# Negative spring constant

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1. Oct 8, 2014

### user977784078

1. The problem statement, all variables and given/known data
A spring system consists of two parallel springs on top and a single spring connected to them on the bottom as shown:

http://i.imgur.com/hgMr12U.png

If the system is extended by 7 cm (i.e. x = 0.07m), F = 5N and k2 = 10 N/m, explain why the value of k1 is unrealistic.

2. Relevant equations

$$F = \frac{2k_1k_2x}{2k_1+k_2}$$
$$k_1 = \frac{F k_2}{2k_2x-2F}$$

3. The attempt at a solution

When I solve for k1 using the given values, I get k1 = -6, but I know that spring constant can't be negative. If spring constant is large and positive, it means the spring is stiff, if it is small, it means the spring is soft, and if it is zero, it means the spring is very soft that it doesn't counteract the force applied to it.

A negative spring constant makes no sense, so what is the correct explanation in this situation?

I have these explanations:

1. Since k2 is small, the bottom spring is soft, so the top springs don't experience any force since the bottom spring is very soft and counteracts the force applied.
2. Since k2 is small, the spring is denatured so the top springs aren't affected by the force.
3. A negative spring constant means that the spring doesn't counteract the force applied to it but instead it produces a force in the direction of the applied force and according to newton's 3rd law, this is not possible unless an external force acts on the spring.
Am I missing something here?

Last edited: Oct 8, 2014
2. Oct 8, 2014

### BvU

You try to explain something unrealistic. The question only asks to explain WHY it is unrealistic.
The 5N would lengthen the lower spring alone by 0.5 m already, so if 0.07 m is determined, the other two don't act as springs.

In your explanation 1 you would still measure an extension of at least 0.5 m.
I don't know what a denatured spring is, but if you mean completely deformed to a taut wire, that would mean the extension when a force of 5 N is applied is still >> 0.5 m , so reasoning number 2 doesn't fly either.
Reasoning 3 is more sensible: if the observation is not some stupid error (0.7 m extension instead of 0.07 m, for example...) but a real measurement, the top two contract by 0.43 m when a force of 5N is applied. So they form some kind of mechanism, but definitely not a spring.

3. Oct 8, 2014

### user977784078

Thank you very much for the great answer BvU. I realize now that I can't explain why the value is negative because it is unrealistic but that I can explain why it is unrealistic.