A particle of mass $m$ is executing a SHM in a straight line under an acceleration $n^2 \times (distance)$. If a periodic force $mk \cos{pt}$ be introduced and the time period of forced vibration becomes increased $2\frac{1}{2}$ times, then show that $$25p^2=4n^2.$$
The equation of motion is $$m\ddot{x}=-mn^2x+mk \cos{pt}$$
The general solution of this differential equation can be obtained as $$x=a\cos{(nt+\epsilon)}+\frac{k}{n^2-p^2}\cos{pt},~~ for~~p\ne n$$ and $$x=a\cos{(nt+\epsilon)}+\frac{k}{2p}t\sin{pt},~~for ~~p=n$$ $a$ and $\epsilon~$ are constants.
How to go further to get the desired answer? Please help.
Without the forced vibration, the period would be $T_1=\frac{2\pi}n$. With the forced vibration the period becomes $T_2=\frac{2\pi}p$. This is a simplification, but close enough to be useful.
Set $\frac{T_2}{T_2}=\frac52$ and square. Squaring is necessary because we know about the magnitude; $p$ or $n$ could be negative.