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?
D. u = 0 . 5 v and f s = 1 . 5   f 0
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 for resonance, appeared frequency should match with any of the natural frequency of the closed organ pipe.If fundamental frequency of the closed organ pipe is f 0 , then its natural frequencies will be f 0 , 3 f 0 , 5 f 0 . . . . 2 n - 1 f 0 So for resonance, f s v v - u = 2 n - 1 f 0 f s = 1 - u v 2 n - 1 f 0 Now from options A , B , C   ∵ u = 0 . 8 v So,   f s = 1 - 0 . 8 2 n - 1 f 0 f s = 2 n - 1 5 f 0 n = 1 ,    f s = f 0 5 n = 2 ,    f s = 3 f 0 5 n = 3 ,    f s = 5 f 0 5 = f 0 n = 4 ,    f s = 7 f 0 5 n = 5 ,     f s = 9 f 0 5 n = 6 ,    f s = 11 f 0 5 So option (A) is correct, (B) and (C) are incorrect Now from option D   ∵ u = 0 . 5 v So, f s = 1 - 0 . 5 2 n - 1 f 0 f s = 2 n - 1 2 f 0 n = 1 ,    f s = f 0 2 n = 2 ,    f s = 3 f 0 2 = 1 . 5 f 0 n = 3 ,     f s = 5 f 0 2 Hence option D is correct.
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