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JEE Main 2024 (31 Jan, Shift 1) · Organ Pipes & Resonance Tube

Organ Pipes & Resonance Tube question from JEE Main 2024 (31 Jan, Shift 1)

Q50JEE Main 2024 · 31 JAN · SHIFT 1Easy

Organ Pipes & Resonance Tube

The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is $60\text{ cm}$ , the length of the closed pipe will be:

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Answer: (a)

First overtone of open pipe: $\frac{2v}{2L_o}=\frac{v}{L_o}$ . Closed fundamental: $\frac{v}{4L_c}$ .
$\frac{v}{4L_c}=\frac{v}{L_o} \Rightarrow L_c=\frac{L_o}{4}=\frac{60}{4}$
$L_c=15\text{ cm}$ , so the answer is (a).

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