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(A) Let the length of both pipes be $L$. The fundamental frequency of an open organ pipe is $
u_{open} = \frac{v}{2L}$.The fundamental frequency of a

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$2$

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(A) Let the length of both pipes be $L$. The fundamental frequency of an open organ pipe is $
u_{open} = \frac{v}{2L}$.The fundamental frequency of a closed organ pipe is $
u_{closed} = \frac{v}{4L}$.The beat frequency is given by $b = |
u_{open} - 
u_{closed}| = |\frac{v}{2L} - \frac{v}{4L}| = \frac{v}{4L} = 4$.If the length of each pipe is doubled,the new length becomes $L' = 2L$.The new frequencies are $
u'_{open} = \frac{v}{2(2L)} = \frac{v}{4L}$ and $
u'_{closed} = \frac{v}{4(2L)} = \frac{v}{8L}$.The new beat frequency is $b' = |
u'_{open} - 
u'_{closed}| = |\frac{v}{4L} - \frac{v}{8L}| = \frac{v}{8L}$.Since $\frac{v}{4L} = 4$,then $\frac{v}{8L} = \frac{4}{2} = 2$.Therefore,the number of beats produced will be $2$.

$A$ closed organ pipe and an open pipe of the same length produce $4$ beats when they are set into vibrations simultaneously. If the length of each of them were twice their initial lengths,the number of beats produced will be

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