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

Two particles $P$ and $Q$ perform $S.H.M.$ of same amplitude $a$ and frequency along the same straight line. At a particular instant,the maximum distance between the two particles is $\sqrt{2} a$. The initial phase difference between them is

On the superposition of two harmonic oscillations represented by $x_1 = a \sin(\omega t + \phi_1)$ and $x_2 = a \sin(\omega t + \phi_2)$,a resulting oscillation with the same time period and amplitude is obtained. The value of $\phi_1 - \phi_2$ is .... $^o$

Four simple harmonic vibrations:
$y_1 = 8 \cos(\omega t)$;
$y_2 = 4 \cos(\omega t + \frac{\pi}{2})$;
$y_3 = 2 \cos(\omega t + \pi)$;
$y_4 = 1 \cos(\omega t + \frac{3\pi}{2})$,
are superposed on each other. The resulting amplitude and phase are respectively:

$A$ point mass is subjected to two simultaneous sinusoidal displacements in $x$-direction,$x_1(t) = A \sin \omega t$ and $x_2(t) = A \sin \left(\omega t + \frac{2 \pi}{3}\right)$. Adding a third sinusoidal displacement $x_3(t) = B \sin (\omega t + \phi)$ brings the mass to a complete rest. The values of $B$ and $\phi$ are

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).