JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Waves & Sound

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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}

)

Answer: 34

Solution

For a closed organ pipe, the fundamental frequency (first resonance) is

f = \frac{v}{4 L}

, where

L

is the length. Solving for

L

L = \frac{v}{4 f} = \frac{340}{4 \times 250} = \frac{340}{1000} = 0.34

m = 34 cm.

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Integer answer

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