$A$ simple harmonic oscillator has an amplitude $a$ and time period $T$. The time required by it to travel from $x = a$ to $x = a/2$ is

$A$ particle executes simple harmonic motion between $x = -A$ and $x = +A$. If time taken by particle to go from $x = 0$ to $x = \frac{A}{2}$ is $2 \text{ s}$, then time taken by particle in going from $x = \frac{A}{2}$ to $x = A$ is (in $\text{ s}$)

$A$ particle executes simple harmonic motion between $x = -A$ and $x = +A$. It starts from $x = 0$ and moves in the $+x$ direction. The time taken for it to move from $x = 0$ to $x = \frac{A}{2}$ is $T_1$,and the time taken to move from $x = \frac{A}{2}$ to $x = \frac{A}{\sqrt{2}}$ is $T_2$. Then:

The period of $S.H.M.$ of a particle is $16 \ s$. The phase difference between the positions at $t = 2 \ s$ and $t = 4 \ s$ will be

The displacement versus time curve for a particle executing $SHM$ is shown in the figure. Identify the points marked at which $(i)$ the velocity of the oscillator is zero,$(ii)$ the speed of the oscillator is maximum.

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?