$A$ particle of mass $2 \, kg$ executing $SHM$ has an amplitude of $20 \, cm$ and a time period of $1 \, s$. Its maximum speed is ......... $m/s$.

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

$A$ particle is making simple harmonic motion along the $x$-axis. If at a distance $x_{1}$ and $x_{2}$ from the mean position the velocities of the particle are $v_{1}$ and $v_{2}$ respectively,the time period of its oscillation is given as -

The maximum velocity of a particle performing simple harmonic motion is $6.28 \text{ cm s}^{-1}$. If the length of its path is $8 \text{ cm}$, then what is its period (in $\text{ s}$)?

$A$ particle performing linear $S.H.M.$ of amplitude $0.1 \,m$ has displacement $0.02 \,m$ and acceleration $0.5 \,m/s^2$. The maximum velocity of the particle in $m/s$ is

$A$ 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