Given data

• Amplitude: $A = 8\text{ cm}$
• Time period: $T = 6\text{ s}$
• We need the time taken to move from maximum displacement to half the amplitude, i.e. from $x=A$ to $x=\frac{A}{2}$.

Use the SHM displacement equation Starting from maximum displacement, we can write $x = A\cos(\omega t)$ where $\omega = \frac{2\pi}{T} = \frac{2\pi}{6} = \frac{\pi}{3}\ \text{rad/s}$

Set the required position We want $x = \frac{A}{2}$ So, $A\cos(\omega t) = \frac{A}{2}$ $\cos(\omega t) = \frac{1}{2}$

Solve for time $\omega t = \frac{\pi}{3}$ since the particle starts from $x=A$ and moves toward the mean position, the first time this happens is at $\cos^{-1}(1/2)=\pi/3$.

Therefore, $t = \frac{\pi/3}{\omega} = \frac{\pi/3}{\pi/3} = 1\text{ s}$

Final answer The required time is $\boxed{1\text{ s}}$

Comparison with stored answer Stored correct answer: $1$

Our derived answer matches the stored answer.