1. Given data

• Time period of SHM: $T = 2\,\text{s}$
• Required displacement from mean position: $x = \dfrac{A}{2}$
• Time taken is given as $\dfrac{1}{a}$ s

We need to find $a$.

2. Write the SHM equation

Taking the particle to start from the mean position,

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

where

$\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi\ \text{rad/s}$

3. Use the condition $x = A/2$

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

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

The first time this happens is when

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

So,

$t = \frac{\pi/6}{\pi} = \frac{1}{6}\,\text{s}$

4. Compare with given form

Given,

$t = \frac{1}{a}\,\text{s}$

Thus,

$\frac{1}{a} = \frac{1}{6}$

which gives

$a = 6$

5. Final answer

$\boxed{6}$

The derived answer matches the stored correct answer.