Two particles undergo $SHM$ along parallel lines with the same time period $(T)$ and equal amplitudes $(A)$. At a particular instant,one particle is at its extreme position while the other is at its mean position. They move in the same direction. They will cross each other after a further time:

The displacement of a particle in a periodic motion is given by $y = 4 \cos^{2}\left(\frac{t}{2}\right) \sin(1000 t)$. This displacement may be considered as the result of the superposition of $n$ independent harmonic oscillations. Here $n$ is

Two mutually perpendicular simple harmonic vibrations have the same amplitude,frequency,and phase. When they superimpose,the resultant form of vibration will be:

The resultant of two rectangular simple harmonic motions of the same frequency and equal amplitudes but differing in phase by $\frac{\pi}{2}$ 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

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: