/s
2d
y
=v×t
1
∴d
y
=
2
v×t
1
=
2
110×0.4
=22metres
We know that the speed of sound in air is approx 330m/s
Hence only Case 1 is correct
(i) speed of sound = 330m/s
(ii) distance of cliff y from man = 198m
Explanation: