In figure, the stiffness of the spring is `k` and mass of the block is `m`. The pulley is fixed. Initially, the block `m` is held such that the elongation in the spring is zero and then released from rest. Find:
a. the maximum elongation in the spring.
b. the maximum speed of the block `m`. Neglect the mass of the spring and that of the string. Also neglect the friction.

Explanation

Let the maximum elongation in the spring be x, when the block is at position 2.
a. The displacement of the block `m` is also `x`.
From conservation of energy, `DeltaK=DeltaU=0`
`=1/2kx^2=mgx`
`impliesx=2mg//k`
b. Let maximum velocity occur when block moves a distance `x`.
Using `DeltaK+DeltaU=0`
`(1/2mv^2-0)+(-mgx+1/2kx^2)=0`
or `1/2mv^2=mgx-1/2kx^2`
Differeciating above equations w.r.t. x
`1/2m*2v*(dv)/(dx)=mg(dx)/(dt)-1/2k2x(dx)/(dt)`
`mv*(dv)/(dt)=mgv-kxv`
For maximum v, `(dv)/(dx)=0impliesx=(mg)/(k)`
So, `mg((mg)/(k))-1/2k((mg)/(k))^2=1/2mv_(max)^2`
`impliesv_(max)=(sqrt(m/k))g`