$A$ particle is performing $S.H.M.$ with an amplitude $4 \,cm$. At the mean position, the velocity of the particle is $12 \,cm/s$. When the speed of the particle becomes $6 \,cm/s$, the distance of the particle from the mean position is:

What is the maximum velocity of a simple harmonic motion with an amplitude of $2 \ m$ and a time period of $2 \ s$?

$A$ particle executes $S.H.M.$ of period $\frac{2 \pi}{\sqrt{3}} \text{ s}$ along a straight line $4 \text{ cm}$ long. The displacement of the particle at which the velocity is numerically equal to the acceleration is (in $\text{ cm}$)

$Assertion :$ For a particle performing $SHM$,its speed decreases as it goes away from the mean position.
$Reason :$ In $SHM$,the acceleration is always opposite to the velocity of the particle.

$A$ spring hangs vertically from the ceiling and a mass is attached to its free end. When the mass is pulled down and released,it oscillates vertically with simple harmonic motion of period $T$. The variation with time $t$ of its distance from the ceiling is as shown. Which statement gives a correct deduction from this graph?

The equation of a particle executing simple harmonic motion is given by $x = \sin \pi (t + 1/3) \, m$. At $t = 1 \, s$,the speed of the particle will be .......... $cm \, s^{-1}$ (Given: $\pi = 3.14$)