Question

Explain analytically how stationary waves are formed. What are nodes and antinodes? Show that the distance between two adjacent nodes or antinodes is $\frac{\lambda }{2}$

Answer 1

Consider two simple harmonic progressive waves of the same amplitude $A$, wavelength $\lambda$ and frequency $n=\omega /2\pi$ travelling along the x-axis in opposite directions. They may be represented by
$y_1=A\sin (\omega t-kx)$ (along the $+$ x-axis) and $.....\left(1\right)$
$y_2=A\sin (\omega t+kx)$ (along the $-$ x-axis) $....\left(2\right)$
where $k=2\pi /\lambda$ is the propagation constant.
By the superposition principle,
$y=y_1+y_2=A\left[\sin \right(\omega t-kx)+\sin (\omega t+kx\left)\right]$
Using the trigonometrical identity,
$\sin C+\sin D=2\sin \left(\frac{C+D}{2}\right)\cos \left(\frac{C+D}{2}\right)$
$y=2A\sin \omega t\cos (-kx)$
$=2A\sin \omega t\cos kx\left[\cos \right(-kx)=\cos (kx\left)\right]$
$=2A\cos kx\sin \omega t....\left(3\right)$
$y=R\sin \omega t....\left(4\right)$
where $R=2A\cos kx....\left(5\right)$
The above equation shows that the resultant disturbance is simple harmonic having the same period as that of the individual waves and the amplitude $R$.
The point at which the particles of the medium are always at rest are called the nodes.
At nodes $R=0$
$\therefore \cos \frac{2\pi x}{\lambda }=0(\because A\ne 0andk=\frac{2\pi }{\lambda })$
$\therefore \frac{2\pi x}{\lambda }=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\frac{7\pi }{2},....$
$\therefore x=\frac{\lambda }{4},\frac{3\lambda }{4},\frac{5\lambda }{4},\frac{7\lambda }{4},....,(2m+1)\frac{\lambda }{4}....\left(6\right)$
where $m=0,1,2,...$ therefore the distance between successive nodes is
$\left[2\right(m+1)+1]\frac{\lambda }{4}-(2m+1)\frac{\lambda }{4}=\frac{\lambda }{2}$.
The points at which the particles of the medium vibrate with the maximum amplitude are called the antinodes.
At antinodes: $R=\pm 2A$
$\therefore \cos \frac{2\pi x}{\lambda }=\pm 1$
$\therefore \frac{2\pi x}{\lambda }=0,\pi ,2\pi ,3\pi ,....$
$\therefore x=0,\frac{\lambda }{2},\lambda ,\frac{3\lambda }{2},...,\frac{m\lambda }{2}(m=0,1,2)....\left(7\right)$
Therefore the distance between successive antinodes is
$\frac{(m+1)}{2}\cdot \lambda -\frac{m\lambda }{2}=\frac{\lambda }{2}$
$\therefore$ Distance between successive nodes
$=$ distance between successive antinodes $=\frac{\lambda }{2}$
From eq. $\left(6\right)$ and $\left(7\right)$, it can be seen that the nodes and the antinodes occur alternately and are equally spaced.