The displacement of simple harmonic oscillator after $$3$$ seconds starting from its mean position is equal to half of its amplitude. The time period of harmonic motion is

For a simple harmonic oscillator starting from the mean position, the displacement as a function of time is:

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

where $$A$$ is the amplitude and $$\omega = \frac{2\pi}{T}$$ is the angular frequency.

Given that at $$t = 3$$ s, the displacement equals half the amplitude:

$$\frac{A}{2} = A\sin(\omega \times 3)$$

Dividing both sides by $$A$$:

$$\sin(3\omega) = \frac{1}{2}$$

The general solution of $$\sin\theta = \frac{1}{2}$$ is $$\theta = n\pi + (-1)^n \frac{\pi}{6}$$, where $$n = 0, 1, 2, \ldots$$

The smallest positive value gives:

$$3\omega = \frac{\pi}{6}$$

(We take the smallest value because $$t = 3$$ s is the first time the oscillator reaches $$A/2$$ after starting from the mean position.)

$$\omega = \frac{\pi}{18}$$ rad/s

The time period is:

$$T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi/18} = 2\pi \times \frac{18}{\pi} = 36 \text{ s}$$

We can verify: at $$t = 3$$ s, $$x = A\sin\left(\frac{\pi}{18} \times 3\right) = A\sin\left(\frac{\pi}{6}\right) = A \times \frac{1}{2} = \frac{A}{2}$$. This confirms our answer.

The correct answer is Option C.