JEE Main 2016 · Physics · Waves and Sound · medium

A pipe closed at one end is found to resonate at two consecutive frequencies of 300 Hz and 500 Hz. If the speed of sound in air is 340 m s^-1, the…

Single correct option

A pipe closed at one end is found to resonate at two consecutive frequencies of

300\ \mathrm{Hz}

and

500\ \mathrm{Hz}

. If the speed of sound in air is

340\ \mathrm{m\,s^{-1}}

, the length of the pipe (neglecting end correction) is

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Answer: (B) $0.85\ \mathrm{m}$

A closed pipe supports only odd harmonics:

f_n=n\dfrac{v}{4L}

with

n=1,3,5,\dots

Consecutive resonances therefore differ by

\dfrac{2v}{4L}=\dfrac{v}{2L}

, which equals

500-300=200\ \mathrm{Hz}

. Hence

L=\dfrac{340}{2\times200}=0.85\ \mathrm{m}

. Check: the fundamental is

\dfrac{340}{4(0.85)}=100\ \mathrm{Hz}

, and

300=3\times100

and

500=5\times100

are indeed the third and fifth harmonics. Treating the spacing as the fundamental, as for an open pipe with

f_1=100

, would give

1.7\ \mathrm{m}

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