1.

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. ThenA. V cannot be equal to vB. `V=v`, if `A=lambda/(2pi)`C. `V-v`, if `A=2 pi lambda`D. `V=v`, if `lambda =A/pi`

Answer» 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)`.


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