Question

Two travelling waves, \( y_1 = A \sin [ k ( x + ct ) ] \) and \( y_2 = A \sin [ k ( x - ct ) ] \) are superposed on a string. The distance between adjacent antinodes is

• A. \( \frac{ct}{\pi} \)
• B. \( \frac{ct}{2\pi} \)
• C. \( \frac{\pi}{2k} \)
• D. \( \frac{k}{\pi} \)
• E. \( \frac{\pi}{k} \)

Hint:

The distance between a node and its adjacent antinode is \(\lambda/4\), while the distance between two adjacent nodes or two adjacent antinodes is always \(\lambda/2\).

Answer 1

Answer: (E)

Concept: When two travelling waves of equal amplitude and frequency move in opposite directions and superpose, they form a stationary (standing) wave.
• Wave Number (\(k\)): The spatial frequency of a wave, related to wavelength (\(\lambda\)) by \( k = \frac{2\pi}{\lambda} \).
• Antinodes: Positions in a standing wave where the amplitude of vibration is maximum.
• Node-Antinode Geometry: In any standing wave, the distance between two adjacent nodes or two adjacent antinodes is exactly half a wavelength (\(\lambda / 2\)).

Step 1: Relate the wave number \(k\) to the wavelength \(\lambda\).
From the standard wave equation form provided, the term \(k\) represents the wave number. The relationship between \(k\) and the wavelength \(\lambda\) is defined as: \[ k = \frac{2\pi}{\lambda} \implies \lambda = \frac{2\pi}{k} \]

Step 2: Calculate the distance between adjacent antinodes.
The distance between two consecutive antinodes in a stationary wave is given by: \[ d = \frac{\lambda}{2} \] Substituting the expression for \(\lambda\) from
Step 1: \[ d = \frac{1}{2} \left( \frac{2\pi}{k} \right) = \frac{\pi}{k} \]