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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
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    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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