$A$ particle executing simple harmonic motion has a maximum acceleration of $24 \ m/s^2$ and a maximum velocity of $16 \ m/s$. What is its amplitude?

$A$ spring executes $S.H.M.$ with a mass of $1 \,kg$ attached to it. The force constant of the spring is $4 \,N/m$. If at any instant its velocity is $20 \,cm/s$, the displacement at that instant is (Amplitude of $S.H.M.$ is $0.4 \,m$)

$A$ 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):

When a particle executes simple harmonic motion,the nature of the graph of velocity as a function of displacement will be:

$A$ particle executing linear $S.H.M.$ has velocities $V_1$ and $V_2$ at distances $x_1$ and $x_2$ respectively, from the mean position. Its angular velocity is:

$A$ particle performs linear $S.H.M.$ At a particular instant,velocity of the particle is $u$ and acceleration is $\alpha$ while at another instant,velocity is $v$ and acceleration is $\beta$ $(0 < \alpha < \beta)$. The distance between the two positions is