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

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

The $S.H.M.$ of a particle is given by the equation $y = 3\sin \omega t + 4\cos \omega t$. The amplitude 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:

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.

For a periodic motion represented by the equation $Y = \sin \omega t + \cos \omega t$,the amplitude of the motion is: