$A$ 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:

Two particles are executing $S.H.M.$ with the same amplitude of $20 \, cm$ and the 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 equal to

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 particles are executing simple harmonic motion of the same amplitude $A$ and frequency $\omega$ along the $x$-axis. Their mean positions are separated by a distance $X_0$ $(X_0 > A)$. If the maximum separation between them is $(X_0 + A)$,the phase difference between their motions is:

Two simple harmonic motions are represented by the equations ${y_1} = 0.1 \sin(100\pi t + \frac{\pi}{3})$ and ${y_2} = 0.1 \cos(\pi t)$. The phase difference of the velocity of particle $1$ with respect to the velocity of particle $2$ is

The displacement of an elastic wave is given by the function $y = 3 \sin \omega t + 4 \cos \omega t$,where $y$ is in $cm$ and $t$ is in $s$. Calculate the resultant amplitude. Also,find the initial phase (epoch).