$A$ particle executes $SHM$ of amplitude $25\, cm$ and time period $3\, s$. What is the minimum time required for the particle to move between two points $12.5\, cm$ on either side of the mean position?

$A$ particle executes $SHM$ with amplitude $0.2 \,m$ and time period $24 \,s$. The time required for it to move from the mean position to a point $0.1 \,m$ from the mean position is (in $\,s$)

The time taken by a particle executing simple harmonic motion of period $T$ to move from the mean position to half the maximum displacement is

Two bodies performing $SHM$ 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 oscillates in a straight line simple harmonically with a period of $8 \ s$ and an amplitude of $4 \sqrt{2} \ m$. The particle starts from the mean position. The ratio of the distance travelled by it in the $1^{\text{st}}$ second of its motion to that in the $2^{\text{nd}}$ second is: $(\sin 45^{\circ} = 1 / \sqrt{2}, \sin \frac{\pi}{2} = 1)$

The equation of $S.H.M.$ is $y = a \sin(2\pi nt + \alpha)$,then its phase at time $t$ is: