Waves

JEE Main level, single correct, medium

10 of 142

Question 10 of 142

A longitudinal wave is represented by y=10sin⁡2π(nt−xλ)cmy=10\sin 2\pi \left(nt-\dfrac{x}{\lambda }\right)cm. The maximum particle velocity will be four times the wave velocity if the determined value of wavelength is equal to

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Answer: (B) 5π5\pi

As we know particle velocity, vp=dydt=(10×2πn)×cos⁡2π(nt−xλ){v}_{p}=\dfrac{dy}{dt}=\left(10\times 2\pi n\right)\times \cos 2\pi \left(nt-\dfrac{x}{\lambda }\right)
Therefore, (vp)max=20πn{\left({v}_{p}\right)}_{max}=20\pi n
Now as we know, v=λnv=\lambda n
As given in the question,
(vp)max=4v⇒20πn=4λn⇒λ=5π{\left({v}_{p}\right)}_{max}=4v \Rightarrow 20\pi n=4\lambda n \Rightarrow \lambda =5\pi

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