In simple harmonic motion,the acceleration of a particle is given by $a = -bx$. What is the time period of the motion?

$A$ 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 $45^{\circ}$.

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

Write the difference between oscillation and vibrations.

The amplitude and the time period in a $S.H.M.$ are $0.5\, cm$ and $0.4\, s$ respectively. If the initial phase is $\pi/2$ radian,then the equation of $S.H.M.$ will be:

$A$ potential is given by $V(x) = k(x+a)^2 / 2$ for $x < 0$ and $V(x) = k(x-a)^2 / 2$ for $x > 0$. The schematic variation of the oscillation period $T$ for a particle performing periodic motion in this potential as a function of its energy $E$ is: