# Very confused about effective spring constant

I know that for springs in parallel, the effective spring constant is $k_1+k_2$ and for springs in series the constant is $1/(1/k_1+1/k_2)$. But there are some weird problems where finding the effective spring constant of some combined springs is not so obvious. For example:

|---------M----------|

where the you have two springs, each with one end at a wall, and one end attached to some mass $M$. The left spring has constant $k_1$, the right spring has constant $k_2>k_1$, both springs have unstretched legnth $L$, and say at equilibrium they are both stretched to $2L$.

If you displace the system x to the left, the left spring has displacement $L-x$ from its natural length, and the right spring has displacement $L+x$. So the forces exerted by each spring are

$F_1=k_1(L-x)$, $F_2=k_2(L+x)$ and the total force is the sum

$F_1+F_2=L(k_1+k_2)+x(k_2-k_1)$

It looks like I did something wrong because this is not of the desired form; if the system acted like one spring, we would instead have $F1+F2=Kx$ where $K$ is the effective spring constant.

So what did I do wrong and how to I correctly derive the effective force constant?

• The right spring has a displacement $x - L$. Also, notice that if force is defined to be in the positive $x$ direction, the mass $M$ is actually attached to the left side of the right spring (in contrast to the left spring), so you have to add a minus sign to the force. Adding those forces yields $(k_1 + k_2)(L - x)$. – kristjan Jan 23 '15 at 19:44
• @kristjan I do believe the right spring has a displacement $x+L$, as correctly stated by the OP. I agree with the observation on the minus sign though. $F_1$ pulls the mass left, $F_2$ pulls it right, so for the overall force pulling leftwards you want $F_1-F_2$ – glS Jan 23 '15 at 19:55
• You are right about the signs. But even correcting for that, you get $F_2-F_1=L(k_1-k_2)+x(k_2+k_1)$ and this is still not of the desired form/ – Joshua Benabou Jan 23 '15 at 19:59
• Well, displacement was perhaps a wrong word, what I meant was stretch from equilibrium position, which actually gives the tension of the spring (the signs are different for the right spring if you use stretch instead of displacement). Note that the sum of stretches must be $0$, from which we derive the second stretch is $x - L$. Now you can apply $F_{sum} = F_1 - F_2$, as second force is pushing to the negative x direction. – kristjan Jan 23 '15 at 20:11
• Thats what I did. I corrected for the signs and you still don't get the desired result. – Joshua Benabou Jan 23 '15 at 20:33