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$$ΔE_e=\frac{1}{2}k(x_f^2-x_i^2)$$
Where $k$ is the spring constant and $x$ is the displacement from equilibrium position.

Are we allowed to select an arbitrary reference level of elastic potential energy? As in, does elastic potential energy have to be strictly $0$ at the point of equilibrium? I think it does, because the equation involves the difference in squares of the displacement. If the equation only involves the change in displacement, then it would be like gravitational energy, where an arbitrary reference point can be made.

(sorry if the question seems simple, i can't find any online resources on this)

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3 Answers 3

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If I understand the question correctly, you are confused because the terms inside the bracket are non-linear.

However, keep in mind that your left hand side measures the change in potential energy. Suppose, you choose a $E_0$ as an arbitrary reference level energy,

$$\Delta E_e \equiv E_f-E_i = (E_f - \color{red}{E_0}) - (E_i - \color{red}{E_0}).$$

Hence, it does not really matter what energy you assign for the equilibrium state.

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Yes. Energy (in classical mechanics) is a quantity that only makes sense for a given reference frame, you measure differences in energy with respect to what 0 energy means to your reference frame. You could indeed shift the total energy of the spring measured in your reference frame as $E_m \mapsto E_m + E_0$. But the key on this is that the equations of motions are invariant. You could have defined your energy to be positive at the equilibrium point but in any case you won't be able to extract it.

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Yes, it is a matter of convenience.

An example is a vertical spring of length $L_0$ with a hook at one end, and the other end fixed on the ceiling.

If we put a weight on it, the new length will be $L_1$. The work done: $\int_{L_0}^{L_1}Fdx = \int_{L_0}^{L_1}kxdx = (k/2)(L_1^2 - L_0^2)$

That can be taken as the elastic energy stored by the spring, even when we know that without the weight of the hook, the length of equilibrium is smaller than $L_0$.

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