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

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

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:

Two particles execute simple harmonic motion $(SHM)$ along close parallel lines. Both particles have the same frequency and same amplitude. When they pass each other moving in opposite directions, their displacement is half their amplitude. Their phase difference 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

$A$ particle is subjected to two mutually perpendicular simple harmonic motions such that its $x$ and $y$ coordinates are given by:
$x = 2 \sin \omega t$
$y = 2 \sin \left( \omega t + \frac{\pi}{4} \right)$
The path of the particle will be: