Homework Help: Frequency of Oscillation of a Mass on a Vertical Spring

1. Dec 10, 2012

cbasst

1. The problem statement, all variables and given/known data
A massless spring hangs from the ceiling with a small object attached to its lower end. The object is initially held at rest in a position yi such that the spring is at its rest length. The object is then released from yi and oscillates up and down, with its lowest position being 10 cm below yi. What is the frequency of the oscillation?

2. Relevant equations
T = $2\pi\sqrt{\frac{m}{k}}$
T = $\frac{2\pi}{\omega}$
F = -kx

3. The attempt at a solution
If I knew the mass and how much the spring was stretched only under the influence of gravity, it would be easy to find k. As it is, I'm not sure what relationship I need to use in order to find the period. Is there something I'm missing? Is there maybe a known (but not explicitly stated in the question) relationship between the force exerted by the spring and the force of gravity on the mass? I'm not really sure how to start this problem.

2. Dec 10, 2012

Delphi51

It would be nice to have the mass all right!
But you don't actually need the mass to find the frequency, just m/k according to your first equation. Can you find m/k?

3. Dec 10, 2012

cbasst

Well, I'm not sure how I would find m/k. I've considered that when the spring is not oscillating (it is at rest), then F=mg, so mg=-kx and then m/k=-x/g. But in this case it seems like I still have an unknown: x. As far as I can tell, I don't know how much the spring is stretched when the force of the spring is equal to the weight of the object, so having m/k=-x/g doesn't seem to be helping at the moment.

4. Dec 10, 2012

Delphi51

It seems to me that when it is in its lowest position (zero acceleration), mg is equal to kx. I would put that into the equation you have in your last post.

There is another spring equation for the energy stored in a spring: E = .½kx²
It probably won't help - it is derived from F = kx averaged over the stretch so not really new information.

5. Dec 10, 2012

cbasst

I thought that at the bottom of the oscillation then the acceleration would be non-zero pointing towards the equilibrium position but the velocity would be equal to zero. I think maybe I am imagining this oscillation as though it had been set up differently than it was. Let's say that the oscillation was set up as so: the mass was allowed to hang on the spring and it was not supported by any other forces. The spring's oscillations were stilled, so that now the mass is hanging by the spring but is not oscillating, thus mg=kx. Now the mass is given a small push upwards and it begins oscillating. Would the assumptions stated in the first sentence of this post be true then?

6. Dec 10, 2012

Delphi51

Oops, you are quite right - there is acceleration at the bottom. Back to the drawing board.
This implies an upward force to compensate for mg down, doesn't it?
Then it is released and falls down 10 cm where it momentarily stops. Can we then use mgx = ½kx² to find m/k?

7. Dec 11, 2012

cbasst

Aha! That worked very nicely. Thanks for the help!

8. Dec 11, 2012

Delphi51

Most welcome! Best of luck to you.