A transverse wave travelling along a stretched string has a speed of $30 \mathrm{~m} / \mathrm{s}$ and a…

A transverse wave travelling along a stretched string has a speed of $30 \mathrm{~m} / \mathrm{s}$ and a frequency of 250 Hz. The phase difference between two points on the string 10 cm apart at the same instant is
  1. $0 \mathrm{ radian}$
  2. $\left(\frac{\pi}{2}\right) \mathrm{ radian}$
  3. $\left(\frac{5 \pi}{3}\right) \text{ radian}$
  4. $\left(\frac{8 \pi}{3}\right) \text{ radian}$

Solution

The wavelength of the wave is given by, $\lambda=\frac{\mathrm{v}}{\mathrm{n}}=\frac{30}{250}=0.12 \mathrm{~m}$
Also, the phase difference ' $\phi$ ' is given by, $\phi=\frac{2 \pi}{\lambda} \times$ path difference $\begin{aligned} & \therefore \quad \phi=\frac{2 \pi(0.1)}{\lambda} \\ & \text {...(path difference }= \\ & 10 \mathrm{~cm}=0.1 \mathrm{~m} \text { ) } \\ & \therefore \quad \phi=\frac{2 \pi}{0.12} \times 0.1 \\ & \therefore \quad \phi=\left(\frac{5 \pi}{3}\right)^{\text{radian}} \end{aligned}$ *

Asked in: MHT CET 2024 (04 May Shift 2)

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