For closed organ pipe
f=(2n+1)4lv
for minimum length, n=0
f0=4lv
⇒l=4f0v
=4×250340=34cm
A tuning fork is vibrating at 250Hz. 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 340ms−1 )
Held on 27 Aug 2021 · Verified 6 Jul 2026.
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Work through every JEE Main Waves & Oscillations PYQ, year by year.