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

$A$ particle is subjected to two mutually perpendicular simple harmonic motions such that its $x$ and $y$ coordinates are given by:
$x = 2 \sin \omega t$
$y = 2 \sin \left( \omega t + \frac{\pi}{4} \right)$
The path of the particle will be:

Two simple harmonic motions are given by $x_{1} = a \sin \omega t + a \cos \omega t$ and $x_{2} = a \sin \omega t + \frac{a}{\sqrt{3}} \cos \omega t$. The ratio of the amplitudes of the first and second motion and the phase difference between them are respectively:

$A$ 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

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:

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