# I In dynamics, do we assume springs always resist motion?

1. Aug 11, 2016

### Leo_m87

I'm making a vibration rig that represents a classic 2 mass, 3 spring system (See attached Image).
(To clarify the image - there will be no forcing frequency. m=m1=m2 and k=k1=k2=k3)

I'm using extension springs in the vibration rig. I plan to displace them in extension at an equilibrium point that will have them remain in tension throughout harmonic motion. The springs will therefore always be in tension and will always only exert a tension spring force on the mass. My question is:

In the equations of motion do we always assume the spring constant will exert force in both tension and compression? In other words, If compressed it pushes away and if stretched it pulls back. If so, how will this affect the system I've proposed above? A system where closing the spring it is still in tension and not opposing the motion.

Thanks for taking the time to read this. Any help is appreciated.

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2. Aug 11, 2016

### BvU

Hello Leo,

You should distinguish between real springs (with thick wires, so they have a minimum length. And with a finite length of the wire they are made up from) and 'ideal physics springs'. The latter theoretical springs do not oppose motion. Instead they oppose displacement, as is is well shown by the spring equation $F = -k\; x$.

Sometimes - such as for your rig - you want to optimize the situation to ensure this $F = -k\; x$ behaviour occurs in a range of x (positive and negative) that is as wide as is reasonable. Meaning you want to pre-stretch real springs somewhat.

3. Aug 11, 2016

### Staff: Mentor

It will just raise the force or lower the force. Lowering the force is the same as applying an additional force in the opposite direction. So, in short, you don't need both tension and compression of the spring to accomplish the same action.

4. Aug 11, 2016

### Leo_m87

Thanks for your reply! This was my intuitive sense of what would happen, but I managed to get confused anyway. Can you just confirm a little further....

If each spring was initially stretched by 10 mm (Apologies to anyone not using metric) there would be spring forces of +k(10) and -k(10) acting on each mass (equilibrium). Then, if the mass is displaced further to the right (+ve direction) by another 5 mm the forces now become +k(15) and -k(5).... Your point is that this has the same net effect as if the compressed spring pushed back ... in other words the other spring being in greater tension is producing the equivalent force?

And most importantly, this set-up won't affect the stiffness matrix that would be formed for the system pictured? Which, before this confusion, I was certain should be:
[2k, -k
-k, 2k]

5. Aug 11, 2016

### Leo_m87

I had a feeling that my use of terminology wouldn't be specific enough! It's easy to be too ambiguous, or use the wrong terms by accident! By motion I meant the motion of the mass during oscillation, which in turn would displace the spring.

My intention is to stretch the springs (real springs) through half of their range of motion, allowing a +/- of the maximum displacement amplitude. That way the spring won't reach it's limit either way and will always be in tension.

What was confusing me is whether each springs inability to resist compressive displacement would alter the stiffness matrix... but I think Chester has cleared that up for me below

6. Aug 11, 2016

### Staff: Mentor

Just so there is no question about it, let x1 and x2 represent the displacements of masses m1 and m2 relative to their initial equilibrium positions, respectfully. Please write down the force balances on m1 and m2.

7. Aug 11, 2016

### Leo_m87

To make my issue clearer I've take the example of a one mass, two spring system. (Attached image).

The first method in the image shows where my normal assumptions break down. The second method gets me the answer that I expect for this system, but when I apply this same approach of including x0 in a system with three springs I run into problems. The problem is the middle spring is extended by x0/2 on either side and I no longer seem to be getting suitable answers when I work through the algebra associated with that approach.

There's a mistake in my logic somewhere... any help would be great! Thanks

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8. Aug 11, 2016

### Staff: Mentor

I don't understand what you did, so I am going to do the force balances on the masses in the original figure myself. The initial tensions in the 3 springs are 10 k, and let $x_1$ represent the displacement of mass m1 to the right and let $x_2$ represent the displacement of mass 2 to the right. So, the force balance on mass 1 is:$$(10k+k(x_2-x_1))-(10k+kx_1)+F_1\sin\omega t=m\frac{d^2x_1}{dt^2}$$
The force balance on mass 2 is:$$(10k-kx_2)-(10k+k(x_2-x_1))+F_2\sin\omega t=m\frac{d^2x_2}{dt^2}$$

9. Aug 11, 2016

### BvU

Newton's law is $F = m\ddot x$ and not $F = - m\ddot x$.

Wouldn't be good if the spring on the right exercised $F = kx$. It should be $F = -kx$.
And for the left spring you have $-F = kx$ -- matter of choosing and using your coordinate system consistently.
For one of the springs x is the extension, for the other it's the compression ! As you work out correctly in the second picture, where yo do write $-kx$.
You make life difficult for yourself this way !

10. Aug 11, 2016

### Leo_m87

Thanks Chester, I completely understand now. The initial tension that you assigned 10k is what I was referring to as Kx0, if that helps you understand what I did in the last picture.

In any event, I now realise my mistake in tackling the 2 mass system - I had incorrectly halved the initial tension for the middle spring and didn't realise why this was incorrect until going over your reply.

Thanks a lot for your help!