The velocity-time diagram of a harmonic oscillator is shown in the figure. The frequency of oscillation is ..... $Hz$.

The amplitude of an oscillating particle is $A$. When the velocity of the particle is one-third of its maximum velocity,determine the position of the particle.

The time period of a particle in simple harmonic motion is $8 \ s$. At $t=0$, it is at the mean position. The ratio of the distances travelled by it in the first and second seconds is

The displacement of a particle executing $S.H.M.$ is $x = a \sin (\omega t - \phi)$. The velocity of the particle at time $t = \frac{\phi}{\omega}$ is (given $\cos 0^{\circ} = 1$):

The velocity vector $v$ and displacement vector $x$ of a particle executing $SHM$ are related as $\frac{v dv}{dx} = -\omega^2 x$ with the initial condition $v = v_0$ at $x = 0$. The velocity $v$,when displacement is $x$,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):