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A particle is performing S.H.M with acce...

A particle is performing `S.H.M` with accerlration `a = 8 pi^(2) - 4 pi^(2) x` where `x` is coordinate of the particle `w.r.t` the origin. The parameters are in `S.I.` units. The particle is at rest at `x = -2` at `t = 0`.

A

coordinate of the particle `w.r.t` origin at any time `t` is `2+4 cos 2pit`

B

coordinate of the particle `w.r.t` origin at any time `t` is `2+4 sin 2pit`

C

coordinate of the particle `w.r.t` origin at any time `t` is `-4+2 cos 2pit`

D

the coordinate cannot be found because mas of the particle is not given.

Text Solution

Verified by Experts

The correct Answer is:
A
`a = 8pi^(2) - 4pi^(2)x = -4x^(2)(x-2) rArr omega x = 2`
Here `a = 0` so mean position at `x = 2`
Let `x = A sin (omegat + phi)`
As particle is at rest at `x = -2` (extreme position) and Amplitude `= 4` as particle start from extreme position. Therefore
`x -2 = -4 cos2pit rArr x = 2-4 cos2pit`
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