$A$ particle of mass $250\,g$ executes a simple harmonic motion under a periodic force $F = (-25\,x)\,N$. The particle attains a maximum speed of $4\,m/s$ during its oscillation. The amplitude of the motion is $...........\,cm$.

$A$ $1.00 \times 10^{-20} \,kg$ particle is vibrating under simple harmonic motion with a period of $1.00 \times 10^{-5} \,s$ and with a maximum speed of $1.00 \times 10^3 \,m/s$. The maximum displacement of the particle from the mean position is .......... $mm$.

$A$ particle executes simple harmonic motion with an amplitude of $4 \, cm$. At the mean position,the velocity of the particle is $10 \, cm/s$. The distance of the particle from the mean position when its speed becomes $5 \, cm/s$ is

$A$ particle is executing a linear simple harmonic motion. Let $V_1$ and $V_2$ be its speeds at distances $x_1$ and $x_2$ from the equilibrium position,respectively. The amplitude of oscillation 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 $\,cm$)

The equation of motion of a particle performing linear $S.H.M$ is $x=5 \sin \left[4 t-\frac{\pi}{6}\right]$,where $x$ is its displacement in $cm$. The velocity of the particle when its displacement is $3 \ cm$ is: (in $cm/s$)