A man beats drum at a certain distance from the mountain . He slowly increase the rate of beating and finds that the echo is not heard distinctly . when the drum beating is at the rate of 40 per minute . He moves by 80m towards the mountain and finds echo is again not heard distinctlywhen the rate of beating of the drum is 1 per second . What is the original distance of the man from the mountain?

Dear Student,

Please find below the solution to the asked query:

Case I. When the rate of drum beating is 40 per minute

Therefore, the interval between two drum beats = 60/40 = 1.5 s
Let the distance between the mountain and the man be x m.
$\text{Speed of sound}, v=\frac{2x}{t}=\frac{2x}{1.5}=\frac{4x}{3} .....\left(1\right)$
Case II. When the man moves 80 m towards the mountain

$$\begin{gathered} \text{Speed of sound}, v=\frac{2(x-80)}{t}=\frac{2(x-80)}{1}=2x-160 .....\left(2\right) \\ \text{From eq. (1) and (2)}, \\ 2x-160=\frac{4x}{3} \\ \Rightarrow 6x-480=4x \\ \Rightarrow 2x=480 \\ \Rightarrow x=240 \text{m} \end{gathered}$$

Hope this information will clear your doubts about the numerical on echo.

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