Question

A closed organ pipe of length $L$ and an open organ pipe contains gases of densities ${\rho }_1$ and ${\rho }_2$ respectively. The compressibility of gases are equal in both the pipes. Both the pipes are vibrating in their first overtone with the same frequency. The length of the open pipe is $\frac{x}{3}L\sqrt{\frac{{\rho }_1}{{\rho }_2}}$where $x$ is ______(Round off to the Nearest Integer)

Answer 1

Step 1: Given Data:

Length of the closed pipe $=L$

Let, the length of the open pipe be $L_2=\frac{x}{3}L\sqrt{\frac{{\rho }_1}{{\rho }_2}}$

Step 2: Formula used:

Speed of sound, $V=\sqrt{\frac{B}{\rho }}=\sqrt{\frac{1}{k\rho }}$………(Where compressibility $k=\frac{1}{B}$)

Frequency of closed pipe, $f=\frac{3}{4L}\sqrt{\frac{1}{k{\rho }_1}}$………….$\left(1\right)$

Frequency of open pipe, $f=\frac{2V}{2L_2}=\frac{1}{L_2}\sqrt{\frac{1}{k{\rho }_2}}$………..$\left(2\right)$

Step 3: Finding the value of $x$:

As the frequencies and compressibility of gases are equal,

Thus $\frac{3}{4L}\sqrt{\frac{1}{k_1{\rho }_1}}=\sqrt{\frac{1}{k_2{\rho }_2}}\frac{1}{L_2}$

$\Rightarrow \frac{3}{4L}\sqrt{\frac{1}{{\rho }_1}}=\sqrt{\frac{1}{{\rho }_2}}\frac{1}{L_2}$

$\Rightarrow \frac{3}{4L}\sqrt{\frac{{\rho }_2}{{\rho }_1}}=\frac{1}{L_2}$

$\Rightarrow L_2=\frac{4}{3}L\sqrt{\frac{{\rho }_1}{{\rho }_2}}$

Now comparing with given equation $\frac{x}{3}L\sqrt{\frac{{\rho }_1}{{\rho }_2}}$ ,we get $x=4$.

Hence, the value of $x$ is $4$.