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

2 Answers
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

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