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MHT CET 2026 (14 May, Shift 1) · Organ Pipes & Resonance Tube

Organ Pipes & Resonance Tube question from MHT CET 2026 (14 May, Shift 1)

Q24MHT CET 2026 · 14 MAY · SHIFT 1Medium

Organ Pipes & Resonance Tube

$n_1$ is the fundamental frequency of the pipe closed at one end and $n_2$ is the fundamental frequency of the pipe open at both ends. If both are joined end to end, the fundamental frequency of closed pipe so formed is

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Answer: (a)

Closed pipe: $L_1 = \frac{v}{4n_1}$ ; open pipe: $L_2 = \frac{v}{2n_2}$ .
Joined: closed pipe of length $L_1 + L_2$ , fundamental $n = \frac{v}{4(L_1 + L_2)}$ .
$n = \frac{v}{\frac{v}{n_1} + \frac{2v}{n_2}} = \frac{n_1 n_2}{n_2 + 2n_1}$
So the answer is (a) $\frac{n_1 n_2}{n_2 + 2n_1}$ .

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