First slide

Standing waves

Question

An open pipe is suddenly closed with the result that, the second overtone of the closed pipe is found to be higher in frequency by 100 Hz, then the first overtone of the original pipe. The fundamental frequency of open pipe will be

Moderate

A

100Hz
B

300Hz
C

150Hz
D

200Hz

Solution

For second overtone of closed pipe
$n_{3} = \frac{5 V}{4 l}$
For first overtone of open pipe
$n_{2} = \frac{2 V}{2 l}$
Given that $\frac{5 V}{4 l} â \frac{2 V}{2 l} = 100 or \frac{V}{2 l} = 200$
Now fundamental frequency of open pipe
$n_{1} = \frac{V}{2 l} = 200$

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