How do you derive the spring constant when two springs are in series?

2 Answers
Dwight
Oct 23, 2017

You add the reciprocals of the individual spring constants to get the reciprocal of the new combined spring constant.

Explanation:

The reciprocal of the new (effective) spring constant is found by adding the reciprocals of the constants for the two connected springs.

#1/k_(eff) = 1/k_1+1/k_2#

In this way, it is exactly like capacitors in series (which ought to be the case, as the role of a capacitor in a circuit is very similar to the role of a spring in a mechanical system).

So, if one spring had a constant of 6 N/m and the other was 3 n/m, the two in series would be

#1/k_(eff) = 1/6+1/3=1/2#

Such that #k_(eff) = 2N/m#.

Check this page for a derivation:

http://scienceworld.wolfram.com/physics/SpringsTwoSpringsinSeries.html

sjc
Oct 23, 2017

see below

Explanation:

Let the two springs have constants #k_1 , k_2#

and extensions #e_1,e_2#

for spring 1

#T_1=k_1e_1--(1)#

for spring 2

#T_2=k_2e_2--(2)#

the force on the springs is the same

so

#T_1=T_2#

#:.k_1e_1=k_2e_2#

#=>e_2=k_1/k_2e_1---(3)#

now overall

#T=k(e_1+e_2)#

where #k# is the spring constant overall

#:.T=k(e_1+k_1/k_2e_1)#

from #(3)#

#=>T=k((k_2+k_1)/k_2)e_1#

using #(1) in order to eliminate #e_1#

#k_1cancel(e_1)=k((k_2+k_1)/k_2)cancel(e_1)#

now rearrange

#k_1k_2=k(k_1+k_2)#

#=>k=(k_1k_2)/(k_1+k_2)#

#1/k=(k_1+k_2)/(k_1k_2)#

#1/k=1/k_1+1/k_2#

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