$A$ particle of mass $0.4 \text{ kg}$ executes simple harmonic motion of amplitude $0.4 \text{ m}$. When it passes through the mean position,its kinetic energy is $256 \times 10^{-3} \text{ J}$. If the initial phase of the oscillation is $\pi / 4$,then the equation of its motion is

$A$ point particle of mass $0.1 \ kg$ is executing $S.H.M.$ with an amplitude of $0.1 \ m$. When the particle passes through the mean position,its kinetic energy is $8 \times 10^{-3} \ J$. Obtain the equation of motion of this particle if the initial phase of oscillation is $45^{\circ}$.

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

Draw a graph of displacement versus time as a function of time in simple harmonic motion.

The motion of a particle as per $x = A \sin \omega t + B \cos \omega t$ is :-