From the given displacement-time graph of an oscillating particle,find the maximum velocity of the particle.

$A$ particle is in simple harmonic motion. The equation of its motion is $x = 5 \sin \left( 4t - \frac{\pi}{6} \right)$,where $x$ is its displacement. If the displacement of the particle is $3$ units,then its velocity is:

$A$ particle performs linear $SHM$. At a particular instant,the velocity of the particle is $u$ and the acceleration is $a_1$. At another instant,the velocity is $V$ and the acceleration is $a_2$ $(0 < a_1 < a_2)$. The distance between the two positions is:

$A$ particle is performing simple harmonic motion with angular frequency $\omega$ and amplitude $A$. If $a$ is acceleration and $v$ is speed at any instant,then the graph showing the correct variation between $v^2$ and $x^2$ is (where $x$ is displacement):

$A$ particle is performing simple harmonic motion with amplitude $A$ and angular velocity $\omega$. The ratio of maximum velocity to maximum acceleration is

The displacement of a particle executing $SHM$ is given by $x = 10 \sin \left(\omega t + \frac{\pi}{3}\right) \ m$. The time period of motion is $3.14 \ s$. The velocity of the particle at $t = 0$ is . . . . . . $m/s$.