Two simple harmonic motions $y_1 = A \sin \omega t$ and $y_2 = A \cos \omega t$ are superimposed on a particle of mass $m.$ The total mechanical energy of the particle is:

$A$ particle is subjected to two simple harmonic motions in the same direction having equal amplitudes and equal frequency. If the resultant amplitude is equal to the amplitude of the individual motion, the phase difference $(\delta)$ between the two motions is

If the displacement $y$ (in $cm$) of a particle executing simple harmonic motion is given by the equation $y = 5 \sin(3 \pi t) + 5 \sqrt{3} \cos(3 \pi t)$,then 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 mutually perpendicular simple harmonic vibrations have the same amplitude,frequency,and phase. When they superimpose,the resultant form of vibration will be:

Four simple harmonic vibrations:
$y_1 = 8 \cos(\omega t)$;
$y_2 = 4 \cos(\omega t + \frac{\pi}{2})$;
$y_3 = 2 \cos(\omega t + \pi)$;
$y_4 = 1 \cos(\omega t + \frac{3\pi}{2})$,
are superposed on each other. The resulting amplitude and phase are respectively: