Period of mass M hanging vertically from a spring

[yasar1967]

What is the period of this motion: a mass M hanging vertically from a spring, of spring constant k, under the influence of gravity.



I figured:
Mg-ky=Ma=Mw^2y

(w being angular freguency)

I know the solution(T=2pi * seqroot(M/k)) but how can I get there?? how can I eliminate Mg? Mg=k*ʌy equation leads to nowhere

 

[jae05]

well say we have the system at rest, then Mg=kx_0, so say now we define our equilibrium point to be L+x_0 where x_0 is the initial extension, and L is the natural length. introducing a small displacement, the spring will then oscillate about L+x_0 under the restoring force -kx. then we see that \ddot{x}=-\frac{k}{M}x. this then gives us \omega=\sqrt{\frac{k}{M}} and thus T=2\pi \sqrt{\frac{M}{k}}

 

[yasar1967]

thank you.