$A$ particle executes simple harmonic motion. Its amplitude is $8 \,cm$ and time period is $6 \,s$. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude,is ............. $s$

The amplitude and the periodic time of a $S.H.M.$ are $5 \, cm$ and $6 \, s$ respectively. At a distance of $2.5 \, cm$ away from the mean position,the phase will be

$A$ particle executes $SHM$ with an amplitude of $20 \, cm$ and a time period of $12 \, s$. What is the minimum time required for it to move between two points $10 \, cm$ on either side of the mean position?

Two bodies performing $S.H.M.$ have the same amplitude and frequency. Their positions and directions of motion at a certain instant are as shown in the figure. The phase difference between them is

$A$ 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:

$A$ simple harmonic oscillator has an amplitude $A$ and time period $6 \pi \text{ s}$. Assuming the oscillation starts from its mean position,the time required by it to travel from $x=A$ to $x=\frac{\sqrt{3}}{2} A$ will be $\frac{\pi}{x} \text{ s}$,where $x=$ . . . . . . .