The differential equation of all 'Simple Harmonic Motions' of given period $\frac{2\pi }{n}$ is $\frac{2\pi }{\omega }$, $x=a\cos (\omega t+\varphi )$

 Given, the equation of simple harmonic motion:
$x=a\cos (\omega t+\varphi )$
Here,$a=$ Amplitude,
$t=$ time period,
$\varphi =$ phase angle.
we law, $\omega =\frac{2\pi }{t}$.

Then,

$x=a\cos \left(\frac{2\pi }{t}·t+\varphi \right)$
for given period of $\frac{2\pi }{n};t=\frac{2\pi }{n}$.
So, we have.
$$\begin{gathered} x=a\cos \left(\frac{2\pi }{2\pi /n}·t+\varphi \right) \\ x=a\cos (n·t+\varphi ) ......\left(1\right) \end{gathered}$$
Differentiate both sides with respect to $t,$,
$\frac{dx}{dt}=-na\sin (nt+\varphi ) \left\{\frac{d}{dx}\left(\cos x\right)=-\sin x\right\}\text{. }$
Again, differentiate the above equation,
$\text{ we have, }\frac{d^{2}x}{d{t}^2}=-n^{2}a\cos (nt+\varphi ) .....\left(2\right) \left\{\frac{d}{dx}\left\{\sin x\right)=\cos x\right\}\text{. }$
Now, substitute $\left(1\right)$ in $\left(2\right),$ we get
$\frac{d^{2}x}{d{t}^2}=-n^{2}x \{\because x=acos(nt+\varphi \left)\right\}\text{. }$
$\Rightarrow \frac{d^{2}x}{d{t}^2}+n^{2}x=0$