Waves

JEE Main level, single correct, medium

8 of 142

Question 8 of 142

In the wave equation y=0.5sin⁡2πλ(400t−x)my=0.5\sin \dfrac{2\pi }{\lambda }\left(400t-x\right)m the velocity of the wave will be :

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Answer: (C) 400ms−1400m{s}^{-1}

Given equation of the wave is y=0.5sin⁡(2πλ400t−2πλx)y=0.5\sin \left(\dfrac{2\pi }{\lambda }400t-\dfrac{2\pi }{\lambda }x\right)
Comparing it with the standard equation of motion y=Asin⁡(ωt−kx)y=A\sin \left(\omega t-kx\right) we get,
ω=2πλ400\omega =\dfrac{2\pi }{\lambda }400 and k=2πλk=\dfrac{2\pi }{\lambda }
Now the velocity of the wave will be,
v=ωk⇒v=400ms−1v=\dfrac{\omega }{k} \Rightarrow v=400m{s}^{-1}

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