$A$ 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 is executing $SHM$ along a straight line. Its velocities at distances $x_1$ and $x_2$ from the mean position are $V_1$ and $V_2$ respectively. Its time period is

$A$ particle is performing simple harmonic motion with an amplitude of $0.06 \,m$ and a time period of $3.14 \,s$. The maximum velocity of the particle is . . . . . . $cm/s$.

$A$ simple pendulum performs $S.H.M$ about $x = 0$ with an amplitude $A$ and time-period $T$. The speed of the pendulum at $x = \frac{A}{2}$ will be

The velocity of a particle executing $SHM$ varies with displacement $(x)$ as $4v^2 = 50 - x^2$. The time period of oscillations is $x/7 \text{ s}$. The value of $x$ is (Take $\pi = 22/7$)

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