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It is not difficult to show from Newton's second law $$m\ddot x = -kx - b\dot x $$ that an underdamped spring has the equation of motion quantified by $$x(t) = c_1e^{-\beta t}\sin\left(\omega t\right) + c_2e^{-\beta t}\cos\left(\omega t\right),$$ where $\omega$ is the oscillation frequency and $\beta$ is a damping frequency. However, $\omega$ is the result of several frequency components given by

$$\omega^2 = \sqrt{\omega_0^2 - \beta^2}.$$

Here, $\omega_0 = \sqrt{k/m}$ is the spring's natural frequency. For a constant $\beta$, the last equation implies that the system will oscillate forever at a single frequency $\omega$.

Intuitively, I feel that real springs will eventually stop oscillating. This would indicate there is some period decay in the system in addition to an amplitude decay. My questions, based on this, are:

  1. Does a spring ever stop, or does it just continue to oscillate at incredibly small amplitudes which are not easily detectable (as predicted by theory)?

  2. The damping frequency $\beta$ should depend on things like area, air density, etc. Is it really constant throughout a spring-mass' motion?

  3. How closely does a physical spring generally follow Hooke's law?

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  • $\begingroup$ Are you asking about the behaviour of real springs, or the behaviour of the mathematical model which you are using? $\endgroup$
    – sammy gerbil
    Jul 10, 2018 at 16:23
  • $\begingroup$ @sammygerbil I'm asking about the behavior of real springs, and if it contradicts the mathematical model. $\endgroup$
    – zh1
    Jul 10, 2018 at 16:24
  • $\begingroup$ Of course the real world (springs, projectiles, bending beams, refracting lenses, etc) departs from a mathematical model of it at some level of accuracy, because models are approximations not reality. It is impossible to specify how much real springs depart from this model, because there are different kinds of spring and different kinds of damping. And at some point oscillations are swamped by air currents or vibrations of the support. $\endgroup$
    – sammy gerbil
    Jul 10, 2018 at 16:37
  • $\begingroup$ Just to reinforce what @sammygerbil wrote, a spring modeled as having force linearly proportional to displacement and with damping linearly proportional to velocity is a spherical cow, or more precisely, a linear cow. Physicists like linear cattle at least as much as they like spherical ones because linear differential equations offer such nice, simple solutions. Real world springs oftentimes behave quite similar to the idealistic Hookian model when the oscillations are neither overly large nor ridiculously small. $\endgroup$
    – David Hammen
    Jul 10, 2018 at 17:13
  • $\begingroup$ Thanks all for your answers. I was hoping there were extended mathematical models to account for the faults you all have mentioned in these comments. I am quite surprised by the number of down votes on this question, and I would definitely reform it to be more useful - but I thought this was an interesting and valid question. $\endgroup$
    – zh1
    Jul 10, 2018 at 17:22

1 Answer 1

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It is easy to see using the solutions you have presented that the spring will continue to oscillate with expontentially small amplitudes for large times.

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  • $\begingroup$ I'm really looking for a physical answer. I understand the theory. I want to know the true behavior of these systems in the real world, considering its deviation from an ideal Hookean spring and (possibly) non-constant damping factor. $\endgroup$
    – zh1
    Jul 10, 2018 at 16:10
  • $\begingroup$ There are a number of directions you could go in. For example, the system could have non-linearities, which become important for large amplitude oscillations (including a pendulum for example). If these can be treated perturbatively, the system will oscillate with a frequency different from the one you have quoted above. Also, note that $\beta$ is not a frequency, $\omega$ is but only as long as $\omega_{0}>\beta$. $\endgroup$
    – 121
    Jul 10, 2018 at 16:29

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