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A closed orgon pipe and an open orgon pipe are tuned to the same fundamental frequency. What is the ratio of their lengths?
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Verified Answer
The correct answer is:
$1: 2$
Avoiding end correction, the length of the closed organ pipe is
$l_1=\frac{\lambda_1}{4}$ or $\lambda_1=4 l_1$
The length of open organ pipe is
$l_2=\frac{\lambda_2}{2}$ or $\lambda_2=2 l_2$
Here, $n_1=n_2$
So, $\frac{v}{\lambda_1}=\frac{v}{\lambda_2}$ or $\frac{v}{4 l_1}=\frac{v}{2 l_2}$
Therefore, $l_1: l_2=1: 2$
$l_1=\frac{\lambda_1}{4}$ or $\lambda_1=4 l_1$
The length of open organ pipe is
$l_2=\frac{\lambda_2}{2}$ or $\lambda_2=2 l_2$
Here, $n_1=n_2$
So, $\frac{v}{\lambda_1}=\frac{v}{\lambda_2}$ or $\frac{v}{4 l_1}=\frac{v}{2 l_2}$
Therefore, $l_1: l_2=1: 2$
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