$A$ particle is executing the motion $x = A \cos (\omega t - \theta)$. The maximum velocity of the particle is

The force ($F$ in newton) acting on a particle of mass $90 \text{ g}$ executing simple harmonic motion is given by $F + 0.04 \pi^2 y = 0$, where $y$ is the displacement of the particle in meters. If the amplitude of the particle is $\frac{6}{\pi} \text{ m}$, then the maximum velocity of the particle is: (in $\text{ m/s}$)

$A$ spring executes $SHM$ with a mass of $10\,kg$ attached to it. The force constant of the spring is $10\,N/m$. If at any instant its velocity is $40\,cm/s$,the displacement will be .... $m$ (where amplitude is $0.5\,m$).

What is the maximum velocity of a simple harmonic motion with amplitude '$a$' and time period '$T$ seconds'?

$A$ particle executing $S.H.M.$ starts from the mean position. Its amplitude is $A$ and time period is $T$. At what displacement is its speed one-fourth of the maximum speed?

The velocity vector $v$ and displacement vector $x$ of a particle executing $SHM$ are related as $\frac{v dv}{dx} = -\omega^2 x$ with the initial condition $v = v_0$ at $x = 0$. The velocity $v$,when displacement is $x$,is