$A$ 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

The displacement of an elastic wave is given by the function $y = 3 \sin \omega t + 4 \cos \omega t$,where $y$ is in $cm$ and $t$ is in $s$. Calculate the resultant amplitude. Also,find the initial phase (epoch).

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

The equations of two waves acting in perpendicular directions are given as $x=a \cos (\omega t+\delta)$ and $y=a \cos (\omega t+\alpha)$,where $\delta=\alpha+\frac{\pi}{2}$. The resultant wave represents:

Two simple harmonic motions are represented as $y_1 = 10 \sin \omega t$ and $y_2 = 10 \sin \omega t + 5 \cos \omega t$. The ratio of the amplitudes of $y_1$ and $y_2$ is

Consider two SHMs along the same straight line $x_1=A_1 \sin \left(\omega t+\phi_1\right)$ and $x_2=A_2 \sin \left(\omega t+\phi_2\right)$,where $A_1$ and $A_2$ are their amplitudes and $\phi_1$ and $\phi_2$ are their initial phase angles. If $R$ is the resultant amplitude,match the conditions in Column-$I$ with the resultant amplitudes in Column-$II$:

Column-$I$	Column-$II$
$A$. $A_1=A_2=A, \delta=0$	$I$. $A_1+A_2$
$B$. $A_1 \neq A_2, \delta=0$	$II$. $0$
$C$. $A_1=A_2=A, \delta=90^{\circ}$	$III$. $2A$
$D$. $A_1=A_2=A, \delta=180^{\circ}$	$IV$. $A\sqrt{2}$