If a simple pendulum oscillates with an amplitude of $50 \,mm$ and time period of $2 \,s$, then its maximum velocity is (in $\,ms^{-1}$)

$A$ $1.00 \times 10^{-20} \,kg$ particle is vibrating under simple harmonic motion with a period of $1.00 \times 10^{-5} \,s$ and with a maximum speed of $1.00 \times 10^3 \,m/s$. The maximum displacement of the particle from the mean position is .......... $mm$.

$A$ 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,

$A$ particle of mass $5 \ g$ performs simple harmonic motion with an amplitude of $10 \ cm$. Its maximum velocity is $100 \ cm/s$. At what distance from the mean position will its velocity be $50 \ cm/s$?

The velocity-time diagram of a harmonic oscillator is shown in the figure. The frequency of oscillation is ..... $Hz$.

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