The displacement of a particle executing simple harmonic motion is given by
$y = A_{0} + A \sin \omega t + B \cos \omega t$
Then the amplitude of its oscillation is given by

The angular frequency of motion whose equation is $4\frac{d^2y}{dt^2} + 9y = 0$ is ($y =$ displacement and $t =$ time).

$A$ 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 $= 0^{\circ}$)

Given below are two statements. One is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A) :$ Knowing initial position $x_0$ and initial momentum $p_0$ is enough to determine the position and momentum at any time $t$ for a simple harmonic motion with a given angular frequency $\omega$.
Reason $(R) :$ The amplitude and phase can be expressed in terms of $x_0$ and $p_0$. In the light of the above statements,choose the correct answer from the options given below $:$

When a particle in linear $S.H.M.$ completes two oscillations,its phase increases by

$A$ particle of mass $0.1 \ kg$ is executing simple harmonic motion of amplitude $0.1 \ m$. When the particle passes through the mean position,its kinetic energy is $8 \times 10^{-3} \ J$. If the initial phase is $45^{\circ}$,the equation of its motion is (Assume $x(t)$ as the position of the particle at time $t$)