Easy MCQ +4 / -1 PYQ · JEE Mains 2025

Displacement of a wave is expressed as $x(t)=5 \cos \left(628 t+\frac{\pi}{2}\right) \mathrm{m}$. The wavelength of the wave when its velocity is $300 \mathrm{~m} / \mathrm{s}$ is :

$(\pi=3.14)$

  1. A 0.33 m
  2. B 0.5 m
  3. C 3 m Correct answer
  4. D 5 m

Solution

<p>$$\begin{aligned} & \mathrm{x}(\mathrm{t})=5 \cos \left[628 \mathrm{t}+\frac{\pi}{2}\right] \mathrm{m} \\ & \text { velocity }\left(\mathrm{v}_\omega\right)=300 \mathrm{~m} / \mathrm{s} \\ & \mathrm{v}_{\mathrm{w}}=\frac{\omega}{\mathrm{K}} \\ & 300=\frac{628}{\mathrm{~K}} \Rightarrow \mathrm{~K}=\frac{628}{300} \\ & \frac{2 \pi}{\lambda}=\frac{628}{300} \Rightarrow \lambda=\frac{2 \times 3.14 \times 300}{628} \\ & \lambda=2 \mathrm{~m} \end{aligned}$$</p>

About this question

Subject: Physics · Chapter: Waves · Topic: Wave Motion and Types

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