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

Two particles executing simple harmonic motion as described by $y_1=30 \sin \left(2 \pi t+\frac{\pi}{3}\right)$ and $y_2=10(\sin 2 \pi t+\sqrt{3} \cos 2 \pi t)$ have amplitudes $A_1$ and $A_2$ respectively. The ratio $A_1: A_2$ 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-$II$
$A$. $A_1=A_2=A, \delta=0$	$I$. $A_1+A_2$
$B$. $A_1 \neq A_2, \delta=0$	$II$. $0$
$C$. $A_1=A_2=A, \delta=90^{\circ}$	$III$. $2A$
$D$. $A_1=A_2=A, \delta=180^{\circ}$	$IV$. $A\sqrt{2}$

The displacement of a particle in a periodic motion is given by $y = 4 \cos^{2}\left(\frac{t}{2}\right) \sin(1000 t)$. This displacement may be considered as the result of the superposition of $n$ independent harmonic oscillations. Here $n$ is

Two particles execute simple harmonic motion along the same straight line with the same amplitude and same frequency. The two particles pass one another when moving in opposite directions each time at a distance of $\frac{1}{\sqrt{2}}$ times the amplitude from their common mean position. The phase difference between the two particles is (in $^{\circ}$)

The amplitude of the vibrating particle due to the superposition of two $SHMs$,$y_1 = \sin \left( \omega t + \frac{\pi}{3} \right)$ and $y_2 = \sin \omega t$ is: