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A 4 kg particle is moving along the x- a...

A `4 kg` particle is moving along the x- axis under the action of the force `F = - ((pi^(2))/(16)) x N` At `t = 2 sec ` the particle passes through the origin and `t = 10sec`, the speed is `4sqrt(2)m//s` The amplitude of the motion is

A

`(32sqrt2)/(pi)m`

B

`(16)/(pi)m`

C

`(4)/(pi)m`

D

`(16sqrt2)/(pi)m`

Text Solution

Verified by Experts

The correct Answer is:
A
`a=-((pi^(2))/(64))x Rightarrow omega=sqrt((pi^(2))/(64))=(pi)/(8) Rightarrow T=(2pi)/(omega)=16sec`
There is a difference of between t=2sec. To=10 sec. Hence particle is again passing through the mean position of SHM where its speed is maximum.
i.e. `V_(max)=Aomega=4sqrt2 Rightarrow A=(4sqrt2)/(pi//8)=(32sqrt2)/(x)m`
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