Question

How are stationary waves formed in closed pipes? Explain the various modes of vibrations and obtain relations for their frequencies. A pipe $30cm$ long is open at both ends. Find the fundamental frequency. Velocity of sound in air is $330{ms}^{-1}$.

Answer 1

The pipe open at one end and close at other end is known as closed pipe. If the waves with some frequency are sent through the closed pipe, the waves get reflected from closed end. When the incident and reflected waves with same frequency and in opposite direction superimposed the stationary waves formed in the closed pipe.

Let, $l$ be the length of pipe $v$ be the velocity of sound.

The first harmonic will form only when there is a node at closed end and anti-node at open end of pipe. Here, the length of pipe is equal to one fourth of the wavelength $\lambda$.

$l=\frac{{\lambda }_1}{4}$

$\lambda =4l$ ...........(1)

The fundamental frequency ${\nu }_1$ will be given by:

${\nu }_1=\frac{v}{{\lambda }_1}$

${\nu }_1=\frac{v}{4l}$ .............from(1)

The third harmonic or first overtone will form only when there is a node at closed end and anti-node at open end of pipe. Here, the length of pipe is equal to three fourth of the wavelength. Hence,

$l=\frac{3{\lambda }_1}{4}$

$\lambda =\frac{4}{3}l$ ...........(2)

The third harmionic frequency ${\nu }_3$ will be given by:

${\nu }_3=\frac{v}{{\lambda }_3}$

${\nu }_3=\frac{v}{\frac{4}{3}l}$ .............from(2)

${\nu }_3=\frac{3v}{4l}$

Similarly, the next overtone will be fifth harmonic with frequency,

${\nu }_5=\frac{5v}{4l}$ .........(3)

Solution for problem :

The fundamental frequency in the pipe open at both ends is given by,

$\nu =\frac{v}{2l}$ ..............(4)

Now, using given data in equation (4) we get,

$\nu =\frac{330}{2\times (30\times {10}^{-2)}}$

$\nu =5.5\times {10}^2$

$\nu =550Hz$.