Here
T=2π√(m/k)=2π or m/k=1
In equilibrium,
mg=kd
d=mg/k=1×g=g meters.

T=2π√(m/k), k=F/x
When the spring is cut into four equal parts,
k^‘=F/(x/4)=4(F/x)=4k
T^‘=2π√(m/k^‘)=2π√(m/4k)=T/2.

T^‘/T=√(l^‘/l)=√(4l/l)=2
T^‘=2T=2×2=4sec.

T∝√l
Percentage increase in time period,
∆T/T×100=1/2×∆l/l×100
=1/2×2%=1%.

When at rest, T=2π√(l/g)
When the rocket moves up with uniform acceleration,
T^‘=2π√(l/(g+a))
Clearly, T^‘<T.