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

Two particles undergo $SHM$ along parallel lines with the same time period $(T)$ and equal amplitudes $(A)$. At a particular instant,one particle is at its extreme position while the other is at its mean position. They move in the same direction. They will cross each other after a further time:

Two particles are executing $S.H.M.$ with the same amplitude of $20 \, cm$ and the same period along the same line about the same equilibrium position. The maximum distance between the two is $20 \, cm$. Their phase difference in radians is equal to

The composition of two simple harmonic motions of equal periods at right angles to each other and with a phase difference of $\pi$ results in the displacement of the particle along

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

$A$ vibratory motion is represented by $x = 2A \cos \omega t + A \cos \left( \omega t + \frac{\pi}{2} \right) + A \cos ( \omega t + \pi ) + \frac{A}{2} \cos \left( \omega t + \frac{3\pi}{2} \right)$. The resultant amplitude of the motion is