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 amplitude of the vibrating particle due to the superposition of two $SHMs$,$y_1 = \sin \left( \omega t + \frac{\pi}{3} \right)$ and $y_2 = \sin \omega t$ is:

Three simple harmonic motions of equal amplitudes $A$ and equal time periods in the same direction combine. The phase of the second motion is $60^{\circ}$ ahead of the first and the phase of the third motion is $60^{\circ}$ ahead of the second. Find the amplitude of the resultant motion.

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:

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:

Two particles execute simple harmonic motion along the same straight line with the same amplitude and same frequency. The two particles pass one another when moving in opposite directions each time at a distance of $\frac{1}{\sqrt{2}}$ times the amplitude from their common mean position. The phase difference between the two particles is (in $^{\circ}$)