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(A) The frequency of the first overtone of a closed organ pipe of length $\ell$ is given by $f = \frac{3V}{4\ell}$.The frequency of the first overtone

Vedclass · Accepted Answer

$3: 4$

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(A) The frequency of the first overtone of a closed organ pipe of length $\ell$ is given by $f = \frac{3V}{4\ell}$.The frequency of the first overtone of an open organ pipe of length $\ell^{\prime}$ is given by $f^{\prime} = \frac{2V}{2\ell^{\prime}} = \frac{V}{\ell^{\prime}}$.Given that the frequencies are identical,we set $f = f^{\prime}$:$\frac{3V}{4\ell} = \frac{V}{\ell^{\prime}}$.Rearranging the terms to find the ratio of lengths $\frac{\ell}{\ell^{\prime}}$:$\frac{\ell}{\ell^{\prime}} = \frac{3}{4}$.Thus,the ratio of their lengths is $3: 4$.

$A$ closed organ pipe and an open organ pipe have their first overtones identical in frequency. Their lengths are in the ratio

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