Springs In Series

  • #1
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?
 

Answers and Replies

  • #2
36
0
Looks OK.
Now try to find a similar expression for N springs..:)
 
  • #3
Integral
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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.
 
  • #4
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.
 
  • #5
Integral
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Last edited by a moderator:
  • #6
Thank you, Integral. I have put your work into my physics notebook (hope you don't mind).
 

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