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

$A$ vibratory motion is represented by $x = 2A \cos \omega t + A \cos \left( \omega t + \frac{\pi}{2} \right) + A \cos ( \omega t + \pi ) + \frac{A}{2} \cos \left( \omega t + \frac{3\pi}{2} \right)$. The resultant amplitude of the motion 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

The equations of two waves acting in perpendicular directions are given as $x=a \cos (\omega t+\delta)$ and $y=a \cos (\omega t+\alpha)$,where $\delta=\alpha+\frac{\pi}{2}$. The resultant wave represents:

Two simple harmonic motions are represented by $y_1 = 5[\sin 2 \pi t + \sqrt{3} \cos 2 \pi t]$ and $y_2 = 5 \sin [2 \pi t + \frac{\pi}{4}]$. The ratio of their amplitudes is

The displacement of a particle is given by $x = 3\sin(5\pi t) + 4\cos(5\pi t)$. The amplitude of the particle is