$A$ 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 performing $S.H.M.$ about its mean position with an amplitude $a$ and periodic time $T$. The speed of the particle when its displacement from the mean position is $\frac{a}{3}$ will be:

If the maximum velocity and maximum acceleration of a particle executing simple harmonic motion are respectively $5 \,m \,s^{-1}$ and $10 \,m \,s^{-2}$, then the time period of oscillation of the particle is

What is the maximum velocity of a simple harmonic motion with an amplitude of $3 \ cm$ and a time period of $6 \ s$?

$A$ particle in $SHM$ is described by the displacement equation $x(t) = A\cos(\omega t + \theta)$. If the initial $(t = 0)$ position of the particle is $1 \, cm$ and its initial velocity is $\pi \, cm/s$,what is its amplitude? The angular frequency of the particle is $\pi \, s^{-1}$.

$A$ 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