Why is a negative spring constant unrealistic in this system?

In summary: Therefore, in summary, the value of k1 is unrealistic because it results in a negative spring constant, which goes against the fundamental characteristics of springs and suggests that there may be external forces at play in the system.
  • #1
user977784078
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Homework Statement


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.

Homework Equations



[tex]F = \frac{2k_1k_2x}{2k_1+k_2}[/tex]
[tex]k_1 = \frac{F k_2}{2k_2x-2F}[/tex]

The Attempt at a Solution


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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?
 
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  • #2
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
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.
 

1. What is a negative spring constant?

A negative spring constant refers to a situation where the force exerted by a spring is in the opposite direction of its displacement. This means that as the spring is stretched or compressed, it exerts a force in the opposite direction, rather than in the same direction as the displacement.

2. How is a negative spring constant measured?

A negative spring constant can be measured by determining the displacement of the spring and the force it exerts in the opposite direction. The negative spring constant is then calculated by dividing the force by the displacement.

3. What are some examples of systems with a negative spring constant?

One example is a metal spring in a pogo stick. As the spring is compressed, it exerts a force in the opposite direction, propelling the rider upwards. Another example is a bungee cord, where the tension force is in the opposite direction of the cord's displacement.

4. What are the implications of a negative spring constant?

A negative spring constant can lead to unstable or oscillatory behavior in a system. It can also result in unexpected or unpredictable movements, as the force and displacement are not in the same direction.

5. Can a spring have a negative spring constant?

Yes, a spring can have a negative spring constant. This is typically observed in systems where the spring is stretched or compressed beyond its equilibrium position, causing the force to be in the opposite direction of the displacement.

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