$A$ 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 executing simple harmonic motion. If the force acting on the particle at a position is $86.6 \%$ of the maximum force on it, then the ratio of its velocity at that point and its maximum velocity is

$A$ simple pendulum performs simple harmonic motion about $X = 0$ with an amplitude $A$ and time period $T$. The speed of the pendulum at $X = \frac{A}{2}$ will be

$A$ particle in $SHM$ is described by the displacement equation $x(t) = A\cos(\omega t + \theta)$. If the initial $(t = 0)$ position of the particle is $1 \, cm$ and its initial velocity is $\pi \, cm/s$,what is its amplitude? The angular frequency of the particle is $\pi \, s^{-1}$.

The graph in the figure represents:

When a particle executes $SHM$,the nature of the graphical representation of velocity as a function of displacement is: