The maximum velocity of a body undergoing $S.H.M.$ is $0.2 \ m/s$ and its acceleration at $0.1 \ m$ from the mean position is $0.4 \ m/s^2$. The amplitude of the $S.H.M.$ is .... $m$

$A$ particle performs linear $S.H.M.$ At a particular instant,velocity of the particle is $u$ and acceleration is $\alpha$ while at another instant,velocity is $v$ and acceleration is $\beta$ $(0 < \alpha < \beta)$. The distance between the two positions is

$A$ 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 $\,cm$)

$A$ 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?

$A$ point mass of $400 \, g$ executes $S.H.M.$ under a force $F = -(10 \, N m^{-1}) x$. If it crosses the centre of oscillation with a speed of $10 \, m s^{-1}$, the amplitude of motion is (in $ \, m$)

$A$ particle is in simple harmonic motion. The equation of its motion is $x = 5 \sin \left( 4t - \frac{\pi}{6} \right)$,where $x$ is its displacement. If the displacement of the particle is $3$ units,then its velocity is: