An object is executing simple harmonic motion with an angular frequency $\omega$. If the maximum velocity is $v_{\max}$,then the maximum acceleration of the object 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 executing simple harmonic motion $\text{(S.H.M.).}$ Its acceleration at a distance of $1 \ cm$ from the mean position is $3 \ cm s^{-2}$. If its velocity is $6 \ cm s^{-1}$ when it is at a distance of $2 \ cm$ from its mean position, then the amplitude of $\text{S.H.M.}$ is,

The displacement of a particle executing $SHM$ is given by $y = 0.25 \sin(200t) \ cm$. The maximum speed of the particle is $......... \ cm \ s^{-1}$.

$A$ particle is executing $S.H.M.$ If its amplitude is $2 \, m$ and periodic time is $2 \, s$,then the maximum velocity of the particle will be:

$A$ particle performs $S.H.M.$ with amplitude $A$. Its speed is tripled at the instant when it is at a distance of $\frac{2A}{3}$ from the mean position. The new amplitude of the motion is