What is the ratio of maximum acceleration to the maximum velocity of a simple harmonic oscillator?

$A$ particle executes simple harmonic motion with a time period $0.6 \,s$ and amplitude $10 \,cm$. Find the mean velocity of the particle over the time interval during which it travels a distance $5 \,cm$ starting from the equilibrium position.

The maximum velocity of a particle performing $S.H.M.$ is $V$. If the periodic time is made $\left(\frac{1}{3}\right)^{rd}$ of its original value and the amplitude is doubled,then the new maximum velocity of the particle will be:

$A$ 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 is executing $SHM$ along a straight line. Its velocities at distances $x_1$ and $x_2$ from the mean position are $V_1$ and $V_2$ respectively. Its time period is

The instantaneous displacement of a particle in $S.H.M.$ is $x = A \cos \left(\omega t + \frac{\pi}{4}\right)$. The time at which the velocity is maximum for the first time is