# Energy Conservation for a spring

1. Feb 28, 2013

### Momentous

1. The problem statement, all variables and given/known data
Consider a mass connected to a spring of sti ness k.
(a) Use conservation of energy to write down a first order differential equation obeyed
by the mass.
(b) Find the time t for the mass to move from the origin at t = 0 out to a position
x assuming that at t = 0 it has initial speed dx/dt = v0.
(c) Use the above to show that the period of motion is T = 2∏sqrt(m/k)

2. Relevant equations

Us = .5kx^2
K = .5mv^2

3. The attempt at a solution

a) I just set up a initial and final conservation of energy

I figured that I'd just set Us = K and make v = (dx/dt). That'd give me a natural log in my answer, which doesn't seem to make much sense.

Everything else doesn't seem to be too hard once I get part a.

2. Feb 28, 2013

### Simon Bridge

I'd just set Us = K and make v = (dx/dt)

That would be: $$\frac{dx}{dt} = \pm \sqrt{\frac{k}{m}}x$$ ... if you are looking for x(t) then the solution should be an exponential, not a log.

Notice that you can have a positive or a negative velocity for any given position.

3. Feb 28, 2013

### Momentous

Wouldn't that set up yield this...

dx/x = ±sqrt(k/m)*dt

and if you integrate...

ln(x) = ±sqrt(k/m)*t

I don't know much about eulers identity, but what if I took an exponential of that answer. Then it'd be

x = e^[±sqrt(k/m)*t]

which (and I know this part is wrong) I'd write as

x = cos[sqrt(k/m)*t]
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If it were second order, then I could get the trig function that I think is supposed to happen from this with an ODE.

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I feel like the problem lies in my first equation. Any thoughts on that?

Last edited: Feb 28, 2013
4. Feb 28, 2013

### Dick

Yes, the problem lies in the first equation. Energy is conserved means Us+K=C where C is some constant. Write that out then take the time derivative of both sides. That will give you a second order ode.

Last edited: Feb 28, 2013
5. Feb 28, 2013

### Simon Bridge

What Dick said - the clue is that you have not accounted for the $\pm$ ... didn't work: I should have just directed you to look again at what "conservation law" means ;)

6. Mar 1, 2013

### Momentous

The only thing I'm worried about is that the question specifically says to write down a first order ODE, but I guess it doesn't make much sense in this case.

So how do I take a time derivative of that? I'm a little confused because of the (dx/dt)^2
I'm guessing I want kx + ma = 0

wouldn't the chain rule get

d/dt(.5kx^2 + .5m(dx/dt)^2)) = kvx + mva = 0

Last edited: Mar 1, 2013
7. Mar 1, 2013

### Simon Bridge

... well done: cancel common factor?
Note: you have written down a 1st order ODE - this is the next step.

You could always have done: $$\frac{dx}{dt} = \pm \sqrt{c-\frac{kx^2}{m}}$$
... but you'd still have to figure how the $\pm$ should be handled.

Last edited: Mar 1, 2013