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:
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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?
Best Answer
The displacement from equilibrium is not L-x but would be (L-x)-L That is the final position which is L-x (assuming your origin is on the left wall) minus the initial position which is L, the equilibrium position. Hence your net force is (k1+k2)x.