The resultant of two rectangular simple harmonic motions of the same frequency and unequal amplitudes but differing in phase by $\frac{\pi}{2}$ 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 simple harmonic motions $y_1 = A \sin \omega t$ and $y_2 = A \cos \omega t$ are superimposed on a particle of mass $m.$ The total mechanical energy of the particle is:

Two particles execute simple harmonic motion along the same straight line with the same amplitude and same frequency. The two particles pass one another when moving in opposite directions each time at a distance of $\frac{1}{\sqrt{2}}$ times the amplitude from their common mean position. The phase difference between the two particles is (in $^{\circ}$)

Two particles $P$ and $Q$ perform $S.H.M.$ of same amplitude $a$ and frequency along the same straight line. At a particular instant,the maximum distance between the two particles is $\sqrt{2} a$. The initial phase difference between them is

The amplitude of the vibrating particle due to the superposition of two $SHMs$,$y_1 = \sin \left( \omega t + \frac{\pi}{3} \right)$ and $y_2 = \sin \omega t$ is: