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 isA. `(32sqrt(2))/(pi)m`B. `(16)/(pi)m`C. `(4)/(pi)m`D. `(16sqrt(2))/(pi)m`

A `4 kg` particle is moving along the x- axis under the action of the

1.	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 isA. `(32sqrt(2))/(pi)m`B. `(16)/(pi)m`C. `(4)/(pi)m`D. `(16sqrt(2))/(pi)m`
Answer» Correct Answer - A `a = -((pi^(2))/(64)) x rArr omega rArr sqrt((pi^(2))/(64)) = (pi)/(8)` `rArr T = (2pi)/(omega) = 16 sec` There is a time difference of `712` between `t = 2sec` to `t = 10sec`, Hence particle is again passing through the mean position of `SHM` where itsw speed is maximum `i.e. V_(max) = A omega = 4sqrt(2)` `rArr A = (4sqrt(2))/(pi//8) = (32sqrt(2))/(pi) m`

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