Solution:
For a closed organ pipe, the frequency of the \(n^{\text{th}}\) overtone is \(f = (2n+1)\frac{v}{4l}\). For \(2^{\text{nd}}\) overtone (\(n=2\)), \(f = 5\frac{v}{4l} = \frac{5 \times 340}{4 \times 2} = 212.5 \text{ Hz}\).
A closed organ pipe of length \(l = 2 \text{ m}\) is vibrating in \(2^{\text{nd}}\) overtone. The frequency of vibration if speed of sound is 340 m/s is
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