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

Two particles are executing $SHM$ in a straight line. The amplitude $A$ and time period $T$ of both particles are equal. At time $t = 0$,one particle is at displacement $x_1 = +A$ and the other is at $x_2 = -A/2$,and they are approaching each other. The time after which they will cross each other is:

Two particles are executing $SHM$ of the same amplitude $A$ and frequency $\omega$ along the $x$-axis. Their mean positions are separated by $X_0$ (where $X_0 > A$). If the maximum separation between them is $X_0 + 2A$,then the phase difference between their motion 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

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