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

The time period of a particle in simple harmonic motion is $8 \ s$. At $t=0$, it is at the mean position. The ratio of the distances travelled by it in the first and second seconds is

$A$ particle is executing simple harmonic motion with amplitude $A$. At a distance $x$ from the mean position,when the particle is moving towards the extreme position,it receives a blow in the direction of motion which instantaneously doubles its velocity. What is the new amplitude of the particle? (Frequency remains constant during the motion)

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 $cm/s$)

The maximum velocity of a particle performing simple harmonic motion is $6.28 \text{ cm s}^{-1}$. If the length of its path is $8 \text{ cm}$, then what is its period (in $\text{ s}$)?

$A$ particle executes a linear $S.H.M.$ In two of its positions,the velocities are $V_1$ and $V_2$,and the accelerations are $a_1$ and $a_2$ respectively $(0 < a_1 < a_2)$. The distance between the positions is