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Damped SHM

  1. Sep 9, 2004 #1
    Hello everybody...

    In my school textbook, I have the following explanation to Damped SHM:

    [tex]\hline [/tex]
    Soving the differential equation:

    [tex] m \frac{d^2x}{dt^2}+\gamma m \frac{dx}{dt}+m\omega _0 ^2x = 0[/tex] (Eq. 1)

    gives

    [tex] x(t)=x_{m}e^{-\frac{\gamma}{2}t}\cos \left( \omega _{\gamma}t + \phi \right), [/tex] (Eq. 2)

    where:

    [tex] \gamma = \frac{b}{m} [/tex]
    and
    [tex] \omega _{\gamma}=\sqrt{\omega _0 ^2 - \frac{b^2}{4m^2}} [/tex]
    and
    [tex] \omega _0 = \sqrt{\frac{k}{m}}. [/tex]

    [tex] \hline [/tex]

    It goes as simply as that. However, I've read on the web that this is an under damped SHM, and that the answer really goes like this:

    [tex]\hline [/tex]

    [tex] x(t)=Ae^{-\frac{\gamma}{2}t}\cos \left( \omega _{\gamma}t + \phi _1 \right)+Be^{-\frac{\gamma}{2}t}\cos \left(- \omega _{\gamma}t + \phi _2 \right) [/tex] (Eq. 3)

    [tex]\hline [/tex]

    My question is: How do I go about finding (Eq. 2) out of (Eq. 3). Is there any special trick?

    Thank you very much. :smile:
     
  2. jcsd
  3. Sep 9, 2004 #2

    Tide

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    Generally, a second order linear differential equation will have two solutions and the particular solution will be a linear superposition of the two. I think the author of your text was just being a little lazy and took both components to have the same amplitude.
     
  4. Sep 9, 2004 #3
    What about the phase? Shall also I set [tex]\phi _1 = \phi[/tex] and [tex]\phi _2 = \frac{\pi}{2} [/tex] in order to get Eq. 2?

    Thanks
     
  5. Sep 9, 2004 #4

    Integral

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    You have not fully specified the problem. Where are, and what are, the initial conditions? The first solution may result in applying the initial conditions, while the second is the full general solution with no initial conditions applied.
     
  6. Sep 9, 2004 #5
    Besides, you can have real solutions as well as imaginary solutions. it depends on the initial conditions and ofcourse you must also know the algorithm for solving second order differential equations. You can have a solution of the form exp(iAt) where A is a constant. But Euler's rule states that this is equal to cos(At) + isin(At). Now in differential equations You also need to implement constants that can be determined by using the initial conditions. A general solution would be of the form : Bcos(At) + Csin(At)

    So basically all depends on the given initial conditions. I just wanted to give some general possibilities concerning solutions of the diff.equation...

    regards
    marlon
     
  7. Sep 9, 2004 #6
    If you wanna see how it is done completely, let me know...

    regards
    marlon
     
  8. Sep 9, 2004 #7

    ehild

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    Yes. You can find the parameters of eq. 2 from those in eq. 3.

    [tex] x_{m}e^{-\frac{\gamma}{2}t}\cos \left( \omega _{\gamma}t + \phi \right)=Ae^{-\frac{\gamma}{2}t}\cos \left( \omega _{\gamma}t + \phi _1 \right)+Be^{-\frac{\gamma}{2}t}\cos \left(- \omega _{\gamma}t + \phi _2 \right) [/tex]

    [tex]x_m\cos \left( \omega _{\gamma}t + \phi \right)=A\cos \left( \omega _{\gamma}t + \phi _1 \right)+B\cos \left(- \omega _{\gamma}t + \phi _2 \right) [/tex]

    [tex]x_m\cos( \omega _{\gamma}t)cos(\phi )-x_m\sin( \omega _{\gamma}t)sin(\phi)=A[cos (\omega _{\gamma}t)cos(\phi_1)-sin( \omega _{\gamma}t)sin(\phi1)]+B[cos( \omega _{\gamma}t)cos(\phi_2)+sin( \omega _{\gamma}t)sin(\phi_2)] [/tex]

    [tex]cos( \omega _{\gamma}t)[x_mcos(\phi)-Acos(\phi_1)-Bcos(\phi_2)]-sin( \omega _{\gamma}t)[x_msin(\phi)+Asin(\phi_1)-Bsin(\phi_2)]=0[/tex]

    The equation can hold for every value of t only if the coefficients of both terms [tex]cos( \omega _{\gamma}t)\mbox { and }sin( \omega _{\gamma}t)[/tex] are zero.

    [tex]x_mcos(\phi)=Acos(\phi_1)+Bcos(\phi_2)[/tex]
    [tex]x_msin(\phi)=Asin(\phi_1)-Bsin(\phi_2)[/tex]

    Solving these equations, we get

    [tex] x_m^2= A^2+B^2+2ABcos(\phi_1-\phi_2)[/tex]
    [tex]tan(\phi)=\frac{Asin(\phi_1)-Bsin(\phi_2)}{Acos(\phi_1)+Bcos(\phi_2)}[/tex]

    I hope I haven't made too many mistakes in the derivations...


    ehild
     
  9. Sep 9, 2004 #8
    Well, I guess [tex]x_msin(\phi)=Asin(\phi_1)-Bsin(\phi_2)[/tex] should actually be [tex]x_msin(\phi)=-Asin(\phi_1)+Bsin(\phi_2)[/tex].

    To be honest, I just could not understand how you found the expression: [tex] x_m^2= A^2+B^2+2ABcos(\phi_1-\phi_2)[/tex]. Anyway, great job!
     
  10. Sep 10, 2004 #9

    ehild

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    You are right, the signs are very confusing, especially if I have to write in TEX... :smile:
    The error was in the formula

    [tex]cos( \omega _{\gamma}t)[x_mcos(\phi)-Acos(\phi_1)-Bcos(\phi_2)]-sin( \omega _{\gamma}t)[x_msin(\phi)-Asin(\phi_1)+Bsin(\phi_2)]=0[/tex]

    I hope it is correct now, (but one never can be sure about signs) so

    [tex]x_msin(\phi)=Asin(\phi_1)-Bsin(\phi_2)[/tex]

    is all right, but there is an other error in [tex]x_m^2[/tex]. The correct form is

    [tex] x_m^2= A^2+B^2+2ABcos(\phi_1+\phi_2)[/tex].

    You can get it by taking the square of both equations

    [tex]x_mcos(\phi)=Acos(\phi_1)+Bcos(\phi_2)[/tex]

    and

    [tex]x_msin(\phi)=Asin(\phi_1)-Bsin(\phi_2)[/tex]

    summing up the squared equations, collecting terms with [tex]A^2 \mbox{, } B^2 \mbox{ and } 2AB [/tex] and using the trigonometric identities [tex]cos^2(\alpha)+sin^2(\alpha)=1 [/tex] and [tex] cos(\alpha +\beta) = cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta)[/tex]. You can use the same method any time when you want to replace the sum or difference of two harmonic motions(signals) of the same frequency with a single one. (You will bump into it when you learn about ac circuits or about interference of waves.)

    ehild
     
  11. Sep 10, 2004 #10
    I get it

    Thank you, ehild. That works!
     
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