Two particles $p$ and $q$ perform $SHM$ with the same amplitude $a$ and the same frequency $f$ along a straight line. The maximum distance between the two particles is $a\sqrt{2}$. The initial phase difference between the particles is:

Four simple harmonic vibrations:
$y_1 = 8 \cos(\omega t)$;
$y_2 = 4 \cos(\omega t + \frac{\pi}{2})$;
$y_3 = 2 \cos(\omega t + \pi)$;
$y_4 = 1 \cos(\omega t + \frac{3\pi}{2})$,
are superposed on each other. The resulting amplitude and phase are respectively:

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

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

When two displacements represented by $y_1 = a \sin(\omega t)$ and $y_2 = b \cos(\omega t)$ are superimposed,the motion is

The $S.H.M.$ of a particle is given by the equation $y = 3\sin \omega t + 4\cos \omega t$. The amplitude is