newtonian mechanics - Why does the acceleration $g$ due to gravity not affect the period of a vertically mounted spring?

For a vertically mounted spring, I was looking at the formula $T= 2\pi \sqrt{m/k}$ for a period. Why doesn't the gravitational acceleration $g$ factor in?

Answer

The effect of gravity is only to shift the equilibrium point, so at equilibrium (at rest), a vertical spring will be extended as compared with the same spring in a horizontal position. But this does not affect the period.

The equation for the dynamics of the spring is $m\frac{d^2x}{dt^2}=-kx+mg$. You can change the variable $x$ to $x'=x+mg/k$ and get $m\frac{d^2x'}{dt^2}=-kx'$. So the dynamics is equivalent to that of spring with the same constant but with the equilibrium point shifted by a distance $mg/k$

Update:

when you replace $x$ in you equation you have $x=x'-mg/k$ so you get $m\frac{d^2(x'-mg/k)}{dt^2}=-k(x'-mg/k)+mg$

On the left side you have $m\frac{d^2(x'-mg/k)}{dt^2}=m\frac{d^2x'}{dt^2}$ because the derivative of a constant ($mg/k$) is zero, and on the right side you get $-kx'$ after distributing.