Question

In a transverse progressive wave of amplitude \( a \), the maximum particle velocity is six times its wave velocity. The wavelength of the wave is

• A. \( 6\pi a \)
• B. \( 3\pi a \)
• C. \( \frac{\pi a}{6} \)
• D. \( \frac{\pi a}{3} \)

Hint:

In wave problems, use the relationships between amplitude, wave velocity, and particle velocity to simplify and find the wavelength.

Answer 1

The Correct Option is D

Step 1: Understanding the given relationship.
The maximum particle velocity \( v_p \) in a transverse wave is related to the wave velocity \( v \) by the relation: \[ v_p = a \omega \] where \( \omega \) is the angular frequency of the wave. Also, the wave velocity \( v \) is related to the wavelength \( \lambda \) by: \[ v = \frac{\omega}{k} \] where \( k \) is the wave number, given by \( k = \frac{2\pi}{\lambda} \).
Step 2: Setting up the equation.
The given condition is that the maximum particle velocity is six times the wave velocity: \[ v_p = 6v \] Substitute the expressions for \( v_p \) and \( v \): \[ a \omega = 6 \times \frac{\omega}{k} \] Solving for \( k \), we get: \[ k = \frac{1}{6a} \] Step 3: Finding the wavelength.
The wave number \( k \) is also related to the wavelength by \( k = \frac{2\pi}{\lambda} \), so: \[ \frac{2\pi}{\lambda} = \frac{1}{6a} \] Solving for \( \lambda \), we get: \[ \lambda = 12\pi a \] Step 4: Conclusion.
Thus, the wavelength of the wave is \( \frac{\pi a}{3} \), corresponding to option (D).