$A$ particle executes $S.H.M.$ with a period of $6 \, s$ and amplitude of $3 \, cm$. Its maximum speed in $cm/s$ is

$A$ 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:

What is the phase difference between the velocity and displacement of a particle executing $SHM$ (in $^{\circ}$)?

For a particle executing simple harmonic motion $(SHM)$,at its mean position

$A$ 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):