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Resonance

A closed orga...

Question

A closed organ pipe of length 99.4 cm is vibrating in its first overtone and is always in resonance with a tuning fork having variable frequency f = (300 - 2t) Hz, where t is time in seconds. If the rate by which radius of organ pipe changes when its radius is 1 cm is $\frac{1}{N} m / s$ , then N is . (speed of sound in organ pipe = 320 m/s)

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Solution

$f = \frac{3 v}{4 (L + 0.6 r)} \frac{d f}{d t} = \frac{3 v (- 0.6 \frac{d r}{d t})}{4 {(L + 0.6 r)}^{2}} \frac{d r}{d t} = \frac{1}{72} m / s$

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