In an open end organ pipe of length ' L ', if the velocity of sound is ' V ', then the fundamental frequency will be (Neglect end correction) (a) V/2L and all harmonics are present. (b) V/4L and all harmonics are present. (c) V/2L and even harmonics are present. (d) V/4L and even harmonics are present.

Answer: (a). Open pipe: antinodes at both ends, L=(λ)/(2), so n=V/2L. Allowed modes L=(pλ)/(2), p=1,2,3, so all harmonics are present. So the answer is (a).

In an open end organ pipe of length ' L ', if the velocity of sound… | MHT CET 2025 (25 Apr, Shift 2)

Q27MHT CET 2025 · 25 APR · SHIFT 2Easy

Organ Pipes & Resonance Tube

In an open end organ pipe of length ' $L$ ', if the velocity of sound is ' V ', then the fundamental frequency will be (Neglect end correction)

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Answer: (a)

Open pipe: antinodes at both ends, $L=\frac{\lambda}{2}$ , so $n=\frac{V}{2L}$ .
Allowed modes $L=\frac{p\lambda}{2}$ , $p=1,2,3\ldots$ , so all harmonics are present.
So the answer is (a).

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