$A$ $S.H.M.$ has amplitude $a$ and time period $T$. The maximum velocity will be

For a simple harmonic motion with amplitude $A$ and time period $T$,at what position is the velocity half of the maximum velocity?

$A$ particle executes simple harmonic motion with an amplitude of $5\, cm$. When the particle is at $4\, cm$ from the mean position,the magnitude of its velocity is equal to the magnitude of its acceleration. Then,its periodic time in seconds is

$A$ particle performing $S.H.M.$ starts from the equilibrium position and its time period is $12 \ s$. After $2 \ s$,its velocity is $\pi \ m/s$. The amplitude of the oscillation is: $[\sin 30^{\circ}=\cos 60^{\circ}=0.5, \sin 60^{\circ}=\cos 30^{\circ}=\sqrt{3}/2]$

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 starts $S.H.M.$ from the mean position as shown in the figure. Its amplitude is $A$ and its time period is $T$. At a certain instant,its speed is half of its maximum speed. What is its displacement at this instant?