Question

The speed of a progressive transverse wave on a string is \(18\ \mathrm{ms^{-1}}\). If the phase difference between two points on the string separated by a distance of \(3\) cm is \(90^\circ\), then the frequency of the transverse wave on the string is

• A. \(300\) Hz
• B. \(225\) Hz
• C. \(75\) Hz
• D. \(150\) Hz

Hint:

For a progressive wave, \[ \boxed{ \phi=\frac{2\pi x}{\lambda} } \] and \[ \boxed{ v=f\lambda. } \] A phase difference of \[ 90^\circ \] corresponds to a path difference of \[ \boxed{\frac{\lambda}{4}.} \]

Answer 1

The Correct Option is D

Step 1: Find the wavelength. The phase difference between two points is \[ \phi = \frac{2\pi x}{\lambda}. \] Given, \[ \phi=90^\circ=\frac{\pi}{2}, \] and \[ x=3\text{ cm}=0.03\text{ m}. \] Hence, \[ \frac{\pi}{2} = \frac{2\pi(0.03)}{\lambda}. \] Therefore, \[ \lambda = 4(0.03) = 0.12\text{ m}. \]

Step 2: Use the wave equation. Wave speed is \[ v=f\lambda. \] Given, \[ v=18\text{ ms}^{-1}. \] Thus, \[ f = \frac{18}{0.12} = 150\text{ Hz}. \]

Step 3: Write the answer. Hence, \[ \boxed{150\text{ Hz}}. \] Thus, \[ \boxed{(D)} \] is the correct answer.