Question

Derive an expression for the equation of stationary wave on a stretched string. Show that the distance between two successive nodes or antinodes is \( \frac{\lambda}{2} \).

Hint:

In stationary waves, nodes and antinodes are fixed points, with the distance between two successive nodes or antinodes being half the wavelength.

Answer 1

A stationary wave is formed when two progressive waves of the same frequency and amplitude, traveling in opposite directions, superpose on a stretched string. The equation of the two waves can be written as: \[ y_1(x,t) = A \sin(kx - \omega t) \] and \[ y_2(x,t) = A \sin(kx + \omega t) \] where: - \( A \) is the amplitude,
- \( k \) is the wave number (\( k = \frac{2\pi}{\lambda} \), where \( \lambda \) is the wavelength),
- \( \omega \) is the angular frequency,
- \( x \) is the position along the string,
- \( t \) is time.
Now, the resultant displacement \( y(x,t) \) of the string is the sum of these two waves:
\[ y(x,t) = y_1(x,t) + y_2(x,t) \] Substitute the expressions for \( y_1(x,t) \) and \( y_2(x,t) \):
\[ y(x,t) = A \sin(kx - \omega t) + A \sin(kx + \omega t) \] Using the trigonometric identity for the sum of sines, we have:
\[ y(x,t) = 2A \sin(kx) \cos(\omega t) \] This is the equation of a stationary wave on the stretched string, where:
- The term \( 2A \sin(kx) \) represents the spatial variation (the shape of the wave along the string),
- The term \( \cos(\omega t) \) represents the time variation.
Identifying Nodes and Antinodes:
- The term \( \sin(kx) \) determines the spatial positions where the displacement is zero. These points are called nodes.
- The term \( 2A \sin(kx) \) reaches its maximum value \( 2A \) at points where \( \sin(kx) = \pm 1 \), and these points are called antinodes.
The nodes occur when \( \sin(kx) = 0 \), which happens at:
\[ kx = n\pi \quad \Rightarrow \quad x_n = \frac{n\pi}{k} \] where \( n \) is an integer. These are the positions of the nodes.
Distance Between Two Successive Nodes:
The distance between two successive nodes is:
\[ \Delta x = x_{n+1} - x_n = \frac{(n+1)\pi}{k} - \frac{n\pi}{k} = \frac{\pi}{k} \] Since the wavelength \( \lambda = \frac{2\pi}{k} \), the distance between two successive nodes is:
\[ \Delta x = \frac{\lambda}{2} \] 
Distance Between Two Successive Antinodes:
The antinodes occur where \( \sin(kx) = \pm 1 \), and the distance between two successive antinodes is also \( \frac{\lambda}{2} \).
Thus, the distance between two successive nodes or two successive antinodes in a stationary wave on a stretched string is \( \frac{\lambda}{2} \).