Question

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

Answer 1

Hint: Here we have to use the principle of superposition. Two waves moving in different directions and having the same frequency and amplitude creates a standing wave. The general wave formula is given by${y_1} = A\sin (\omega t - kx)$. Here, y = Displacement y of a particle, A= Amplitude, $\omega $= Angular Frequency. t = Time period, k = Wavenumber, x = x direction. Nodes and Antinodes define the position of a wave at a particular and amplitude. The distance between them can be calculated by the general wave formula mentioned in the previous line.

Formula Used:
The formula used in the question is given below
${y_1} = A\sin (\omega t - kx)$
Here
Y= Axis
A= Amplitude
$\omega $= Angular Frequency.
t = Time period.
K = Wavenumber
x = x direction
Complete step by step answer:
Step 1: Write the general equation of wave and apply the principle of superposition.
${y_1} = A\sin (\omega t - kx)$
${y_2} = A\sin (\omega t + kx)$
The above two equations are the general equations of the wave
Now apply the principle of superposition
$y = {y_1} + {y_2}$
Put the value of ${y_1}$and${y_2}$in the above equation
$y = A\left[ {\sin (\omega t - kx) + \sin (\omega t + kx)} \right]$
Apply the trigonometric property:
$\operatorname{Sin} (C + D) = 2\operatorname{Sin} (\dfrac{{C + D}}{2}).\operatorname{Cos} (\dfrac{{C - D}}{2})$
= $2\operatorname{Sin} (\dfrac{{\omega t - kx + \omega t + kx}}{2}).\operatorname{Cos} (\dfrac{{\omega t - kx - \omega t - kx}}{2})$
Solve
= \[2\operatorname{Sin} (\dfrac{{2\omega t}}{2})\cos (\dfrac{{ - 2kx}}{2})\]
Put the above value in the wave equation,
$y = 2A\operatorname{Sin} \omega t\cos ( - kx)$
Here,$\operatorname{Cos} ( - kx)\operatorname{Cos} (kx) = \operatorname{Cos} (Kx)$
$y = R\sin \omega t$
$R = 2A\cos kx$
$y = R\operatorname{Sin} (\omega t)$
The above equation represents Simple Harmonic motion. The absence of x in the equation shows that the resultant wave is neither travelling forwards or backward. Hence it is called a stationary wave.
Step 2: Identify the distance between two nodes
Nodes are the points at which the particle of the medium is always at rest.
At node R=0
$\operatorname{Cos} \dfrac{{2\pi x}}{\lambda } = 0$; … Where K=$\dfrac{{2\pi }}{\lambda }$; A$ \ne 0$
$
  \dfrac{{2\pi x}}{\lambda } = \dfrac{\pi }{2},\dfrac{{3\pi }}{2},\dfrac{{5\pi }}{2}...... \\
    \\
$
Find out the value of x
$x = \dfrac{\lambda }{4},\dfrac{{3\lambda }}{4},\dfrac{{5\lambda }}{4}....(2n + 1)\dfrac{\lambda }{4}$
Subtract ${x_2}$ from${x_1}$:
${x_2} - {x_1} = \dfrac{{5\lambda }}{4} - \dfrac{{3\lambda }}{4} = \dfrac{\lambda }{2}$
The distance between two successive nodes is $\dfrac{\lambda }{2}$
Step 3: Identify the distance between antinodes.
Antinodes are the points at which the particle of the medium vibrates with maximum amplitude.
$R = \pm 2A$ ….(Antinode)
$\operatorname{Cos} \dfrac{{2\pi x}}{\lambda } = \pm 1$
The value ranges from 0, π, 2 π, 3 π ….
$
  \operatorname{Cos} \dfrac{{2\pi x}}{\lambda } = 0,\pi ,2\pi ,3\pi ...... \\
  x = 0,\dfrac{\lambda }{2},\lambda ,\dfrac{{3\lambda }}{2},......,\dfrac{{n\lambda }}{2}(n = 0,1,2) \\
$
Solve the difference
\[{x_2} - {x_1} = \lambda - \dfrac{\lambda }{2} = \dfrac{\lambda }{2}\]
The distance between two successive antinodes is \[\dfrac{\lambda }{2}\]

Note: Here there is an extensive use of trigonometric properties in each step. Make sure to apply the properties at the appropriate equations Go step by step as it is a long process. Make sure to define mathematical relations clearly.