Two closed organ pipe A and B have the same length. A is wider than B. They resonate in the fundamental mode at frequencies `V_(A) and V_(B)` respectively , thenA. `n_(A) = n_(B)`B. `n_(A) gt n_(B)`C. `n_(A) lt n_(B)`D. Either (b) or ( c) depending on the ratio of their diameters

Two closed organ pipe A and B have the same length. A is wider than B.

1.	Two closed organ pipe A and B have the same length. A is wider than B. They resonate in the fundamental mode at frequencies `V_(A) and V_(B)` respectively , thenA. `n_(A) = n_(B)`B. `n_(A) gt n_(B)`C. `n_(A) lt n_(B)`D. Either (b) or ( c) depending on the ratio of their diameters
Answer» Correct Answer - c In closed organ pipe. First resonance occurs at `lambda//4.` So, in fundamental mode of vibration of organ pipe `lambda/4=(l+0.3d)` where, `0.3` d is necessary end correction. Frequency of vibration, `n=v/lambda = v/(4(l+0.3d))` As / is same, wide pipe A will resonate at a lower Frequency, I.e. `n_(A) lt n_(B).`

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Two closed organ pipe A and B have the same length. A is wider than B. They resonate in the fundamental mode at frequencies `V_(A) and V_(B)` respectively , thenA. `n_(A) = n_(B)`B. `n_(A) gt n_(B)`C. `n_(A) lt n_(B)`D. Either (b) or ( c) depending on the ratio of their diameters

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