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Q.
A closed pipe of length $10 \,cm$ has its fundamental frequency half that of the second overtone of an open pipe. The length of the open pipe
Waves
Solution:
$l=0.1 \,m$
Fundamental frequency of pipe $\left(f_{1}\right)=\frac{v}{4 l}$ or $f_{1}=\frac{v}{0.4}$
Frequency of $2^{\text {nd }}$ overtone of open pipe $2$ :
$f_{2}=\frac{3 v}{2 l} $
$2 f_{1}=f_{2} $
$2 \times \frac{v}{0.4}=\frac{3 v}{2 l}$
$l=0.3 \,m$