Question

The first overtone frequency of a closed organ pipe $P_1$, is equal to the fundamental frequency of an open organ pipe $P_2$. If the length of the pipe $P_1$, is 30 cm, what will be the length of $P_2$ ?

Answer 1

According to the questions

$f_1$ first overtone of a closed organ pipe

$P_1=\frac{3v}{4I}=\left(\frac{3\times V}{4\times 30}\right)$

$f_2$ =fundamental frequency of a open orgain pipe

$P_2=\left(\frac{V}{2I_2}\right)$

Here given, $\left(\frac{3\times V}{4\times 30}\right)=\frac{V}{2I_2}$

$\Rightarrow I_2=20cm$

$\therefore$ Length of the pipe $P_2$ will be 20 cm.