Question

A closed organ pipe, of length $1.5\text{m}$ and filled with a gas, is resonated in its fundamental mode with the help of a tuning fork. Another open pipe of same length but filled with air resonates with the same fork in its fundamental mode. The room temperature is ${30}^{\circ }\text{C}$. If the speed of sound in air at ${30}^{\circ }\text{C}$ is $340\text{m/s}$, then the speed of sound in the first gas at ${30}^{\circ }\text{C}$ will be

• A. $340\text{m/s}$
• B. $680\text{m/s}$
• C. $510\text{m/s}$
• D. $322.5\text{m/s}$

Answer 1

The correct option is B $680\text{m/s}$

Let the velocity of sound in the gas be $v_g$.
Given that
Length of pipe $\left(L\right)=1.5\text{m}$
Velocity of sound in air $\left(v_a\right)=340\text{m/s}$
Modes of vibration of an air column in a closed organ pipe are given by
$f_{nc}=(2n+1)\frac{v}{4L}$ where $n=0,1,\mathrm{2...}$
Fundamental frequency of closed organ pipe filled with gas,
$\left(f_{0c}\right)=\frac{v_g}{4L}$


The modes of vibration of an open organ pipe are given by
$f_{no}=\frac{nv}{2L}$ where $n=1,2,\mathrm{3...}$
Fundamental frequency of open organ pipe filled with air, $\left(f_{1o}\right)=\frac{v_a}{2L}$
Since both resonates at the same frequency,
$f_{0c}=f_{1o}$
$\Rightarrow \frac{v_g}{4L}=\frac{v_a}{2L}$
$\Rightarrow v_g=2\times v_a=2\times 340=680\text{m/s}$