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(A) Let $l_1$ be the length of the closed organ pipe and $l_2$ be the length of the open organ pipe.For a closed organ pipe,the fundamental frequency

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$1:2$

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(A) Let $l_1$ be the length of the closed organ pipe and $l_2$ be the length of the open organ pipe.For a closed organ pipe,the fundamental frequency is $n_1 = \frac{v}{4l_1}$.For an open organ pipe,the fundamental frequency is $n_2 = \frac{v}{2l_2}$.Given that the fundamental frequencies are the same,$n_1 = n_2$.Therefore,$\frac{v}{4l_1} = \frac{v}{2l_2}$.Simplifying this,we get $\frac{1}{4l_1} = \frac{1}{2l_2}$,which implies $\frac{l_1}{l_2} = \frac{2}{4} = \frac{1}{2}$.Thus,the ratio of the lengths of the closed pipe to the open pipe is $1:2$.

$A$ closed organ pipe and an open organ pipe are tuned to the same fundamental frequency. What is the ratio of their lengths?

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