Distance between two consecutive points 60∘ out of phase is given by Δx=2πλΔϕ
Putting the values, we have
6=2πλ(3π)
⇒λ=36m
Now, velocity of wave is v=fλ=(500Hz)(36m)
=18000ms−1=18kms−1
The distance between two consecutive points with phase difference of 60∘ in a wave of frequency 500Hz is 6.0m. The velocity with which wave is travelling is ______ kms−1.
Held on 25 Jan 2023 · Verified 6 Jul 2026.
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Work through every JEE Main Waves & Oscillations PYQ, year by year.