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The displacemeent of a particle is given...

The displacemeent of a particle is given at time t by `x = A sin(-2omega t) + B sin^(2)omegat`. Then,

A

the motion of the particle is SHM with an amplitude of `sqrt(A^(2) + B^(2)/4)`

B

the motion of the particle is not SHM, but oscillatory with a time period of `T = pi/omega`.

C

the motion of the particle is oscillatory with a time period of `T = pi/(2 omega)`.

D

the motion of the particle is a not periodic

Text Solution

Verified by Experts

The correct Answer is:
A
The displacement of the particle is given by `x = A sin(-2omega t) + B sin^(2) omega t`
`-A sin 2omega t + B/2 (1- cos2 omega t)`
`=(-A sin 2omegat + B/2 cos 2omegat) + B/2`
This motion represents SHM with an amplitude `sqrt(A^(2) + (B^(2)//4))`, and mean position of B/2.
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