Wave Mechanics Question 854
Question: A man, standing between two cliffs, claps his hands and starts hearing a series of echoes at intervals of one second. If the speed of sound in air is 340 ms-1, the distance between the cliffs is
[KCET 2004]
Options:
A) 340 m
B) 1620 m
C) 680 m
D) 1700 m
Show Answer
Answer:
Correct Answer: A
Solution:
Total time taken for both the echoes $ t=t _{1}+t _{2}=2 $ sec but $ t=\frac{2d _{1}}{v}+\frac{2d _{2}}{v} $
$ \Rightarrow t=\frac{2}{v}( d _{1}+d _{2} ) $
$ \Rightarrow (d _{1}+d _{2})=\frac{v\times t}{2}=\frac{340\times 2}{2} $ =340m.
BETA
