Answer
(a) $f = 0.464~Hz$
$A = 0.343~m$
$T = 2.16~s$
(b) The time it takes for the system to return to the point of the collision is $1.08~s$
Work Step by Step
(a) We can use conservation of momentum to find the speed of the two blocks just after the collision.
$(m+m)v_2 = mv_1$
$v_2 = \frac{v_1}{2}$
$v_2 = \frac{2.00~m/s}{2}$
$v_2 = 1.00~m/s$
Let $M$ be the the total mass of the two blocks. We can find the frequency.
$f = \frac{1}{2\pi}~\sqrt{\frac{k}{M}}$
$f = \frac{1}{2\pi}~\sqrt{\frac{170.0~N/m}{20.0~kg}}$
$f = 0.464~Hz$
We can find the amplitude.
$\frac{1}{2}kA^2 = \frac{1}{2}Mv_{max}^2$
$A = \sqrt{\frac{M}{k}}~v_{max}$
$A = \sqrt{\frac{20.0~kg}{170.0~N/m}}~(1.00~m/s)$
$A = 0.343~m$
We can find the period.
$T = \frac{1}{f}$
$T = \frac{1}{0.464~Hz}$
$T = 2.16~s$
(b) For the system to return to the point of the collision, the blocks need to move through half of one cycle. We can find the time this takes.
$t = \frac{T}{2} = \frac{2.16~s}{2} = 1.08~s$
