The displacement of a particle is given by $x = 3\sin(5\pi t) + 4\cos(5\pi t)$. The amplitude of the particle 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 $S.H.M.$ of a particle is given by the equation $y = 3\sin \omega t + 4\cos \omega t$. The amplitude is

Two particles are oscillating in $SHM$ along two very close parallel paths such that they have the same mean position. The equations of $SHM$ for the two particles are $x_1 = A \sin \omega t$ and $x_2 = A \sin(\omega t + \phi)$ respectively. If the maximum distance between them is $\frac{6A}{5}$,then $\phi$ is equal to ..... $^o$.

Two simple harmonic motions,as shown,are at right angles. They are combined to form Lissajous figures.
$x(t) = A \sin(at + \delta)$
$y(t) = B \sin(bt)$
Identify the correct match below.

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: