$A$ 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

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:

When two displacements represented by $y_1 = a \sin(\omega t)$ and $y_2 = b \cos(\omega t)$ are superimposed,the motion is

Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to $A$ and $T$,respectively. At time $t=0$,one particle has displacement $A$ while the other one has displacement $\frac{-A}{2}$ and they are moving towards each other. If they cross each other at time $t$,then $t$ is

Two particles are executing simple harmonic motion of the same amplitude $A$ and frequency $\omega$ along the $x$-axis. Their mean positions are separated by a distance $X_0$ $(X_0 > A)$. If the maximum separation between them is $(X_0 + A)$,the phase difference between their motions is:

The equation of $SHM$ is given as:
$x = 3 \sin(20\pi t) + 4 \cos(20\pi t)$,
where $x$ is in $cm$ and $t$ is in $seconds$. The amplitude is ..... $cm$.