A particle moves along X axis such that its acceleration is given by a = – $\mathrm{\beta }$(x – 2),where $\mathrm{\beta }$ is a positive constant and x is the position co-ordinate. 

The time period of oscillations is T and 
 the distance of equilibrium position from the origin of coordinate system is d . Then

Motion is simple harmonic. The origin of the co-ordinate system is not the equilibrium position for the particle. In equilibrium force on the particle shall be zero. Thus x = 2 is the equilibrium position.
$\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}=-\mathrm{\beta }(\mathrm{x}-2)$
Let (x – 2) = X
Then $\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}=\frac{{\mathrm{d}}^2\mathrm{X}}{{\mathrm{dt}}^2}$
Hence, $\frac{{\mathrm{d}}^2\mathrm{X}}{{\mathrm{dt}}^2}=-\mathrm{\beta X}$
$\therefore {\mathrm{\omega }}^2=\mathrm{\beta }\Rightarrow \mathrm{T}=2\mathrm{\pi }\sqrt{\frac{1}{\mathrm{\beta }}}$