Two closed organ pipes having lengths 20,cm a | vedclass

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(A) For a closed organ pipe,the fundamental frequency is given by $n = \frac{v}{4\ell}$,which implies $n \propto \frac{1}{\ell}$.Since the pipe with t

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$205\,Hz, 200\,Hz$

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(A) For a closed organ pipe,the fundamental frequency is given by $n = \frac{v}{4\ell}$,which implies $n \propto \frac{1}{\ell}$.Since the pipe with the shorter length has a higher frequency,let $n_1$ be the frequency of the pipe with length $\ell_1 = 20\,cm$ and $n_2$ be the frequency of the pipe with length $\ell_2 = 20.5\,cm$.Given $n_1 - n_2 = 5\,Hz$,we have $n_1 = n_2 + 5$.Using the relation $\frac{n_1}{n_2} = \frac{\ell_2}{\ell_1}$,we substitute the values:$\frac{n_2 + 5}{n_2} = \frac{20.5}{20}$$20(n_2 + 5) = 20.5 n_2$$20 n_2 + 100 = 20.5 n_2$$0.5 n_2 = 100$$n_2 = \frac{100}{0.5} = 200\,Hz$.Therefore,$n_1 = 200 + 5 = 205\,Hz$.The frequencies are $205\,Hz$ and $200\,Hz$.

Two closed organ pipes having lengths $20\,cm$ and $20.5\,cm$ produce $5\,beats/sec$. Determine the frequency of both organ pipes.

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