There are two walls X and Y which are 120m apart from each other. A boy standing 40m away from wall X claps his hands once. If the speed of sound in air is 330m s-¹, then what is the time interval between the first and the second echo that he hears?
Answers
Answer:
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Let d
1
and d
2
be the distances of cliffs from the boy .
for first echo , which is heard after 2s ,
total distance travelled by sound =2d
1
,
given v=330m/s ,
we have , distance=speed×time ,
2d
1
=330×2 ,
or d
1
=660/2=330m ,
for second echo , which is heard after 2.5s ,
total distance travelled by sound =2d
2
,
given v=330m/s ,
we have , distance=speed×time ,
2d
2
=330×2.5 ,
or d
2
=825/2=412.5m ,
therefore , distance between cliffs , d=d
1
+d
2
=330+412.5=742.5m
Concept:
- Sound waves
- Echoes
Given:
- Distance between the two walls D = 120 m
- Distance between wall X and the boy x = 40 m
- Distance between wall Y and the boy y = 120-40 = 80 m
- Speed of sound in air v = 330 m/s
Find:
- The time interval between the first and the second echo
Solution:
The first echo will come from the sound being reflected off wall X.
The second echo will come from the sound being reflected off wall Y.
For the first echo to reach the boy, the distance it has to cover = 2x = 2* 40 = 80
The time for the first echo to reach the boy = 2x/v = 80/330 = 8/33 s
For the second echo to reach the boy, the distance it has to cover = 2y = 2* 80 = 160
The time for the first echo to reach the boy = 2x/v = 160/330 = 16/33 s
The time interval = 16/33 = 8/33 = 8/33 s
The time interval is 8/33 s.
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