Two particles are executing $S.H.M.$ of same amplitude and frequency along the same straight line path. They pass each other when going in opposite directions,each time their displacement is half of their amplitude. What is the phase difference between them?

$A$ particle in $S.H.M.$ is described by the displacement function $x(t) = a\cos (\omega t + \theta )$. If the initial $(t = 0)$ position of the particle is $1 \, cm$ and its initial velocity is $\pi \, cm/s$. The angular frequency of the particle is $\pi \, rad/s$,then its amplitude is

The periodic time of a particle doing simple harmonic motion is $4 \,s$. The time taken by it to go from its mean position to half the maximum displacement (amplitude) is

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

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?

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