Simple Harmonic Motion - Mass on a Spring

1. Feb 6, 2014

For a mass on a spring (vertical set up) undergoing SHM, we equate the restoring force, -kx, to -ω^2 x, coming to a conclusion that ω = $\sqrt{\frac{k}{m}}$. My question is, is the restoring force |mg - T| Where T is the tension in the spring? Because this seems to be the net force. I am used to equating tension to kx, not the net force.

2. Feb 6, 2014

nasu

The oscillation is around the equilibrium position.
In the equilibrium position the spring is stretched by an amount x_o, just enough for the elastic force to be equal to the weight.
So we have
kx_o=mg.

Now if we move it a little from this equilibrium position, let say by pulling it down, the spring will be stretched by an extra amount, x. x is measured from the equilibrium position.
So the elastic force will be
F=k(x_o+x)
and the net force will be
F_net=F-mg= k(x_o+x) -mg= kx_o+kx-mg = kx.
So the net force depends only on the displacement from the equilibrium position. And this is the restoring force.

3. Feb 6, 2014

sophiecentaur

That looks to be along the right lines to me, except why do you have a Modulus sign there? The restoring force needs to be given a sign to tell you which way it acts. At equilibrium, T = mg. and the equilibrium position is not directly related to the actual spring length. Oscillation is about this position and the tension will increase of decrease according to the 'restoring force'.
I can see your confusion but you just need to relate the physical situation to the maths describing it.