The displacement of an object attached to a spring and executing simple harmonic motion is given by $$x = 2 \times 10^{-2} \cos \pi t$$ metres. The time at which the maximum speed first occurs is

The SHM is described by $$x = 2 \times 10^{-2} \cos \left( \pi t \right)$$ metres, which has the standard form $$x = A \cos \left( \omega t \right)$$.

Comparing, the amplitude is $$A = 2 \times 10^{-2}\,\text{m}$$ and the angular frequency is $$\omega = \pi \text{ rad s}^{-1}$$.

Velocity in SHM is the time derivative of displacement: $$v = \frac{dx}{dt} = -A \omega \sin \left( \omega t \right)$$.

The speed (magnitude of velocity) is $$|v| = A \omega \left| \sin \left( \omega t \right) \right|$$.
Its maximum value occurs when $$\left| \sin \left( \omega t \right) \right| = 1$$, i.e. when $$\omega t = \frac{\pi}{2} + n\pi$$ for integer $$n$$.

The first positive time after $$t = 0$$ is obtained by taking $$n = 0$$:
$$\omega t = \frac{\pi}{2} \;\;\Rightarrow\;\; t = \frac{\pi/2}{\omega} = \frac{\pi/2}{\pi} = 0.5 \text{ s}.$$

Therefore, the maximum speed is reached for the first time at $$0.5 \text{ s}$$.

Option A which is: 0.5 s