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Question
A cylindrical tube, open at both ends, has a fundamental frequency v. The tube is dipped vertically in water so that half of its length is inside the water. The new fundamental frequency is
Options
v/4
v/2
v
2v
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Solution
υ
If v is the velocity of the wave and L is the length of the pipe,
then the fundamental frequency for an open organ pipe is \[\nu = \frac{v}{2L}\]
For a closed organ pipe of length L' = L/2, the fundamental frequency is \[\nu = \frac{v}{4L'} = \frac{v \times 2}{4 \times L} = \frac{v}{2L} = v\]
(When the pipe is dipped in water, it behaves like a closed organ pipe that is half the length)
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