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The Law of Mr Hooke and His Wacky Spring Constant

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Hello girls and boys,


I was wondering if someone could help me understand a seemingly simple concept. How is the spring constant in hooke's law (F=kx) affected by changes in length and/or addition of additional springs?


I've noticed sometimes MCAT problems address this issue, for example:




---- <-- let this be a spring

X <---let this be the mass





In the case above, what happens to the spring constant? Using my intuition, I managed to get the answer right (the spring constant becomes 2K), but I wanted to understand WHY this is the case... so that I can always apply it to different scenarios.



another example:


spring 1 spring 2

------- ------- X


Two springs are attached one after another. What happens now?



Are there any other variations of this to think about?


Thanks y'all. D-day in <3 weeks!

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I'm not sure I totally understand what you're indicating in the first scenario. I see two (I assume identical) springs attached to the same object from either end. Each spring applies the same resisting force (F = kx) to the object's displacement. It just happens that because you have two springs, the total resistive force is twice that (because they're identical)... I wouldn't call the spring constant itself doubled unless you take the whole system to be one entirely new spring with a new constant Keq (K-equivalent).


Having two springs with the object in the middle (assuming the object is moving in the same axis as the springs) is no different from this scenario:


----------[ big ]



(Pulling or pushing a spring results in the same *resistive* force F = kx... so two springs = 2*F = 2kx) or keq = 2k.


If you had one spring twice as strong as the other, your total k-eq would be 3x the weaker one. You basically just add up the k's for your k-eq.




The second scenario is actually quite a bit more tricky. First, you must know that a spring exerts the same resistive force on either end. Therefore, the force you apply with result in both springs compressing in relation to their own spring constant. That is to say:


F-applied = F-spring1 = k1x1 = F-spring2 = k2x2


Think about it intuitively: if you had a hard spring on the left (i.e. a high spring constant k1) and a soft spring (low spring constant k2) on the right... and you tried to push the object left, this is what you expect to happen:


spring 1 spring 2

========== ----------- X <- apply a Force


spring 1 spring 2

======== ----- X result


In other words, the high k1 results in a low x1 (displacement of hard spring) and a high x2 (displacement of soft spring). It doesn't matter the order.


In fact, this relationship can be seen from the above equation


F-spring1 = F-spring2, i.e. k1x1 = k2x2 OR:


k1/k2 = x2/x1 (the ratio of the spring constants is inversely proportional to the ratio of the displacements).


So what does that mean for the equivalent spring constant? Well, it's a little complicated, but if you test it out with some numbers (or derive it properly), you'll find that


1/k-eq = 1/k1 + 1/k2


The k-eq will always be less than the k of each spring


Just think, you are displacing the same amount for each spring as it would be normally displaced, so once you add up the displacements (x1 + x2), for the same amount of force you obviously have less k-eq.




Hopefully I have not confused you further! :)

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Thanks everyone for the help! I am doing quite a bit of reviewing right now, so will check out the links and read throgh estairella's lovely post soon. :)


I'm getting encouraged finally, on the AAMC tests I've scored a 12 or 11 on every one of my last 4 PS sections... hope this holds up when the real thing comes around.

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