For a simple harmonic motion with a time period T , determine the...

Let the equation of simple harmonic motion starting from the mean position be $x = A \sin(\omega t)$.

Time taken to reach $x = A/2$ from the mean position is $t_1$:

$\dfrac{A}{2} = A \sin(\omega t_1) \Rightarrow \sin(\omega t_1) = \dfrac{1}{2} \Rightarrow \omega t_1 = \dfrac{\pi}{6}$

$t_1 = \dfrac{\pi}{6\omega} = \dfrac{\pi}{6} \times \dfrac{T}{2\pi} = \dfrac{T}{12}$

Time taken to reach $x = A$ from the mean position is $t_2$:

$A = A \sin(\omega t_2) \Rightarrow \sin(\omega t_2) = 1 \Rightarrow \omega t_2 = \dfrac{\pi}{2}$

$t_2 = \dfrac{\pi}{2\omega} = \dfrac{\pi}{2} \times \dfrac{T}{2\pi} = \dfrac{T}{4}$

Time required to increase the displacement from half the amplitude to the full amplitude is:

$\Delta t = t_2 - t_1 = \dfrac{T}{4} - \dfrac{T}{12} = \dfrac{3T - T}{12} = \dfrac{2T}{12} = \dfrac{T}{6}$