$A$ 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:

The equation of motion of a particle performing linear $S.H.M$ is $x=5 \sin \left[4 t-\frac{\pi}{6}\right]$,where $x$ is its displacement in $cm$. The velocity of the particle when its displacement is $3 \ cm$ is: (in $cm/s$)

The amplitude of an oscillating particle is $A$. When the velocity of the particle is one-third of its maximum velocity,determine the position of the particle.

$A$ particle is executing the motion $x = A \cos (\omega t - \theta)$. The maximum velocity of the particle is

$A$ body is performing simple harmonic motion with amplitude $A$ and time period $T$. The variation of its acceleration $(f)$ with time $(t)$ is shown in the figure. If at time $t$,the velocity of the body is $v$,which of the following graphs correctly represents the variation of velocity $(v)$ with time $(t)$?

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 $\text{ m/s}$)