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Question

(C) For an organ pipe of length $L$ closed at one end, the fundamental frequency is given by $
u = \frac{v}{4L}$, where $v$ is the speed of sound.Whe

Vedclass · Accepted Answer

$2
u, 4
u$

Vedclass · Answer

(C) For an organ pipe of length $L$ closed at one end, the fundamental frequency is given by $
u = \frac{v}{4L}$, where $v$ is the speed of sound.When the pipe is cut into two equal parts, each part has a length of $L' = \frac{L}{2}$.One part remains closed at one end, so its fundamental frequency is $
u_1 = \frac{v}{4L'} = \frac{v}{4(L/2)} = \frac{2v}{4L} = 2
u$.The other part is now open at both ends, so its fundamental frequency is $
u_2 = \frac{v}{2L'} = \frac{v}{2(L/2)} = \frac{v}{L} = 4 	imes (\frac{v}{4L}) = 4
u$.Thus, the frequencies are $2
u$ and $4
u$.

An organ pipe closed at one end produces a fundamental note of frequency '$\nu$'. The pipe is cut into two pipes of equal length. The fundamental frequencies produced in the two pipes are

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