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?

The initial phase of a body executing $SHM$ is $\frac{\pi}{4}$. What will be its phase at the end of $10$ oscillations?

$A$ particle executing $SHM$ of amplitude $4 \, cm$ and $T = 4 \, s$. The time taken by it to move from $+2 \, cm$ to $+2\sqrt{3} \, cm$ is

$A$ particle executes simple harmonic motion (amplitude $= A$) between $x = -A$ and $x = +A$. The time taken for it to go from $x = 0$ to $x = A/2$ is ${T_1}$ and to go from $x = A/2$ to $x = A$ is ${T_2}$. Then:

$A$ particle executes simple harmonic motion represented by the displacement function $x(t) = A \sin (\omega t + \phi)$. If the position and velocity of the particle at $t = 0 \, s$ are $2 \, cm$ and $2 \omega \, cm \, s^{-1}$ respectively,then its amplitude is $x \sqrt{2} \, cm$,where the value of $x$ is ..... .