$A$ particle executes a simple harmonic motion of time period $T$. Find the time taken by the particle to go directly from its mean position to half the amplitude.

What is the phase difference between the simple harmonic motions $x = a \sin(\omega t - \alpha)$ and $y = b \cos(\omega t - \alpha)$?

$A$ particle is executing simple harmonic motion. If the minimum time taken by the particle to move from extreme position to half of the amplitude is $t_1$,and the minimum time taken by the particle to move from mean position to half of the amplitude is $t_2$,then

The periodic time of a body executing simple harmonic motion is $3 \, s$. After how much interval from time $t = 0$,will its displacement be half of its amplitude?

The periodic time of a particle doing simple harmonic motion is $4 \,s$. The time taken by it to go from its mean position to half the maximum displacement (amplitude) is

$A$ particle executing linear $S$.$H$.$M$. has a period of $3 \ s$ and an amplitude of $6 \ cm$. The time required by it to travel a distance of $3 \ cm$ from the positive extreme position is:
$[\sin 30^{\circ} = \cos 60^{\circ} = \frac{1}{2}, \sin 60^{\circ} = \cos 30^{\circ} = \frac{\sqrt{3}}{2}]$ (in $s$)