Physics - Sound Wave Question with Solution | TestHub

Let the length of the closed pipe be $L_1$ and the length of the open pipe be $L_2$. Let $v$ be the speed of sound in air.

The fundamental frequency of the closed pipe is given by:
$n_1=\frac{v}{4L_1}\Rightarrow L_1=\frac{v}{4n_1}$

The fundamental frequency of the open pipe is given by:
$n_2=\frac{v}{2L_2}\Rightarrow L_2=\frac{v}{2n_2}$

When the two pipes are joined end to end, the new pipe formed is closed at one end and open at the other. The total length of this new closed pipe is $L=L_1+L_2$.

The fundamental frequency of the new closed pipe is:
$n=\frac{v}{4L}$

Substituting the values of $L_1$ and $L_2$:
$n=\frac{v}{4(L_1+L_2)}$

$n=\frac{v}{4\left(\frac{v}{4n_1}+\frac{v}{2n_2}\right)}$

$n=\frac{1}{4\left(\frac{1}{4n_1}+\frac{2}{4n_2}\right)}$

$n=\frac{1}{\frac{1}{n_1}+\frac{2}{n_2}}$

$n=\frac{n_1{n}_2}{n_2+2n_1}$

Answer: $\frac{n_1{n}_2}{n_2+2n_1}$