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

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:

Two particles are executing $SHM$ in a straight line. The amplitude $A$ and time period $T$ of both particles are equal. At time $t = 0$,one particle is at displacement $x_1 = +A$ and the other is at $x_2 = -A/2$,and they are approaching each other. The time after which they will cross each other 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

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