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Derivations of Harmonic Oscillator Laws

  1. Oct 18, 2012 #1
    When people talk about harmonic oscillators it seems to me that they always assume either that the relationship of force and displacement is linear, or that it behaves in some sinusoidal fashion. Do you always have to assume one to be able to arrive at the other? Or is there something I'm missing? I feel like the starting point is always assuming something based on experiments; is that correct? Or is it just that the actual derivation of harmonic oscillators formulas are beyond the scope of the lectures I've been taking?
     
    Last edited: Oct 18, 2012
  2. jcsd
  3. Oct 18, 2012 #2
    You have highlighted a problem that always confused me.... are all oscillations sinusoidal?
    I have learned that oscillations arise whenever an object is displaced from its equilibrium positionand there is a RESTORING force acting back towards the equilibrium position.
    When analysing oscillations it makes sense to choose the simplest situation as the starting point and the simplest situation from a physics point of view is to assume that the restoring force depends on displacement and is proportional to displacement.
    i.e F = -kx this means that acceleration will be given by F = ma so ma = kx or a = -kx/m.
    If you then use the fact that a = dv/dt and v = dx/dt you will end up with the sinusoidal expressions for x, v and a. This is SIMPLE analysis and the corresponding motion is called SIMPLE Harmonic motion.
    If the oscillation arises from a relationship that is not F = -kx then it is not Simple harmonic motion but can be solved by combinations of sinusoidal oscillations.
    hope this helps
     
  4. Oct 18, 2012 #3
    We do not really assume anything, the simple harmonic oscillator is defined as a system where the restoring force is F=-kx. This is a second order ODE that has sinusoidal solutions. There are no actual simple harmonic oscillators, but we can use this model to approximate the behavior of some oscillatory systems in some region where the restoring force is approximately F=-kx.
     
  5. Oct 18, 2012 #4
    I am a physics student and naturally look at the relationship between forces and displacement (F = -kx being the simplest?)
    A mathematician could pose the question....what are the characteristics of motion for which displacement (x) varies with time periodically as x = ACos(ωt) or x = ASin(ωt)
    This analysis produces sinusoidal solutions for velocity and acceleration (this is the beauty of sine waves)....exactly the same as the physicists F = -kx analysis.
     
  6. Oct 18, 2012 #5

    AlephZero

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    No. They are only sinusoidal when the restoring force is proportional to the displacement.

    If the restoring force is a function of x (and nothing else) but it is NOT a linear function, the motion of the system is usually much more complicated. As well as a non-sinusoidal "shape" of oscillation, the oscillation frequency depends on the amplitude. In fact there amy be several different shapes (with different amplitudes) possible at the same frequency, or severa different shapes with different frequencies possible at the same amplitude.

    Google "Duffing equationr" for the next-simplest example of what can happen, if the force = ##kx + ax^3##
     
  7. Oct 18, 2012 #6
    Aleph zero.... I did not make it clear enough !!!! I meant that SHM only occurs when F = -kx, ie force is proportional to displacement.
    Oscillations (in general) occur when there is an F that depends on displacement in some way.
    I think my wording is clear enough that sinusoidal oscillationsoccur when Force is proportional to displacement.
    My last sentence also makes it clear that if F is not proportional to displacement then the resulting oscillation is NOT SHM
     
  8. Oct 18, 2012 #7
    One important property of the SHM oscillator is that the system is conservative.

    Sinusoidal solutions lead to conservative fields and the existance of conservative potential fields.

    This allows the use of conservation of total energy (swapping incessantly between KE and PE)

    Non sinusoidal functions introduce the possibility of hysteresis and energy loss from the system at every cycle.
     
  9. Oct 18, 2012 #8

    sophiecentaur

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    This thread demonstrates a number of misconceptions and over-simplifications, due to inaccurate definitions that are often used in the teaching this subject (also due to inaccurate memory, too). If you stick to the strict definitions, there are no contradictions in any of this.

    If you take a mass, suspended on a system involving two springs of different stiffnesses, that will not necessarily involve hysteresis or energy loss but the resulting waveform will not be sinusoidal. Each issue needs to be considered separately.
    The only oscillator that will produce a sine wave is a 'simple' one in which the restoring force is proportional to displacement (a Simple Harmonic Oscillator) other 'Harmonic Oscillators' are not simple. There are no practical oscillators that are purely sinusoidal.
     
  10. Oct 18, 2012 #9
    Since you seem to have picked my post out for your ire, perhaps you would be good enough to demonstrate what you think is in error?

    Yes any compound oscillator is... well... more complicated.
     
  11. Oct 18, 2012 #10

    sophiecentaur

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    I am so sorry. I must be sounding more stroppy in my old age than I meant to. I intended to convey no 'ire'.

    I wrote "thread" and was not aiming at your contribution in particular, except in respect of the blue bit, which seems to imply more than is really there. Non linearity does not necessarily imply energy loss (hysteresis) - although it is often associated with it.

    Looking back at the thread, (and others), there is the general use of the term SHM to describe all oscillations and the assumption that all oscillations are sine waves. Also there are assumptions about the consequences of linearity and energy loss. I was just pointing this out. Generally, when people use the correct definitions and terms, there are fewer problems in understanding the subject and each other.
     
  12. Oct 18, 2012 #11
    Well since we are having a colourful day I have picked out one word in magenta from the blue bit.

    This clearly demonstrates that I appreciated your point and said the same thing so all is well.

    :approve:

     
  13. Oct 18, 2012 #12

    AlephZero

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    IMO It's not actually wrong (i.e. all the statements are true).

    But the first 3 points gives me the impression you are saying that SHM oscillators have those properties, but other types of oscillators do not. That implication isn't correct, and could be misleading.

    I'm not sure what the last point has to do with the price of fish in this thread, but in any case there isn't a one-to-one conrrespondence between hysteresis and energy loss (unless hysteresis means something different for you than it does for me).
     
  14. Oct 19, 2012 #13

    sophiecentaur

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    Here's another blue bit. I only know the word 'hysteresis' when it refers to an area that is enclosed by a stress/strain curve (or involving equivalent magnetic / electric quantities). That area will always represent energy loss because more work is done over one path than the return path. Are there some other uses that don't correspond to this?

    Thinks . . . I guess you could refer to the use of the term in electronics systems (Schmidt Triggers etc.) - but they are active and, by their nature, involve the transfer of energy and the term has been 'borrowed' to describe the shape of the transfer function.
     
  15. Oct 19, 2012 #14
    Force = mass times acceleration.

    It is possible to develop SHM using acceleration rather than the restoring force definition and this leads to a very simple derivation.

    However this needs very basic calculus (understanding what a derivative is)

    I do not know of any way to develop SHM without some calculus without simply defining it as sinusoidal motion.

    I will post it if there is the demand.

    The necessary calculus is understanding the relationships from kinematics that


    [tex]\begin{array}{l}
    velocity = v = \frac{{dx}}{{dt}} \\
    acceleration = f = \frac{{dv}}{{dt}} = \frac{{{d^2}x}}{{d{t^2}}} = v\frac{{dv}}{{dx}} \\
    {v^2} = {u^2} + 2fx \\
    \end{array}[/tex]
     
    Last edited: Oct 19, 2012
  16. Oct 19, 2012 #15

    sophiecentaur

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    I don't think you need to be apologetic for introducing calculus into this. :smile: It's as fundamental as arithmetic when discussing most matters in Science. There is a basic, entry level, requirement of knowledge and skill involved if people want to get involved in any field. Calculus, spelling and ability to read music are just examples of things which one can't do without for the fullest experience of the world.

    I had a girl friend who was subjected to a 'Non Mathematical" A level Physics course, back in the 60's. Disaster.
     
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