$A$ particle is performing $SHM$ with a path length of $8 \ cm$. If it takes $1 \ s$ to reach from the mean position to the extreme position,what is its velocity at the equilibrium position?

For a particle performing $S.H.M.$,which of the following statements is $INCORRECT$?

$A$ particle performs simple harmonic motion with amplitude $A$. Its speed is trebled at the instant that it is at a distance $\frac{2A}{3}$ from the equilibrium position. The new amplitude of the motion is

$A$ 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 $\sqrt{\alpha} \ cm$,where $\alpha = $ . . . . . . .

$A$ particle is performing simple harmonic motion with an amplitude of $0.06 \,m$ and a time period of $3.14 \,s$. The maximum velocity of the particle is . . . . . . $cm/s$.

$A$ particle executes simple harmonic motion with amplitude $A$ and period $T$. If it is halfway between the mean position and the extreme position,then its speed at that point is: