Question

obtain a relation between the period of a simple harmonic motion and acceleration at unit displacement

Answer 1

The displacement for a simple harmonic motion

$$x(t)=x_0\cos\omega t=x_0\cos\frac{2\pi t}{T}$$

The acceleration

$$a(t)=x''(t)=-\frac{4\pi^2}{T^2}x_0\cos\frac{2\pi t}{T}=-\frac{4\pi^2}{T^2}x(t)$$

So

$$T=2\pi\sqrt{x/a}$$