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Deriving SHM equations

  1. Jan 24, 2006 #1
    Using Newton's 2nd Law for a damped oscillator:

    [tex] ma = -kx - \alpha x [/tex]

    which is a second order linear DE. To solve it we use the trial integrating factor [itex] e^{\lambda x} [/tex] to come to the root equation

    [tex] mx^2 + \alpha x + k = 0[/tex] where we can find our two solutions to be

    [tex] r_{1} \ and \ r_{2} \frac{-\alpha}{2m} +/- \frac{1}{2m} \sqrt{\alpha^2 - 4km} [/tex]

    And the function [itex] x(t) [/itex] is supposed to be represented by
    [tex] x(t) = Ae^{r_1 t } + Be^{r_2 t} [/tex] which can be solved with th given initial conditions [itex] x(0) = 0 \ and \ v(0) = v_0[/itex].

    I'm supposed to end up with the product of two exponentials which dissolve to a cosine and sin function.
    Four lines up is where I'm lost.. I'm not sure how to jump to the two exponential functions. What I have is:

    [tex] x(t) = Ae^{r_1 t}(\cos(x(\alpha^2 - 4km)) +/- i \sin(x(\alpha^2 - 4km)) [/tex] which just smells really wrong. I can solve the particular equations once I can get to the general solution, but I'm stuck there. I know how the damping is determined and everything.. I just cant get the general equation :(
    Last edited: Jan 24, 2006
  2. jcsd
  3. Jan 27, 2006 #2
    The differential equation modelling this system is given by:

    [tex] m\Ddot{x} + 2\beta \dot{x} + \omega^2 x = 0 [/tex]

    With [itex] \beta \ and \ \omega [/itex] defined as

    [tex] \beta = \frac{b}{2m} \ and \ \omega = \sqrt{\frac{k}{m}} [/tex]

    with m being mass, k the characteristic constant of the physical system, and b the damping coefficient. The equation is solved via the auxilliary equation which simplifies to

    [tex] r_{1,2} = -\beta \pm \sqrt{\beta^2 - \omega^2} [/tex]

    The damping coefficient is less than the frequency of the system in underdamped harmonic motion, thus [itex] \beta < \omega [/itex] and imaginary numbers appear. We introduce another simplification, [itex] \omega_1 [/itex] defined by

    [tex] \omega_1 = \sqrt{\omega^2 - \beta^2} [/tex]

    Thus [tex] r_{1,2} = -\beta \pm \imath \omega_1 [/tex] and [itex]x(t)[/itex] can be expressed by the exponentials:

    [tex] x(t) = \exp{\left((-\beta + \imath \omega_1)t}\right) + \exp{\left((-\beta - \imath \omega_1)t}\right) } [/tex] and by Euler's formula we simplify to the general solution to underdamped harmonic motion:

    Equation A
    [tex] x(t) = e^{-\beta t}\left(A\cos(\omega_1 t) + B\sin(\omega_1 t) ) [/tex]
    Last edited: Jan 27, 2006
  4. Jan 31, 2006 #3
    [tex] \frac{T}{2} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} a_n a_m\cos(\omega _n t) \cos(\omega_m t) {\left(\frac{2\pi}{L} x\right)}^2 n m \int_0^L \cos \left(\frac{2\pi n}{L} x\right) \cos\left(\frac{2\pi m}{L} x\right) dx [/tex]
    Last edited: Jan 31, 2006
  5. Jan 31, 2006 #4
    This is not really introductory physics, but anywho, have you taken differential equations yet? This is a classic example probably found in your book if you have one on O.D.E. That post number 3 looks scary as hell to me, what is that?
    Last edited: Jan 31, 2006
  6. Jan 31, 2006 #5
    iii) (challenging) now consider a wave made from a superposition of modes:

    [tex] \sum_{n=1}^{\infty} a_n\cos(\omega_n t)\sin(\frac{2\pi n}{L} x) [/tex]

    calculate the total energy of this wave and comment on your answer.
    I get the KE and PE each to be 0 by the expression in post 3. We have arguments for it being 0 and arguments against it being 0.
  7. Feb 2, 2006 #6
    [tex] \alpha ^{x+n} 2 dy + \left( \alpha ^3 x^2 - 3 \alpha ^n y) dx = 0 [/tex]
  8. Feb 3, 2006 #7
    [tex] f = \frac{1}{2L}\sqrt{\frac{mg}{\mu}} [/tex]
  9. Feb 7, 2006 #8

    [tex] \left(\frac{ia - 1}{ia+1}\right)^{ib} = exp(-2b cot^{-1} a)[/tex] and

    [tex] \sum_{n=1}^{N} \cos(2n-1) x = \frac{\sin 2n x}{2\sin x}} [/tex]
    Last edited: Feb 7, 2006
  10. Feb 8, 2006 #9


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    Homework Helper
    Gold Member

    Whozum, are those questions with which you need help?

    If so, for the second one, you could multiply numerator and denominator by 2sin(x) and write 2SinACosB as Sin(A+B) + Sin(A-B).
  11. Feb 9, 2006 #10
    No, I'm just using this as a tex notepad since no one was helping when I needed it. I'm good with all these.

    [tex] v = \lambda f [/tex]

    [tex] \lambda = 4L [/tex]

    [tex] v = 4L f [/tex]

    [tex] f = \frac{v}{4L} [/tex]

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