Question 87 of 142

The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is $60cm$, the length of the closed pipe will be :

A$60cm$B$45cm$C$30cm$D$15cm$

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Answer: (D) $15cm$

$\dfrac{\lambda }{4}={L}_{1}2\left(\dfrac{\lambda }{2}\right)={L}_{2}$
$\text{As},v=f\lambda \Rightarrow {f}_{2}=\dfrac{2v}{2{L}_{2}}$
$\Rightarrow v={f}_{1}\left(4{L}_{1}\right)\Rightarrow {f}_{2}=\dfrac{v}{{L}_{2}}$
$\Rightarrow {f}_{1}=\dfrac{v}{4{L}_{1}}$
As per question, ${f}_{1}={f}_{2}$
$\dfrac{v}{4{L}_{1}}=\dfrac{v}{{L}_{2}}$
$\Rightarrow {L}_{2}=4{L}_{1}$
$\Rightarrow 60=4\times {L}_{1}$
$\Rightarrow {L}_{1}=15cm$