$A$ particle executing $S.H.M.$ with an amplitude of $4 \, cm$ and a time period $T = 4 \, s$. The time taken by it to move from the positive extreme position to half the amplitude is ..... $s$.

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

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 phase (at a time $t$) of a particle in simple harmonic motion tells:

$A$ 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

Two particles are oscillating along two close parallel straight lines side by side,with the same frequency and amplitudes. They pass each other,moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is