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Q.

A tuning fork is vibrating at 250 Hz. The length of the shortest closed organ pipe that will resonate with the tuning fork will be ____________cm. (Take speed of sound in air as 340 ms^-1)

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answer is 34.

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Detailed Solution

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The resonant frequency of a closed organ pipe of length L is $f = \frac{nv}{4 L}$
Here, n =odd positive integer v = speed of sound in air
For L to be minimum, $n = 1 ∴ 250 = \frac{v}{4 L} \Rightarrow 250 = \frac{340}{4 L} \Rightarrow L = \frac{34}{4 \times 25} = 0 .34 m \Rightarrow L = 34 cm$

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