Question

A one meter long (both ends open) organ pipe is kept in a gas that has double the density of air at STP. Assuming the speed of sound in air at STP is $300m/s,$ the frequency difference between the fundamental and second harmonic of this pipe is __$Hz$.

Answer 1

Step 1: Given data:

Velocity of sound in air $=300m/s$

${\rho }_{air}=2{\rho }_{pipe}$

Step 2: Formula used:

We know, $v=\sqrt{\frac{B}{\rho }}$

where, $v$ is velocity of sound in medium

$B$ is Bulk's modulus

$\rho$ is density of the medium

Assuming Bulk's modulus similar in both the cases,

$\begin{array}{rcl}\frac{v_{pipe}}{v_{air}}& =& \sqrt{\frac{{\rho }_{air}}{2{\rho }_{pipe}}}\\ & \Rightarrow & v_{pipe}=\frac{v_{air}}{\sqrt{2}}\\ & \Rightarrow & v_{pipe}=\frac{300}{\sqrt{2}}\\ & & \end{array}$

length of pipe be $L=1m$

Fundamental mode, $\begin{array}{rcl}L& =& 1=\frac{\lambda }{2}\\ & \Rightarrow & \lambda =2m\end{array}$

Second harmonic,$\begin{array}{rcl}L& =& 1=\lambda \\ & \Rightarrow & \lambda =1m\end{array}$

Step 3: Finding the difference in frequencies:

Let, $f\text{'}$be the frequency of the second harmonic

$F$ be the frequency of the fundamental mode

$\begin{array}{rcl}∆f& =& f\text{'}-F\\ & =& v_{pipe}\left[\frac{1}{\lambda \text{'}}-\frac{1}{\lambda }\right]\\ & =& \frac{300}{\sqrt{2}}\left[1-\frac{1}{2}\right]\\ & =& \frac{300}{2\sqrt{2}}\\ & =& 106.06Hz\end{array}$

Thus, the difference between the fundamental and second harmonic of this pipe is $106.06Hz$