In my book Advanced Acoustics there is a line-

A particle undergoing SHM is called a linear harmonic oscillator

If I say that the word linear is used for the 2 reasons-

  1. The motion of the particle can be defined by a linear differential equation $\frac{d^2x}{dt^2}+\omega ^{2}x=0$
  2. Restoring force depends upon the first power of $x$(displacement)

But if we see the case of Damped oscillation the equation of motion is $\frac{d^2x}{dt^2}+2k\frac{dx}{dt}+\omega_{0} ^{2}x=0$ which is also a linear differential equation & the restoring force is dependent on first power of $x$

Now my question is why only Simple Harmonic Oscillator is termed as linear harmonic oscillator & Damped Oscillator not ?

  • 1
    $\begingroup$ I'm voting to close this question as off-topic because it is about vocabulary, not physics. $\endgroup$
    – garyp
    Sep 24, 2017 at 20:14
  • $\begingroup$ I think linear means here that the force depends linearly on the position. ($\rightarrow$ the energy depends quadratically). $\endgroup$
    – peterh
    Sep 25, 2017 at 2:31

2 Answers 2


For simplicity.

It can be called Damped Harmonic Oscillator or linear damped oscillator, see this paper called On the first integrals of linear damped oscillators. And when the damped oscillator is not linear, that's often explicitly stated, as in the paper On the amplitude decay of strongly non-linear damped oscillators.

There's actually a book called Damped Oscillations of Linear Systems (eprint).


A particle that oscillate is called an oscillator. Quite often it is not a physical particle but some mathematical quantity which oscillate in a similar manner. Such quantities are also called oscillator s or equivalent oscillators. If the potential energy V( x) is strictly parabolic then the oscillator is called linear harmonic oscillator. If the potential energy curve is not strictly parabolic (except near the minimum) then the oscillator is called non-linear. In cse of SHM PE curve is strictly parabolic but approximately in case of damped oscillatory motion. In such situation for concrete calculations we must include terms higher than quadratic in the Taylor's series expansion of V(x)

  • $\begingroup$ I believe that's not correct: the damping is not conservative, so how can you associate it with an influence over the potential $V(x)$? $\endgroup$
    – stafusa
    Sep 25, 2017 at 15:06

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.