Written by Sunny Lu.
Consider what happens when an object attached to two Hookean springs is released.
An object attached to two springs, each attached to a stationary wall.
We’ll first consider what happens if we assume that the stiffness coefficient of both springs are the same, e.g. k. Now, we can consider the force applied by each spring. We can let the equilibrium positions of the springs be x1 and x2 respectively, and the net force exerted on the block is -k(p-x1)-k(p-x2), where p is the position of the block. Simplifying, we get kx1+kx2-2kp. Clearly, if we wish to simplify the scenario to the case of one spring, then we will have -k’(p-x’), with stiffness k’ and equilibrium position a. Clearly, both kx1+kx2-2kx’ and -k’(p-x’) are linear functions, and therefore for them to be equivalent their slope and intercepts must be equal, and hence -k’=-2k and k’x’=kx1+kx2. Solving for k’ and x’ now yields k’=2k and P=(x1+x2)/2. Therefore, the net force applied by the two springs is equivalent to the force applied by another spring of double the stiffness, with its equilibrium point at the midpoint of the two original equilibrium points.
Now let us consider the scenario in which the original springs have different stiffness, k1 and k2. Using a similar logic, we can confirm that the net force exerted on the block is now:
which is equivalent to a spring of stiffness k1+k2 and equilibrium position (k1x1+k2x2)/(k1+k2), i.e. the weighted average of x1 and x2 with weights k1 and k2, which conforms to our intuition.
Extending this to more springs or the third dimension is left as an exercise to the reader.