The displacement of a particle varies according to the relation $x = 4(\cos \pi t + \sin \pi t)$. The amplitude of the particle is

When two displacements represented by $y_1 = a \sin(\omega t)$ and $y_2 = b \cos(\omega t)$ are superimposed,the motion is

Two particles execute simple harmonic motion $(SHM)$ along close parallel lines. Both particles have the same frequency and same amplitude. When they pass each other moving in opposite directions, their displacement is half their amplitude. Their phase difference 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

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).

$A$ body of mass '$m$' performs linear $S$.$H$.$M$. given by the equation $x = P \sin \omega t + Q \sin \left(\omega t + \frac{\pi}{2}\right)$. The total energy of the particle at any instant is