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A series of $n$ springs, where the $i$-th spring has spring constant $k_i$, is said to be equivalent to one single spring with spring constant $K$ if $1/K = \sum_{i=1}^n 1/k_i$.

But what does equivalent mean in this case?
It's derived only in the equilibrium state with one external force acting at the end.

Would this "equivalent spring" show the same behaviour like the series of springs while oscillating? (non-equilibrium state)

Under which assumptions does this equivalence hold?

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Basically what it's saying is that if you pull this series of springs out by a distance $\Delta x$, then the restoring force is $-K \cdot \Delta x$. From this fact, all the equations related to SHM follow. $T=2\pi \sqrt{\frac{m}{k}}$, $U_s = \frac{kx^2}{2}$, etc. Therefore, you could imagine this whole complicated arrangements of springs as a simple spring with spring constant $K$. I don't think this "equivalence" works if the springs are not ideal or the springs have nonnegligible mass.

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  • $\begingroup$ Does this mean the end of the series of springs would oscillate the same as the end of a normal spring? And if so, why? I have no way of explaining this. $\endgroup$
    – Nico G.
    Commented Jan 12, 2022 at 12:39
  • $\begingroup$ TL;DR - yes. Since we are dealing with ideal springs, the springs can only oscillate if you put a mass on the end of it (ideal springs have no mass and forces cannot be exerted on massless objects). So, let's imagine we put a mass on the end of the series of springs. Then, the restoring force is given as $-Kx$. Because of this fact, we can derive that the position of the mass, $x(t)$, is given as $x(t)=A\cos (\sqrt{\frac{K}{m}}\cdot t)$ However, this is just the same as if it were a regular spring with spring constant $K$! The masses in each setup would have the same position vs time graph. $\endgroup$
    – Eli Yablon
    Commented Jan 13, 2022 at 3:19

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