$A$ particle performing $SHM$ has a time period of $\frac{2 \pi}{\sqrt{3}} \,s$ and a path length of $4 \,cm$. The displacement from the mean position at which the magnitude of acceleration is equal to the magnitude of velocity is (in $\,cm$)

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 linear $S.H.M.$ When the displacement of the particle from the mean position is $3 \ cm$ and $4 \ cm$,the corresponding velocities are $8 \ cm/s$ and $6 \ cm/s$ respectively. Its periodic time is

$A$ particle executes simple harmonic motion with amplitude $A$ and period $T$. If it is halfway between the mean position and the extreme position,then its speed at that point is:

$A$ particle is performing simple harmonic motion about $x = 0$ with an amplitude $a$ and periodic time $T$. The speed of the particle at $x = \frac{a}{3}$ will be

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