1
$\begingroup$

It was stated during a physics lecture that a simple harmonic oscillation undergoing light damping would have a period that is slightly greater than it would be without damping but the new period would remain constant throughout. I don't understand it as it seems counter-intuitive that the period would remain constant. :(

$\endgroup$
6
  • 1
    $\begingroup$ If your intuition disagrees with the math, then your intuition is wrong! But the statement in the lecture is only true for some types of damping - for example when the damping force is proportional to velocity, which is the only situation that is considered in a first course in dynamics. $\endgroup$
    – alephzero
    Mar 8 '17 at 16:02
  • 1
    $\begingroup$ @alephzero "Don’t be too proud of this technological terror you’ve constructed" (your math model). Just because you've modeled a physical process with equations doesn't mean that it's right. In all likelihood you've missed some detail, made a bad assumption. Intuition can sometimes be a very good attribute. It leads us to question, challenges the status quo, and sometimes uncovers new discovery. Isn't science all about skepticism? $\endgroup$
    – docscience
    Mar 8 '17 at 16:08
  • $\begingroup$ Nonlinear systems are much more common than linear systems. Heed the nonlinearity of nature. $\endgroup$
    – docscience
    Mar 8 '17 at 16:10
  • $\begingroup$ @docscience I'm sorry but lightly-damped systems are very well understood, either as exact solutions or via perturbation theory. Maybe you'd care to expand you comment to an answer as I'm not sure what you're getting at. $\endgroup$ Mar 8 '17 at 16:20
  • 2
    $\begingroup$ Same question as Physical interpretation of time constancy in damped harmonic oscillator $\endgroup$ Mar 8 '17 at 17:34
2
$\begingroup$

The period stays constant because the solutions can be rescaled at will. So if you have some valid motion $\theta(t)$ describing the pendulum's angle $\theta$ as a function of time $t$, then you can scaled this motion up or down by any factor $s$ and still get a valid motion $s \theta(t)$. This scaling property follows from the niceness (specifically, linearity) of the underlying model. If you remove simplifying assumptions then you lose the scaling property.

Why does the scaling property mean the period is constant? Suppose you have a pendulum and release it from an angle $\theta_1$, then the pendulum will swing away and swing back to a maximum angle $\theta_2$ at some later time $T$. The motion after the time $T$ is the same as if you had simply released the pendulum from the angle $\theta_2$ But now you can use the scaling property to get the future motion of the pendulum. For example, the motion from $T$ to $2T$ must be $\dfrac{\theta_2}{\theta_1} \theta(t-T)$. In fact, this equation must hold for all $t>T$, therefore the pendulum must alway have zero velocity at integer multiples of $T$, and so it shouldn't be hard to believe that the period must be $T$ for all time.

$\endgroup$
1
  • $\begingroup$ This is a nice answer about the mathematical features of the model that are related to constant frequency, but it does not address the physics, or physical intuition, or limitations to the model. I'll admit that the question of the physical "why" is hard to address. $\endgroup$
    – garyp
    Feb 20 '18 at 14:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.