Question

An air column vibrates in a pipe closed at one end in its third overtone due to a tuning fork of frequency $595\text{Hz}$. $P_o$ is the mean pressure at any point in the pipe and $\Delta P_o$ is the maximum amplitude of pressure variation. Speed of sound in air is $340\text{m/s}$ and end correction has a negligible effect. If the amplitude of pressure variation at the middle of the pipe is $\frac{\Delta P_o}{\sqrt{n}},$ then $n$ is:
[Given $\cos \left(\frac{595\pi }{340}\right)=\frac{1}{\sqrt{2}}$ ]

• A. $2$
• B. $1$
• C. $3$
• D. $4$

Answer 1

The correct option is A $2$

Given, speed of sound $\left(v\right)=340\text{m/s}$
Frequency, $f=595\text{Hz}$
So, wavelength $\left(\lambda \right)=\frac{v}{f}=\frac{340}{595}\text{m}$

Modes of vibration of an air column in a closed organ pipe are given by

$f_n=\frac{(2n-1)v}{4L}$

Now, for third overtone,
$f_4=\frac{7v}{4L}$

From the data given in the question,

$L=\frac{7v}{4f_4}=\frac{7\times 340}{4\times 595}=1\text{m}$

The general equation of pressure variation for a stationary sound wave in a closed organ pipe is given by
$\Delta p=-\Delta p_o\cos \left(kx\right)\sin \left(\omega t\right)$

So, pressure amplitude $=|-\Delta p_o\cos \left(\frac{2\pi }{\lambda }x\right)|$

Pressure amplitude at the middle of the pipe

$=|-\Delta p_o\cos (\frac{2\pi }{\left(340/595\right)}\times \frac{1}{2})|$

$=|-\Delta p_o\cos \left(\frac{595\pi }{340}\right)|=\frac{\Delta p_o}{\sqrt{2}}$

Comparing this with $\frac{\Delta p_0}{\sqrt{n}}$,
we get, $n=2.$