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All else being equal, is an optimal series of ideal springs stronger and more reliable then a single optimal and ideal spring of equal length?

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What do you consider a pro and what a con? Just asking for advantages and disadvantages doesn't make a good question , but if you specify what sort of actual characteristics you're looking for, this would be fine. –  David Z Mar 30 '12 at 7:28
    
Defining pro vs con is left as an exercise to the respondent. Better information will be collected by not dictating the criteria. Additionally, I am not qualified to make such a judgement. Hence, the question. –  user1066701 Mar 30 '12 at 9:15
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Without more specific criteria, though, this question is too vague to be appropriate for this site. You could get completely opposite answers that would be equally valid (or invalid), which is the kind of thing the Stack Exchange system is built to discourage –  David Z Mar 30 '12 at 9:47
    
Perhaps some assistance clarifying the question would be more appropriate then dismissing it. –  user1066701 Mar 30 '12 at 18:13
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Well, as I said: what constitutes an advantage and what a disadvantage? What criteria do you want to use to evaluate the two systems? What's the context of the question (i.e. what prompted you to ask it)? For example, are you trying to solve some larger problem with this? –  David Z Mar 30 '12 at 18:44
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closed as not constructive by David Z Mar 30 '12 at 9:43

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1 Answer

An ideal spring is characterized by its constant in Hooke's Law, $k$. When you start putting them in combination, the resultant system behaves as a spring with a different constant, I'll call $k'$. The combinatory laws and justifications for them can be established easily.

Series

The principle of an ideal spring is that when its end is displaced by change in position $\Delta x$ the force from the spring changes by $\Delta F = -k \Delta x$. If you attach one spring onto the end of another one, then for every $\Delta x$ the end of the spring system moves, each spring expands or contracts correspondingly by $\Delta x /2$. Then, since they are connected in series, the force maintained by both springs as well as the spring system are all equal.

Thus the new spring constant of the attached springs is $k'=k/n$ where $n$ is the number of springs connected.

Parallel

Here, all springs are perturbed by the same $\Delta x$ and the forces from all of the individual spring add to produce the force of the spring system. Thus, the new spring constant of the system will be $k'=n k$.

Answer

Thus, it would be better to have your ideal springs strung together in series if you desired a lower spring constant. There is also the fact that real springs will break at some terminal value of $\Delta x$ and this will be increased in the case that they are strung together, but the breaking force would be the same. Nonetheless, these are not real springs, they are ideal springs, so the only difference is that the spring constant is decreased according to the above equation and argument.

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I think he wanted the real-life issues -- breaking stress &c. A stronger spring is not necessarily better. From this POV, to me it seems that it doesn't make a difference, like you've said regarding breaking stress. But there may be extra factors. –  Manishearth Mar 30 '12 at 3:03
    
Manishearth is correct. Real world comparison is appreciated. –  user1066701 Mar 30 '12 at 5:34
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