The given equation is y=2sin(ωt−kx)
The maximum particle velocity =Aω
And the wave velocity =kω
For wave velocity to be equal to maximum particle velocity
kω=Aω
⇒k=A1
Since, k=λ2π
λ=2πA=4πcm.
JEE Main 2022 — Physics Waves & Oscillations
A transverse wave is represented by y=2sin(ωt−kx)cm. The value of wavelength (in cm) for which the wave velocity becomes equal to the maximum particle velocity, will be
Held on 26 Jul 2022 · Verified 6 Jul 2026.
4π
2π
π
2
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