An open organ pipe and closed organ pipe of same length produce 2 beats per second, when they are set into vibrations together, in fundamental mode. The length of open pipe is made half and that of cl

Question

An open organ pipe and closed organ pipe of same length produce 2 beats per second, when they are set into vibrations together, in fundamental mode. The length of open pipe is made half and that of closed pipe is doubled. The number of beats produced per second will be (neglect end correction)

TestHub · Accepted Answer

Let the initial length of both pipes be $L$ .

Fundamental frequency of the open pipe, $n_{o} = \frac{v}{2 L}$

Fundamental frequency of the closed pipe, $n_{c} = \frac{v}{4 L}$

Given that the number of beats produced is $2$ Hz:

$n_{o} - n_{c} = 2$

$\frac{v}{2 L} - \frac{v}{4 L} = 2$

$\frac{v}{4 L} = 2 \Rightarrow \frac{v}{L} = 8$

When the length of the open pipe is halved, its new length is $L_{o}^{'} = \frac{L}{2}$ .

New fundamental frequency of the open pipe, $n_{o}^{'} = \frac{v}{2 L _{o}^{'}} = \frac{v}{2 ( L /2 )} = \frac{v}{L} = 8$ Hz.

When the length of the closed pipe is doubled, its new length is $L_{c}^{'} = 2 L$ .

New fundamental frequency of the closed pipe, $n_{c}^{'} = \frac{v}{4 L _{c}^{'}} = \frac{v}{4 ( 2 L )} = \frac{v}{8 L} = \frac{1}{8} (\frac{v}{L}) = \frac{1}{8} \times 8 = 1$ Hz.

The new number of beats produced per second is:

$n_{o}^{'} - n_{c}^{'} = 8 - 1 = 7$

Answer: 7

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