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A source, approaching with speed u towards the open end of a stationary pipe of length L , is emitting a sound of frequency f s . The farther end of the pipe is closed. The speed of sound in air is v and f 0 is the fundamental frequency of the pipe. For which of the following combination(s) of u and f s , will the sound reaching the pipe lead to a resonance?

Options

  1. Au = 0 . 8 v and f s = f 0
  2. Bu = 0 . 8 v and f s = 2 f 0
  3. Cu = 0 . 8 v and f s = 0 . 5   f 0
  4. Du = 0 . 5 v and f s = 1 . 5   f 0

Correct answer

A. u = 0 . 8 v and f s = f 0

Step-by-step solution

Consider a tuning fork as a source of sound which is moving towards the stationary pipe with velocity u . According to the Doppler effect , f ' = f s v - v o v - v s Where, v = is speed of sound v s = is velocity of source v o = is velocity of observer f ' = is appeared frequency f s = is actual frequency So the appeared frequency of the sound will be, f ' = f s v - 0 v - u Since pipe is closed at one end so it will be like a close organ pipe and we know that close organ pipe has only odd harmonics. And

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