Two particles $P$ and $Q$ perform $SHM$ with the same amplitude $a$ and frequency $v$ 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 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 resultant of two rectangular simple harmonic motions of the same frequency and unequal amplitudes but differing in phase by $\frac{\pi}{2}$ 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

Two simple harmonic motions are represented as $y_1 = 10 \sin \omega t$ and $y_2 = 10 \sin \omega t + 5 \cos \omega t$. The ratio of the amplitudes of $y_1$ and $y_2$ is

$A$ particle is subjected to two simple harmonic motions as:
$x_1 = \sqrt{7} \sin(5t) \ cm$
and $x_2 = 2\sqrt{7} \sin(5t + \frac{\pi}{3}) \ cm$
where $x$ is displacement and $t$ is time in seconds.
The maximum acceleration of the particle is $x \times 10^{-2} \ ms^{-2}$. The value of $x$ is