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Question
A closed organ pipe can vibrate at a minimum frequency of 500 Hz. Find the length of the tube. Speed of sound in air = 340 m s−1.
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Solution
Given:
Speed of sound in air v = 340 ms−1
Frequency of closed organ pipe f = 500 Hz
Length of tube L = ?
Fundamental frequency of closed organ pipe is given by :
\[f = \frac{v}{4L}\]
\[\therefore L = \frac{v}{4f}\]
\[ \therefore L = \frac{340}{4 \times 500} = 0 . 17 \text{ m } = 17 \text { cm }\]
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