Question

A simple harmonic progressive wave is given by equation \( y = a \sin 2\pi (nt - \frac{x}{\lambda}) \). If the wave velocity is equal to \( \frac{1}{4} \times \) (maximum particle velocity), then the wavelength ' \( \lambda \) ' is (Given \( a = \) amplitude, \( n = \) frequency, \( t = \) time, \( y = \) displacement, \( x = \) distance )

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

Hint:

- $v = n\lambda$ - $v_{\max} = \omega a = 2\pi n a$

Answer 1

The Correct Option is B

Concept:
• Wave velocity: \[ v = n\lambda \]
• Maximum particle velocity: \[ v_{\max} = \omega a = 2\pi n a \]

Step 1: Given relation.
\[ v = \frac{1}{4} v_{\max} \]

Step 2: Substitute expressions.
\[ n\lambda = \frac{1}{4}(2\pi n a) \]

Step 3: Simplify.
\[ n\lambda = \frac{1}{2}\pi n a \]

Step 4: Cancel $n$.
\[ \lambda = \frac{\pi a}{2} \]

Step 5: Check options carefully.
Correct expression matches option (B).