Question

The air column in a pipe closed at one end is made to vibrate in its second overtone by tuning fork of frequency $440Hz$. The speed of sound in air is $330m/s$. End corrections may be neglected. Let $p_0$ denote the mean pressure at any point in the pipe and $\Delta ;p_0$ the maximum amplitude of pressure variation.
(a) Find the length L of the air column.
(b) What is the amplitude of pressure variation at the middle of the column?
(c) What are the maximum and minimum pressures at the open end of the pipe?
(d) What are the maximum and minimum pressures at the closed end of the pipe?

Answer 1

(a) Frequency of second overtone of the closed pipe
$=5\left(\frac{v}{4L}\right)=440$
$\therefore L=\frac{5v}{4\times 440}m$
Substituting $v=$ speed of sound in air $=330m/s$
$L=\frac{5\times 330}{4\times 440}=\frac{15}{16}m$
$\lambda =\frac{4L}{5}=\frac{4\left(\frac{15}{16}\right)}{5}=\frac{3}{4}m$
(b) Open end is displacement antinode.
Therefore, it would be a pressure node.
or at $x=0;\Delta p=0$
Pressure amplitude at x = x, can be written as,
$\Delta p=\pm ;\Delta ;p_0\sin kx$
where, $k=\frac{2\pi }{\lambda }=\frac{2\pi }{3/4}=\frac{8\pi }{3}{m}^{-1}$
Therefore, pressure amplitude at
$x=\frac{L}{2}=\frac{15/16}{2}m$
or $(15/32)m$will be
$\Delta ;p=\pm ;\Delta p_0\sin \left(\frac{8\pi }{3}\right)\left(\frac{15}{32}\right)$
$=\pm ;\Delta ;p_0\sin \left(\frac{5\pi }{4}\right)$
$\Delta ;p=\pm ;\frac{\Delta ;p_0}{\sqrt{2}}$
(c) Open end is a pressure node i.e. $\Delta p=0$
Hence, $p_{\text{max}}=p_{\text{min}}=$ Mean pressure $(p_0)$
(d) Closed end is a displacement node or pressure antinode.
Therefore, $p_{\text{max}}=p_0+\Delta p_0$
and $p_{\text{min}}=p_0-\Delta p_0$