A man is standing between a stationary source and cliff. When he starts moving along line joining him and source, he hears 10 beats per second. The velocity of man is [frequency of source = 600 Hz , v

Question

A man is standing between a stationary source and cliff. When he starts moving along line joining him and source, he hears 10 beats per second. The velocity of man is [frequency of source = 600 Hz , velocity of sound = 330 m s - 1 ]

TestHub · Accepted Answer

As man is moving towards cliff, so apparent frequency heard directly towards source will be $f_{2}$

$f_{2} = (\frac{330 - v}{330 - 0}) 600$

similarly, apparent frequency heard by observer after reflection from the cliff will be $f_{3} = (\frac{330 + v}{330 - 0}) 600$

Now these two frequency $f_{2}$ and $f_{3}$ will produce 10 beats per second therefore $f_{3} - f_{2} = 10$ by solving we get $v = 2.75 m / s$

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