The displacement of a particle, executing simple harmonic motion with time period T, is expressed as $$x(t) = A\sin \omega t$$, where A is the amplitude. The maximum value of potential energy of this oscillator is found at $$t=T/2\beta$$. The value of $$\beta$$ is_____.

Potential energy in SHM is maximum at extreme positions:

$$x=\pm A$$

Given

$$x(t)=A\sin\omega t$$

For maximum potential energy,

$$\sin\omega t=\pm1$$

First time this happens is

$$\omega t=\frac{\pi}{2}$$

So

$$t=\frac{\pi}{2\omega}$$

Now

$$\omega=\frac{2\pi}{T}$$

Substitute:

$$t=\frac{\pi}{2(2\pi/T)}$$

$$=\frac{T}{4}$$

Given

$$t=\frac{T}{2\beta}$$

So

$$\frac{T}{2\beta}=\frac{T}{4}$$

Cancelling T,

$$2\beta=4$$

$$β=2$$