Question

The air column in an open organ pipe is made to vibrate in its first overtone by a tuning fork of frequency $680\text{Hz}$. Speed of sound in air is $340\text{m/s}$ and ${10}^5{\text{N/m}}^2$ is the mean pressure at any point in the pipe. If $10{\text{N/m}}^2$ is the maximum amplitude of pressure variation, then

• A. Length of organ pipe is $1\text{m}$.
• B. Amplitude of pressure variation at $x=\frac{1}{3}$ is $8.6{\text{N/m}}^2$.
• C. Pressure at the end of the pipe is ${10}^5{\text{N/m}}^2$.
• D. Maximum pressure in the pipe is $({10}^5+10){\text{N/m}}^2$.

Answer 1

Answer: (B), (C), (D)

Maximum pressure in the pipe is $({10}^5+10){\text{N/m}}^2$.

Given that,
Frequency of tuning fork $\left(f\right)=680\text{Hz}$
Speed of the sound $\left(V\right)=340\text{m/s}$
Mean Pressure $\left(P_0\right)={10}^5{\text{N/m}}^2$
Max. amplitude of pressure $\left(△P_o\right)=10{\text{N/m}}^2$

Modes of vibration of air column in an open organ pipe are given by
$f_n=\frac{nv}{2L}$
For first overtone,
$f_2=\frac{v}{L}$
$\Rightarrow \lambda =L$

From the data given in the question,
$f=f_2$
$\Rightarrow 680=\frac{340}{L}$
$\Rightarrow L=\frac{340}{680}=0.5\text{m}$

Stationary wave can be represented as
$\Delta p=\Delta p_o\sin \left(kx\right)\cos \left(\omega t\right)$
$\Rightarrow \Delta p=10\sin \left(4\pi x\right)\cos \left(\omega t\right)[\because k=\frac{2\pi }{\lambda }]$

For amplitude of pressure variation at $x=\frac{1}{3}$,

$\Delta p\left(x\right)=|10\sin \left(4\pi x\right)|$
$\Rightarrow \Delta p\left(x\right)=\left|10\sin \left(\frac{4\pi }{3}\right)\right|$
$\Rightarrow \Delta p\left(x\right)=5\sqrt{3}=8.6{\text{N/m}}^2$

As pressure node is formed at the end of the pipe, the pressure at the end of pipe is equal to the mean pressure.
$\Rightarrow p_{end}={10}^5{\text{N/m}}^2$

Maximum pressure in the pipe is
$p_{max}=p_o+\Delta p_o=({10}^5+10){\text{N/m}}^2$
Options B, C, D are correct.