A particle executes simple harmonic motion between x = – Aandx = +...

The position of a particle performing an SHM is given by:

$x = Asin ( \omegat + ϕ )$

Since at $t = 0s$, the particle is at $x = 0$, therefore the initial phase $ϕ = 0$.

At time $t = 2s$, the particle is at $x = \dfrac{A}{2}$.

$∴ \dfrac{A}{2} = Asin\left(2\omega\right) ⇒ \omega = \dfrac{\pi}{12}{s}^{- 1}$

Therefore, time period

$T = \dfrac{2\pi}{\omega} = 24s$

The total time taken from $x = 0$ to $x = A$ is $\dfrac{T}{4} = 6s$.

Since the time taken from $x = 0$ to $x = \dfrac{A}{2}$ is $2s$, therefore the time taken from $x = \dfrac{A}{2}$ to $x = A$ is $4s$.