JEE Advanced 2013 Paper 2

v=μ0 sin ωt (suppose t1 is the time of collision)

μ0

2

=μ0cosωt1⇒t1=

π

3ω

Now the particle returns to equilibrium position at time t2=2t1i.e

2π

3ω

with the same mechanical energy
i.e. its speed will µ0 .
Let t3 is the time at which the particle passes through the equilibrium position for the second time.
∴ t3=

T

2

+2t1
=

π

ω

+

2π

3ω

=

5π

3ω

=

5π

3

√

m

k

Energy of particle and spring remains conserved.