$A$ 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:

$A$ simple pendulum performs simple harmonic motion about $X = 0$ with an amplitude $A$ and time period $T$. The speed of the pendulum at $X = \frac{A}{2}$ will be

$A$ $S.H.M.$ has amplitude $a$ and time period $T$. The maximum velocity will be

$A$ simple pendulum performs simple harmonic motion about $x=0$ with an amplitude $a$ and time period $T$. The speed of the pendulum at $x=a/2$ will be

$A$ particle performs simple harmonic oscillation of period $T$ and the equation of motion is given by $x = a \sin(\omega t + \pi/6)$. After the elapse of what fraction of the time period will the velocity of the particle be equal to half of its maximum velocity?

$A$ particle is performing $SHM$ with a path length of $8 \ cm$. If it takes $1 \ s$ to reach from the mean position to the extreme position,what is its velocity at the equilibrium position?