Question

A simple harmonic progressive wave is given by \( Y = Y_0 \sin 2\pi \left( nt - \frac{x}{\lambda} \right) \). If the wave velocity is \(\left( \frac{1}{8} \right)^{\text{th}}\) the maximum particle velocity then the wavelength is

• A. \(\frac{\pi Y_0}{2}\)
• B. \(\frac{\pi Y_0}{4}\)
• C. \(\frac{\pi Y_0}{8}\)
• D. \(\frac{\pi Y_0}{16}\)

Hint:

For a sinusoidal wave, maximum particle velocity = amplitude × angular frequency. Wave velocity = frequency × wavelength. Keep symbols distinct to avoid confusion.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a wave equation and a relation between wave velocity and maximum particle velocity. We need to find wavelength in terms of amplitude \(Y_0\).

Step 2: Key Formula or Approach:
From \(Y = Y_0 \sin(2\pi n t - \frac{2\pi}{\lambda} x)\), wave velocity \(v = n\lambda\). Maximum particle velocity \(v_{p,\text{max}} = \omega Y_0 = 2\pi n Y_0\).
Given \(v = \frac{1}{8} v_{p,\text{max}}\).

Step 3: Detailed Explanation:
Substitute: \(n\lambda = \frac{1}{8} (2\pi n Y_0)\). Cancel \(n\) (assuming \(n \neq 0\)): \[ \lambda = \frac{1}{8} \cdot 2\pi Y_0 = \frac{\pi Y_0}{4}. \]

Step 4: Final Answer:
Wavelength \(\lambda = \frac{\pi Y_0}{4}\), which corresponds to option (B).