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

Two particles execute $SHM$ of same amplitude of $20 \, cm$ with same period along the same line about the same equilibrium position. The maximum distance between the two is $20 \, cm$. Their phase difference in radians is

$A$ particle is subjected to two simple harmonic motions in the same direction having equal amplitudes and equal frequency. If the resultant amplitude is equal to the amplitude of the individual motion, the phase difference $(\delta)$ between the two motions is

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:

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