Waves

JEE Main level, numerical answer, medium

32 of 142

Question 32 of 142

The distance between two consecutive points with phase difference of 60∘60^{\circ} in a wave of frequency 500Hz500Hz is 6.0m6.0m. The velocity with which wave is travelling is ______ kms−1km{s}^{-1}.

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Answer: 18

Distance between two consecutive points 60∘60^{\circ} out of phase is given by Δx=λ2πΔϕ\Delta x=\dfrac{\lambda }{2\pi }\Delta \phi
Putting the values, we have
6=λ2π(π3)6=\dfrac{\lambda }{2\pi }\left(\dfrac{\pi }{3}\right)
⇒λ=36m\Rightarrow \lambda =36m
Now, velocity of wave is v=fλ=(500Hz)(36m)v=f\lambda =(500Hz)(36m)
=18000ms−1=18kms−1=18000m{s}^{-1}=18km{s}^{-1}

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