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JEE Main 2014 · 6 Apr · Shift 1, Physics question 15

A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.

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Answer and worked solution

Official answer: C

A pipe closed at one end has only odd harmonics:
f=(2n−1)v4L=(2n−1)3404×0.85=(2n−1)×100 Hz.f=(2n-1)\frac{v}{4L}=(2n-1)\frac{340}{4\times0.85}=(2n-1)\times100\text{ Hz}.
Below 1250 Hz: 100, 300, 500, 700, 900, 1100 Hz. That is 6 modes.

Topics:Organ pipes and resonance
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