Equation of SHM

A particle is executing simple harmonic motion (S.H.M.). Its acceleration at a distance of $1 \mathrm{cm}$ from the mean position is $3 \mathrm{cm} {\mathrm{s}}^{-2}.$ If its velocity is $6 \mathrm{cm} {\mathrm{s}}^{-1}$ when it is at a distance of $2 \mathrm{cm}$ from its mean position, then the amplitude of S.H.M. is, 

A simple harmonic oscillator has an amplitude of $0.5 \mathrm{m}$ and a time period of $2$ seconds. What is the magnitude of acceleration when it is displaced from the mean position by $0.25 \mathrm{m}?$

Which of the following equation represents a simple harmonic motion? ($\omega$ is angular frequency, A is amplitude of oscillation and $i=\sqrt{-1}$)

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

A simple harmonic oscillator of frequency $1 \mathrm{Hz}$ has a phase of $1$ radian. By how much should the origin be shifted in time so as to make the phase of the oscillator vanish. (time in seconds).