# Period of mass M hanging vertically from a spring

1. Apr 9, 2008

### 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

2. Apr 9, 2008

### 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 $$T=2\pi \sqrt{\frac{M}{k}}$$

3. Apr 10, 2008

### yasar1967

thank you.

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