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

Answer 1

The phase difference between the two points is $\frac{5\pi}{3}$ radians.

Answer 2

Solution:

1. Calculate the wavelength ($\lambda$):

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

2. Find the phase difference ($\phi$) between two points 0.10 m apart:

   $$\phi = \frac{2\pi}{\lambda} \times x = \frac{2\pi}{0.12} \times 0.10 = \frac{0.20\pi}{0.12} = \frac{5\pi}{3} \, \text{radians}$$

Minimal Explanation:
Wavelength is $0.12\,\text{m}$. Phase difference is calculated by $\phi = \frac{2\pi}{\lambda}x = \frac{5\pi}{3}$ radians.