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StephenPrivitera
Oct8-03, 10:51 PM
Two springs are in series. You know the constants for each individual spring. How do you find the spring constant for the spring system?
|-OOOOO---o---OOOOOOOOOO-| --->F
In the middle of the spring system (where the "o" is) the net force is zero and we have
k1x1=k2x2 disregarding sign
F=k3x3
x3 =x2+x1
F=k3(x2+x1)
F=k3(k1x1/k2+x1)
F=k3*x1/k2*(k1+k2)
but F=k1x1+k2x2
F=2k1x1
2k1x1=k3*x1/k2*(k1+k2)
2k1k2=k3(k1+k2)
k3=2k1k2/(k1+k2)
Is this right?
Looks OK.
Now try to find a similar expression for N springs..:)
Integral
Oct9-03, 10:16 AM
Looks to me like you have an extra factor of 2 in there. Springs in series add as Resistors in parallel. So
1/KT = 1/K1+ 1/K2
OR
KT= (K1+K2)/(K1K2)
Also Springs in parallel add as resistors in series.
KT=K1+K2
I do not have time to do the derivation now. Should be able to post it later to day. A key is that in the series case the force seen by each spring is equal, while in the parallel case the extensions are equal.
StephenPrivitera
Oct9-03, 01:12 PM
Originally posted by Integral
Looks to me like you have an extra factor of 2 in there. Springs in series add as Resistors in parallel. So
1/KT = 1/K1+ 1/K2
Ah, I see my mistake. I have F=k1x1+k2x2 but F=k2x1=k1x1. I need to be more careful.
Integral
Oct9-03, 06:15 PM
OK, that looks better!
Here (http://home.comcast.net/~rossgr1/Math/Spring_constant.PDF) is my derivation. As long as I was at it I did the case for parallel springs also.
BTW: it would be trivial to extend this to multiple springs.
StephenPrivitera
Oct9-03, 10:33 PM
Thank you, Integral. I have put your work into my physics notebook (hope you don't mind).
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