A particle of mass m is attached to one end of a mass-less spring of force constant k, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time `t=0` with an initial velocity `u_0`. when the speed of the particle is `0.5u_0`, it collides elastically with a rigid wall. After this collision

Options

A. the speed of the particle when it returns to its equilibrium position is `u_0`.
B. the time at which the particle passes through the equilibrium position for the first time is `t = pi sqrt((m)/(k))`
C. the time at which the maximum compression of the spring occurs is `t = (4 pi)/(3) sqrt((m)/(k))`
D. the time at which the particle passes through the equilibrium position for the second time is `t = (5 pi)/(3) sqrt((m)/(k))`.

Answer

A, D

Explanation

`v = u_(0) sin omega t`
(suppose `t_(01)` is the time of collision)
`(u_(0))/(2) = u_(0) cos omega t_(1) rArr t_(1) = (pi)/(3 omega)`
Now the particle returns to equilibrium position at time `t_(2) = 2t_(1)` i.e., `(2 pi)/(3 omega)` with the same mechanical energy i.e., its speed will be `u_(0)`. Let `t` be the time at which the particle passes through the equilibrium position for the second time.
Energy of particle and spring remains conserved.
`t_(2)=(T)/(2)+2t_(1) =(pi)/(omega)+(2pi)/(3 omega)=(5 pi)/(3 omega) =(5pi)/(3) sqrt((m)/(k))`.