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=(4pi)/3 sqrt(m/k)`
D. The time at which the particle passes through the equilibrium position for the second time is `t=(5pi)/3sqrt(m/k)`

Answer

A, D

Explanation

(a,d) The particle collides elastically with rigid wall. Here
`e=V/(0.5u_0)=1 :. V=0.5u_0`

i.e. the particle rebounds with same speed. Therefore
the particle will return to its equilibrium position with
speed `u_0.` option (a) is correct.
The velocity of the particle becomes `0.5u_0` after time t.
Then using the equation `V=V_(max) cos wt` we get
`0.5u_0=u_0cos wt`
`:. pi/3=(2pi)/txxT :. t=T/6`
The time period `T=2pisqrt(m/k).Therefore t=pi/3sqrt(m/k)`
The time taken by the particle to pass through the
equilibrium for the first time `=2t=(2pi)/3sqrt(m/k).` Therefore
option (b) is incorrect
The time taken for the maximum compression
`=pi/3sqrt(m/k)+pi/3sqrt(m/k)+pi/3sqrt(m/k)=pisqrt(m/k)[1/3+1/3+1/2]`
`=(7pi)/6sqrt(m/k).` Therefore option c is incorrect.
The time taken for particle to pass through the
equilibrium position second time
`=2[pi/3sqrt(m/k)]+pisqrt(m/k)=pisqrt(m/k)(2/3+1)=5/3pisqrt(m/k)`
option (d) is correct.