$A$ body is executing simple harmonic motion with an angular frequency $2\,rad/s$. The velocity of the body at $20\,mm$ displacement,when the amplitude of motion is $60\,mm$,is ...... $mm/s$.

$A$ 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

The velocity of a particle executing simple harmonic motion along the $x$-axis is described by the equation $v^2 = 50 - x^2$, where $x$ represents displacement. If the time period of the motion is $\frac{x}{7} \ \text{s}$, the value of $x$ is . . . . . . .

The velocity of a particle executing a simple harmonic motion is $13 \ m/s$, when its distance from the equilibrium position $(Q)$ is $3 \ m$ and its velocity is $12 \ m/s$, when it is $5 \ m$ away from $Q$. The frequency of the simple harmonic motion is

$A$ body is vibrating in simple harmonic motion with an amplitude of $0.06\, m$ and frequency of $15\, Hz$. The maximum velocity and maximum acceleration of the body are:

$A$ particle is performing $S.H.M.$ with an amplitude $4 \,cm$. At the mean position, the velocity of the particle is $12 \,cm/s$. When the speed of the particle becomes $6 \,cm/s$, the distance of the particle from the mean position is: