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

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 particles executing simple harmonic motion as described by $y_1=30 \sin \left(2 \pi t+\frac{\pi}{3}\right)$ and $y_2=10(\sin 2 \pi t+\sqrt{3} \cos 2 \pi t)$ have amplitudes $A_1$ and $A_2$ respectively. The ratio $A_1: A_2$ is

$A$ particle is subjected to two mutually perpendicular simple harmonic motions such that its $x$ and $y$ coordinates are given by:
$x = 2 \sin \omega t$
$y = 2 \sin \left( \omega t + \frac{\pi}{4} \right)$
The path of the particle will be:

$A$ particle executes two types of $SHM$. $x_1 = A_1 \sin \omega t$ and $x_2 = A_2 \sin [\omega t + \frac{\pi}{3}]$.
$(a)$ Find the displacement at time $t = 0$.
$(b)$ Find the maximum speed of the particle.
$(c)$ Find the maximum acceleration of the particle.

Three simple harmonic motions in the same direction having the same amplitude $a$ and same period are superposed. If each differs in phase from the next by $45^\circ$,then: