Newtons second law for coupled oscillators

[pinsky]

Hello there!

Could someone please help me with setting the starting equations for coupled oscillators. I'm having serious troubles with setting the +- signes right (yes, more than with the differental equations :) ). OFF TOPIC: any reading materials about problems with signs in physics will be appreciated.

So, I've drawn 3 situations here (no need to count the equillibrium), and I need help in writing Newtons second law just for the blue (or cyan if you like) body. No need to bother with solving the equations, i just need the things that go after ma=... ... .. for all three cases.

And could you please write it with all the plusses and minuses (even if, for example, -(-x) occures).

Thanks

[PLAIN]http://img683.imageshack.us/img683/3959/Newtonzacoupledoscillat.gif

 

[nasu]

It seems that you have the right directions for the elastic forces.
Now you just need to choose the positive direction.
Let say is positive to the right. Then for the first (non-equilibrium) situation, N's 2nd law will be something like
$$F_k - F_{k'} = ma$$
Is this what you are looking for?
Or you mean how to express the forces in terms of displacements?

 

[Delta2]

1. -k(-x)-(-k'(-y-(-x)))
2. -k(-x)-(-k'(y-(-x)))
3. -kx-(-k'(-y-x))

 

[AlephZero]

I think you are getting confused because you are labelling the diagrams with things like "-x". The displacement isn't "-x", it is just x, and x happens to be a negative number.

The same equation applies to all your diagrams

M d²x/dt² = -kx + k'(y-x)

It doesn't matter whether x and y are positive or negative.

 

[pinsky]

Thank you both for a fast replay.

So, in my textbook, we got the solutions.

ma1=-kx-k'(x-y) (for the first body)

ma2=-ky+k'(x-y) (for the second body)

a1 = d²x/dt²
a2 = d²y/dt²

and i just didn't see how they got it. Let's say situation 1. (in my picture) as Delta² wrote it.

We begin with:

F_k-F_k'=ma

F_k=-k(-x)

Becouse the x shift was in the negative direction.

I can't quite see the F_k'. If you could, please, explain that. I kind of got it as (x-y) and it made sort of sence. I explained it to myself as $$F=-\Delta x$$ where delta x is the relative change.


What bothers me the most about signs, is the two ways one can set the equations (that is sort of a citation from my professor). He always sais that it is better not to bother with the signs until the very end. I don't have a clue what that means.

For example, when we dealt with momentum, we always wrote

$$m_1 v_0 = m_1 v_1 + m_2 v_2$$

regardles of the direction.

 

[pinsky]

Thank you AlephZero.

I way writing the replay so i didn't see your post.

But can you explain how to observe the system to set the equation like that?

In which positions should the object be? What should be the relations of the forces?

 

[Delta2]

Dont bother if x or y are negative or positive, you know that whatever x is then the force F_k will be F_k=-kx.

F_k' is just abit more tricky. if z is the total displacement of the k' spring then simply F_k'=-k'z. BUT we have to be carefull how to calculate z. It turns out that z=x-y and that is because the y body is in the other end of the spring than the body x, so a y displacement towards that end is like a -y displacement towards the side of body x.
Total force is in any case F_k+F_k'=-kx-k'(x-y).

For body y similarly the total force will be -ky-k'z'. But z' here is y-x because we have to measure the total displacement from the end of where body y is lying. A displacement of x from the other end is like a displacement of -x from the end of y.

 

[pinsky]

Thank you delta.

I'll have to think this through a bit, and solve a few more examples, but that's basically the answer i was looking for, i just didn't know how to verbalise it.

 

[neelakash]

In fact, trying to solve these questions writing laws of mechanics can be extremely confusing. And the books somehow do not highlight this point. In your figure, I hope you are more confused about the signs for the middle spring.

you need to remember that the direction of tension at the two ends of the spring actually depends on the OVERALL stretch of this spring. Say, for this middle spring (2), $$l\rightarrow l+\delta$$. The tension force's direction depends on $$\delta$$ and according to that you need to write the equations. Notice that the sign of the stretch $$\delta$$ depends on spring 1 and spring 3.

You are free to take any convention: both x1 and x2 in the same direction / opposite direction / x1>x2 or x2>x1 etc. Mentally assume one such case; say x2>x1. Then you know the tensions on spring 1 and 3. Now, from the relations of $$\delta$$,x1 and x2, find out for your case if $$\delta$$ is +ve or -ve and accordingly assign tension on the two sides of spring 2.

It might be of help to do a single problem for all possible x1 and x2 combinations. That might give you the feel of the problem.