Question

A sine wave has an amplitude A and wavelength `lambda`. Let V be the wave velocity and v be the maximum velocity of a particle in the medium. Then
A. V cannot be equal to v
B. `V=v`, if `A=lambda/(2pi)`
C. `V-v`, if `A=2 pi lambda`
D. `V=v`, if `lambda =A/pi`

Answer 1

Correct Answer - B
Let the equation of wave by `y=A sin (omega t-kx)`
where, `omega=2pi n` and `k=(2pi)/lambda`.
Wave velocity, `V=n lambda =omega /(2pi)xx(2pi)/k=omega/k`
Maximum particle velocity `v= A omega`
For `V=v, omega/k=A omega` or `A=1/k=lambda/(2pi)`.