Waves

JEE Main level, single correct, medium

33 of 142

Question 33 of 142

A travelling wave is described by the equation y(x,t)=[0.05sin⁡(8x−4t)]my\left(x,t\right)=\left[0.05\sin \left(8x-4t\right)\right]m. The velocity of the wave is:
[All the quantities are in SI unit]

Show answer and solution

Answer: (C) 0.5ms−10.5m{s}^{-1}

The given wave equation is of the form:
y=Asin⁡(kx−ωt)y=A\sin \left(kx-\omega t\right)
with k=8k=8 and ω=4\omega =4
Since, k=2πλk=\dfrac{2\pi }{\lambda } and ω=2πf\omega =2\pi f
Then, ωk=fλ=v\dfrac{\omega }{k}=f\lambda =v
⇒v=48=0.5ms−1\Rightarrow v=\dfrac{4}{8}=0.5m{s}^{-1}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Standing waves on a string simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Waves (Travelling Waves), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.