Since, y=kt2
∴dydt=2kt
d2ydt2=2k...(i)(∵k=1m/s2given)
a=2m/s2
We know that T=2π√lg
∴T21T22=g2g1
As the point of suspension accelerate in upward direction, a Pseudo force acts in opposite of the the motion's direction.
Therefore, here g2 is effective acceleration due to gravity because of motion of point of suspension.
[∵g1=10m/s2g2=g+2=12m/s2]
⇒T21T22=1210=65=x10
∴x=12