$A$ simple harmonic oscillator has a period of $0.01 \, s$ and an amplitude of $0.2 \, m$. The magnitude of the velocity in $m \cdot s^{-1}$ at the centre of oscillation is:

$A$ particle executing simple harmonic motion has a maximum speed of $40 \,ms^{-1}$ and maximum acceleration of $60 \,ms^{-2}$. The period of oscillation 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 $cm/s$)

$A$ particle starts $S.H.M.$ from the mean position as shown in the figure. Its amplitude is $A$ and its time period is $T$. At a certain instant,its speed is half of its maximum speed. What is its displacement at this instant?

$A$ 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,

Two particles $P$ and $Q$ start from the origin and execute simple harmonic motion along the $X$-axis with the same amplitude but with periods $3 \ s$ and $6 \ s$, respectively. The ratio of the velocities of $P$ and $Q$ when they meet is