Two particles are executing $SHM$ in a straight line. The amplitude $A$ and time period $T$ of both particles are equal. At time $t = 0$,one particle is at displacement $x_1 = +A$ and the other is at $x_2 = -A/2$,and they are approaching each other. The time after which they will cross each other is:

$A$ body of mass '$m$' performs linear $S$.$H$.$M$. given by the equation $x = P \sin \omega t + Q \sin \left(\omega t + \frac{\pi}{2}\right)$. The total energy of the particle at any instant is

Two particles are executing $SHM$ of the same amplitude $A$ and frequency $\omega$ along the $x$-axis. Their mean positions are separated by $X_0$ (where $X_0 > A$). If the maximum separation between them is $X_0 + 2A$,then the phase difference between their motion is:

The equations of two waves acting in perpendicular directions are given as $x=a \cos (\omega t+\delta)$ and $y=a \cos (\omega t+\alpha)$,where $\delta=\alpha+\frac{\pi}{2}$. The resultant wave represents:

Two particles are executing $S.H.M.$ with the same amplitude of $20 \, cm$ and the same period along the same line about the same equilibrium position. The maximum distance between the two is $20 \, cm$. Their phase difference in radians is equal to

Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to $A$ and $T$,respectively. At time $t=0$,one particle has displacement $A$ while the other one has displacement $\frac{-A}{2}$ and they are moving towards each other. If they cross each other at time $t$,then $t$ is