1. Write the SHM equation

Since the oscillator starts from the mean position, we can write displacement as

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

where:

• $A$ = amplitude
• $\omega$ = angular frequency
• $t$ = time

2. Use the given condition

After $3$ s, the displacement is half the amplitude:

$x = \frac{A}{2}$

So,

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

Dividing by $A$,

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

3. Find the possible angle

For

$\sin\theta = \frac{1}{2}$

possible values are:

$\theta = \frac{\pi}{6},\; \frac{5\pi}{6},\; \text{etc.}$

Taking the smallest positive value (normally used unless stated otherwise),

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

So,

$\omega = \frac{\pi}{18}$

4. Relate angular frequency to time period

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

Thus,

$\frac{2\pi}{T} = \frac{\pi}{18}$

Solving,

$T = 36\text{ s}$

5. Check options

• A: $6$ s
• B: $8$ s
• C: $12$ s
• D: $36$ s

Hence, the correct option is:

$\boxed{36\text{ s}}$

6. Comparison with stored correct answer

Stored correct answer = D

Derived answer = D

They match.