$A$ particle is executing $SHM$ with an amplitude $A$. The distance travelled by the particle in one time period is

$A$ point particle of mass $200 \text{ g}$ is executing $S.H.M.$ of amplitude $0.2 \text{ m}$. When the particle passes through the mean position,its kinetic energy is $16 \times 10^{-3} \text{ J}$. The equation of motion of this particle is (Initial phase of oscillation $= 0^{\circ}$)

Why is the periodic time $T = \frac{2\pi}{\omega}$ taken for the function $f(t) = A \cos \omega t$?

$A$ particle is executing simple harmonic motion with an amplitude $A$ and time period $T$. The displacement of the particle after $2T$ time from its initial position is

$Assertion :$ In simple harmonic motion,the motion is to and fro and periodic.
$Reason :$ Velocity of the particle $(v) = \omega \sqrt {A^2 - x^2}$ (where $x$ is the displacement and $A$ is the amplitude).

Out of the following functions representing the motion of a particle,which represent $SHM$?
$(A)\; y = \sin \omega t - \cos \omega t$
$(B)\; y = \sin^3 \omega t$
$(C)\; y = 5 \cos \left( \frac{3\pi}{4} - 3\omega t \right)$
$(D)\; y = 1 + \omega t + \omega^2 t^2$