Displacement of a wave is expressed as x(t)=5 (628 t+(π)/(2)) m . The wavelength of the wave when its velocity is 300 ~m / s is :

Correct answer: B. 3 m aligned & x(t)=5 [628 t+(π)/(2)] m & velocity (v_ω)=300 m / s & v_ w = ω K & 300= 628 K K=(628)/(300) & (2 π)/(λ)=(628)/(300) λ=(2 × 3.14 × 300)/(628) & λ=2 m aligned

JEE Main 20254 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 4 April
Shift: Evening Shift
Subject: Physics
Chapter: Waves and Sound

JEE Main 2025 · 4 April · Evening Shift · Physics · Waves and Sound

Displacement of a wave is expressed as x(t)=5 (628 t+(π)/(2)) m ....

Displacement of a wave is expressed as

x(t)=5 \cos \left(628 t+\frac{\pi}{2}\right) m

. The wavelength of the wave when its velocity is

300 \mathrm{~m} / \mathrm{s}

is :

Options

A5 m
B3 m
C0.5 m
D0.33 m

Correct answer

B. 3 m

Step-by-step solution

\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}

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