For a periodic motion represented by the equation $Y = \sin \omega t + \cos \omega t$,the amplitude of the motion 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-$II$
$A$. $A_1=A_2=A, \delta=0$	$I$. $A_1+A_2$
$B$. $A_1 \neq A_2, \delta=0$	$II$. $0$
$C$. $A_1=A_2=A, \delta=90^{\circ}$	$III$. $2A$
$D$. $A_1=A_2=A, \delta=180^{\circ}$	$IV$. $A\sqrt{2}$

Two particles are executing $SHM$ of the same amplitude $A$ and frequency $\omega$ along the $x$-axis. Their mean positions are separated by $X_0$ (where $X_0 > A$). If the maximum separation between them is $X_0 + 2A$,then the phase difference between their motion is:

Two simple harmonic motions are represented by the equations $y_{1} = 10 \sin(3 \pi t + \frac{\pi}{3})$ and $y_{2} = 5(\sin 3 \pi t + \sqrt{3} \cos 3 \pi t)$. The ratio of the amplitude of $y_{1}$ to $y_{2}$ is $x : 1$. The value of $x$ is ...... .

Two simple harmonic motions are represented by $y_1 = 5[\sin 2 \pi t + \sqrt{3} \cos 2 \pi t]$ and $y_2 = 5 \sin [2 \pi t + \frac{\pi}{4}]$. The ratio of their amplitudes is

$A$ particle executes two types of $SHM$. $x_1 = A_1 \sin \omega t$ and $x_2 = A_2 \sin [\omega t + \frac{\pi}{3}]$.
$(a)$ Find the displacement at time $t = 0$.
$(b)$ Find the maximum speed of the particle.
$(c)$ Find the maximum acceleration of the particle.