$A$ particle is executing simple harmonic motion in one-dimension. If the amplitude of oscillations is $0.2 \,cm$ and if its velocity at the mean position is $5 \,m/s$, then the angular frequency of the oscillation 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 $\text{ m/s}$)

$A$ particle performs simple harmonic motion with amplitude $A$. Its speed is increased to three times at an instant when its displacement is $\frac{2A}{3}$. The new amplitude of motion is $\frac{nA}{3}$. The value of $n$ is . . . . . . .

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

The displacement of a particle executing $SHM$ is given by $x = 10 \sin \left(\omega t + \frac{\pi}{3}\right) \ m$. The time period of motion is $3.14 \ s$. The velocity of the particle at $t = 0$ is . . . . . . $m/s$.

$A$ particle executes $S.H.M.$ with a period of $6 \, s$ and amplitude of $3 \, cm$. Its maximum speed in $cm/s$ is