For a particle executing $S.H.M.$ having amplitude $A$,the speed of the particle is $\left(\frac{1}{3}\right)^{rd}$ of its maximum speed when the displacement from the mean position is

For a simple harmonic motion given by the equation $x = 5 \sin (4t - \frac{\pi}{6}) \, m$,what is the velocity when its displacement is $3 \, m$ (in $, m/s$)?

$A$ 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:

The force ($F$ in newton) acting on a particle of mass $90 \text{ g}$ executing simple harmonic motion is given by $F + 0.04 \pi^2 y = 0$, where $y$ is the displacement of the particle in meters. If the amplitude of the particle is $\frac{6}{\pi} \text{ m}$, then the maximum velocity of the particle is: (in $\text{ m/s}$)

$A$ 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):

$A$ spring executes $S$.$H$.$M$. with mass $10 \text{ kg}$ attached to it. The force constant of spring is $10 \text{ N/m}$. If at any instant its velocity is $40 \text{ cm/s}$, the displacement at that instant is (Amplitude of $S$.$H$.$M$. is $0.5 \text{ m}$) (in $\text{ m}$)