Waves

JEE Main level, single correct, medium

9 of 142

Question 9 of 142

A transverse wave is represented by y=2sin⁡(ωt−kx)cmy=2\sin \left(\omega t-kx\right)cm. The value of wavelength (in cmcm) for which the wave velocity becomes equal to the maximum particle velocity, will be

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Answer: (A) 4π4\pi

The given equation is y=2sin⁡(ωt−kx)y=2\sin \left(\omega t-kx\right)
The maximum particle velocity =Aω=A\omega
And the wave velocity =ωk=\dfrac{\omega }{k}
For wave velocity to be equal to maximum particle velocity
ωk=Aω\dfrac{\omega }{k}=A\omega
⇒k=1A\Rightarrow k=\dfrac{1}{A}
Since, k=2πλk=\dfrac{2\pi }{\lambda }
λ=2πA=4πcm\lambda =2\pi A=4\pi cm.

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