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3rd Harmonic Content for a Simple Pendulum?

  1. Sep 10, 2004 #1

    cj

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    A simple pendulum of length l oscillates
    with an amplitude of 45°.

    What is the approximate amount of 3rd harmonic
    content in the oscillation of the pendulum?

    NOTE: the numerical answer is apparently 0.0032.
    I need to figure out how this was arrived at.

    [tex]\hline [/tex]
    As a starting point I'm using a power series:

    [tex] m \frac{d^2x}{dt^2}+kx = \epsilon(x) = \epsilon_2x^2 + \epsilon_3x^3 + ...[/tex]

    ... and looking at the cubic term, so that

    [tex] m \frac{d^2x}{dt^2}+kx = \epsilon_3x^3[/tex]

    Specifically, I'm told to use the trial solution:

    [tex]x = A cos \omega t + B cos 3 \omega t[/tex]

    to find the ratio:

    [tex]\frac{B}{A}[/tex] where A = the amplitude (45°)

    where B is, ultimately, approximately equal to:

    [tex] -\frac { \lambda A^3}{32\omega _0 ^2}[/tex]

    and

    [tex] \epsilon_3/m = \lambda[/tex]

    [tex]\hline [/tex]

    Here's where I'm stuck. What values do I use for:

    [tex]\omega_0, \epsilon_3, m[/tex]

    None of these are given in the problem statement??
    Where the heck did B/A = 0.0032 come from?
     
  2. jcsd
  3. Sep 10, 2004 #2

    Tide

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    First you have to justify dropping the quadratic term. If you work with angle instead of displacement then what you really have is a [itex]\sin \theta[/itex] which resolves the problem.
     
  4. Sep 10, 2004 #3

    cj

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    Is justification needed? I'm thinking the task is to
    simply, and arbitrarily, examine the 3rd harmonic as
    a way of getting familiar with the method of success
    approximations. Maybe I'm wrong?

    As far as amplitude is concerned,

    [tex]A = 45^\circ = \pi/4[/tex] radians, and

    A also can be expressed as:

    [tex]A = l\theta[/tex]

    For small angles,

    [tex]sin\theta = \theta[/tex]

    ... but I'm not sure where these equivalences are leading.



     
  5. Sep 10, 2004 #4

    Tide

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    Certainly you have to justify it since [itex]\left( \frac {\pi}{4} \right)^2>\left( \frac {\pi}{4} \right)^3[/itex]. Otherwise you might as well toss out the kx term too.

    With regard to your other question, all the coefficients in the ODE are known if you pay attention to its origin - namely Newton's Laws of Motion. For example, k = mg.
     
  6. Sep 10, 2004 #5

    krab

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    I get generally [itex]B/A=A^2/192[/itex], so for [itex]A=\pi/4[/itex], indeed, it is 0.00321...
     
  7. Sep 11, 2004 #6

    cj

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    My interpretation was that specific terms weren't being thrown-out -- rather that the contribution of the cubic term was simply being solely examined.

     
  8. Sep 11, 2004 #7

    cj

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    ... implying that

    [tex] \frac { \lambda}{\omega _0 ^2} = \frac {1}{6}[/tex]

    I know that

    [tex]\omega _0^2 = g/l[/tex]

    but am still at a loss to see the solution since

    [tex]\lambda = \frac { \epsilon _3}{m}[/tex]

    which seems to be too many unknowns??


     
  9. Sep 11, 2004 #8

    pervect

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    First, let's write the equations of motion for a pendulum in terms of the generalized coordinate x, which is the angle of the pendulum.

    If we use the Lagrangian method, we get L = T-V = .5*m*(l*xdot)^2+m*g*l*cos(x)

    So the equations of motion are d/dt(diff(L,xdot)) = diff(L,x)

    [tex]
    m l^2 \frac{d^2x}{dt^2} = -m g l sin(x)
    [/tex]

    or, to put it in the required form, we rearrange variables and do a taylor series expansion of sin(x) around x=0

    [tex]
    m \frac{d^2x}{dt^2} + \frac{m*g}{l}(x-\frac{x^3}{6})
    [/tex]

    We can get the same equations of motion without the Lagrangian method by the usual "balance of force" methods, which requires ascii diagrams to do well, for those not familiar with the Lagrangian approach.

    This should clear up the origin of most of the terms, except for the natural frequency 'w0', which should be easily found by solving the differential equation

    m x'' + kx = 0

    giving the usual result w0=sqrt(k/m)
     
  10. Sep 11, 2004 #9

    Tide

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    As I indicated earlier, if you set this problem up correctly you won't even have the quadratic term as pervect has now shown you. But, in general, you are not free to simply dismiss it looking for the third harmonic since ALL of the coefficients will ultimately go into evaluating the amplitude of each harmonic. You certainly would not dismiss the quadratic terms in solving a cubic polynomial equation though things would definitely be easier if we could!
     
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