Question

A simple harmonic progressive wave is represented as \( Y = A \sin 2\pi \left( \frac{nt - x}{\lambda} \right) \) cm. If the maximum particle velocity is four times the wave velocity, then the wavelength of the wave is

• A. \( \frac{\pi A}{4} \)
• B. \( 4\pi A \)
• C. \( 2\pi A \)
• D. \( \frac{\pi A}{2} \)

Hint:

In wave problems, use the relation between particle velocity and wave velocity to find the wave number and wavelength.

Answer 1

The Correct Option is D

Step 1: Understanding the relationship between particle velocity and wave velocity.
The maximum particle velocity \( v_{\text{particle}} \) is related to the wave velocity \( v_{\text{wave}} \) by: \[ v_{\text{particle}} = 4v_{\text{wave}} \] The maximum particle velocity is given by \( v_{\text{particle}} = A \omega \), where \( \omega \) is the angular frequency. The wave velocity is given by \( v_{\text{wave}} = \frac{\omega}{k} \), where \( k \) is the wave number.
Step 2: Using the given condition.
Since \( v_{\text{particle}} = 4v_{\text{wave}} \), we have: \[ A \omega = 4 \times \frac{\omega}{k} \] Solving for \( k \), we find: \[ k = \frac{1}{2A} \] Step 3: Conclusion.
Thus, the wavelength \( \lambda = \frac{2\pi}{k} = \frac{\pi A}{2} \), which corresponds to option (D).