First slide

Reflection and refraction of waves

Question

A man standing in front of a mountain beats a drum at regular intervals. The rate of drumming is generally increased and he finds that the echo is not heard distinctly when the rate becomes 40 per minute. He then moves nearer to the mountain by 90 m and finds that echo is again not heard when the drumming rate becomes 60 per minute. The distance between the mountain and the initial position of the man is

Difficult

A

205 m
B

300 m
C

180 m
D

270 m

Solution

For not hearing the echo the time interval between the beats of drum must be equal to time of echo.
$\Rightarrow t_{1} = \frac{2 d}{v} = \frac{60}{40} = \frac{3}{2}$ ......(i)
and $t_{2} = \frac{2 (d - 90)}{v} = \frac{60}{60} = 1$
$\Rightarrow 2 d - 180 = v$ ......(ii)
Form (i), we get $2 d = \frac{3}{2} v$ . Substituting in (ii), we get
$\Rightarrow \frac{3}{2} v - 180 = v \Rightarrow 180 = \frac{v}{2} \Rightarrow v = 360 {ms}^{- 1}$
$\Rightarrow \frac{2 (d)}{360} = \frac{3}{2} \Rightarrow d = 270 m$ .

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