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:

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 simple harmonic motions are represented by the equations $x_{1}=5 \sin \left(2 \pi t+\frac{\pi}{4}\right)$ and $x_{2}=5 \sqrt{2}(\sin 2 \pi t+\cos 2 \pi t)$. The ratio of the amplitude of $x_{1}$ and $x_{2}$ is

If the displacement $y$ (in $cm$) of a particle executing simple harmonic motion is given by the equation $y = 5 \sin(3 \pi t) + 5 \sqrt{3} \cos(3 \pi t)$,then the amplitude of the particle 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: