Question

A transverse wave travelling along a stretched string has a speed of 30 m/s and a frequency of 250 Hz. The phase difference between two points on the string 10 cm apart at the same instant is

• A. π/3 radian
• B. 4π/3 radian
• C. 5π/3 radian
• D. π/6 radian

Answer 1

Answer: (C)

5π/3 radian

Answer 2

1. Calculate the wavelength:

   $$\lambda = \frac{v}{f} = \frac{30}{250} = 0.12\,\text{m}$$

2. Compute the phase difference:

   $$\phi = \frac{2\pi}{\lambda} \times \text{distance} = \frac{2\pi}{0.12} \times 0.1 = \frac{0.2\pi}{0.12} = \frac{5\pi}{3}\,\text{rad}$$

Core Explanation:
Wavelength $\lambda = 0.12\,\text{m}$. The phase difference between two points 0.1 m apart is $\phi = \frac{2\pi}{0.12}(0.1) = \frac{5\pi}{3}\,\text{rad}$.