Question

Explain the formation of stationary waves in an air column enclosed in open pipe. Derive the equations for the frequencies of the harmonics produced. An open organ pipe 85 cm long is sounded. If the velocity of sound is \[340m{{s}^{-1}}\], what is the fundamental frequency of vibration of the air column?

Answer 1

Hint: Stationary or standing waves are the superposition of waves and its reflective wave. In an open pipe, both ends are open and the antinodes will be at the ends of the pipe. From the wave equations, we can find out the formula of frequencies of vibrations. By using the fundamental frequency formula, we can find out the frequency of vibration in the air column.

Complete step by step answer:
Standing waves are produced by the superposition of waves. When the waves meet together, they will form either constructive or destructive interference. For a standing wave or stationary wave, two identical but oppositely moving waves have to interfere. It is just like a wave and the reflection of the wave. Standing waves do not transfer any energy. Standing waves produce only at the natural frequencies of the material. If a system is disturbed, it will produce many vibrations of different frequencies. Only the standing wave will continue. All others will die out.

A resonant air column is a type of longitudinal wave system. An open pipe resonator contains a tube in which both ends are open. Each pattern of the standing wave is known as harmonic. The smallest frequency or the largest wavelength of a standing wave is the fundamental pattern and it is known as first harmonic. For an open pipe, the standing waves have antinodes at the two ends of the open pipe.

To find the mathematical formulas of standing waves, we have to consider a wave travelling along the positive x-direction and the reflected wave of the same amplitude and wavelength in the negative x-direction.

A sinusoidal wave can be written as,

\[{{y}_{1}}(x,t)=a\sin (kx-\omega t)\]

The reflected wave can be written as,

\[{{y}_{2}}(x,t)=a\sin (kx+\omega t)\]

According to the superposition, we can write the resultant wave as,

\[{{y}_{1}}(x,t)+{{y}_{2}}(x,t)=a\sin (kx-\omega t)+a\sin (kx+\omega t)\]

We can alter this equation by using the following formula.

\[\sin (A+B)+\sin (A-B)=2\sin A\cos B\]

Therefore, the resultant wave will be,

\[y(x,t)=2a\sin kx\cos \omega t\]……………………..(1)

According to the standing waves, when the amplitude is zero, then the points are known as nodes and when the amplitude is maximum, then the points are known as antinodes.

The stationary waves are characterised by normal modes or natural frequencies.

At the position of the nodes, amplitudes will be zero. Thus,

\[\sin kx=0\]
So,
\[kx=n\pi ;n=0,1,2,3\]
As we know, \[k=\dfrac{2\pi }{\lambda }\]

Therefore, the x will be,

\[x=\dfrac{n\lambda }{2};n=0,1,2,3,...\]
As we know, the distance between any two nodes is \[\dfrac{\lambda }{2}\]
At the position of antinodes, the amplitude will be the largest value.

\[|\sin kx|=1\]
Which implies,

\[kx=(n+\dfrac{1}{2})\pi ;n=0,1,2,3...\]
As we know, \[k=\dfrac{2\pi }{\lambda }\]

Therefore, the x will be,

\[x=\left[ n+\dfrac{1}{2} \right]\dfrac{\lambda }{2};n=0,1,2,3...\]
As we know, the distance between any two antinodes is \[\dfrac{\lambda }{2}\]. So the length (L) can be equated with the x.

\[L=\dfrac{n\lambda }{2};n=1,2,3,...\]
\[\lambda =\dfrac{2L}{n};n=1,2,3,...\]

So, the frequencies of stationary waves can be written as,

\[\nu =\dfrac{nv}{2L};n=1,2,3,...\]

For fundamental mode,

\[L=\dfrac{{{\lambda }_{1}}}{2}\]
\[2L={{\lambda }_{1}}\]

So, the fundamental frequency can be written as,

\[{{\nu }_{1}}=\dfrac{v}{2L}\]

The frequency of the nth harmonic is

\[{{\nu }_{n}}=\dfrac{nv}{2L}\]

Here we have to find the fundamental frequency of the organ pipe of 85 cm. The velocity of the sound is \[340m{{s}^{-1}}\].

Since it is a fundamental frequency, the equation will be,
\[{{\nu }_{1}}=\dfrac{v}{2L}\]

We can assign the given values to the equation. Therefore,
\[{{\nu }_{1}}=\dfrac{340}{2\times 0.85}=200Hz\]

Note: Open pipes are extensively used in some organ pipes, flutes and oboes. For an open tube, the amplitudes will be maximum at both ends. So maximum air displacement occurs at the ends of the pipe. The resonant frequencies of a pipe closed at one end and open at another side.

\[{{\nu }_{n}}=\dfrac{nv}{4L},n=1,3,5....\]

That means for the fundamental frequency of both open pipes is twice of the one side open pipe. Moreover, it has a different spectrum of overtones. Both sides of the opened pipe have odd and even multiples of the fundamental frequency.