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Mass-Spring System Beats and Resonance

  1. Dec 20, 2009 #1
    Beats and Resonance

    In the Beat not have friction force, correct ?

    [tex]m \frac{d^2x}{dt^2} + kx = F_o cos(wt)[/tex]

    We can write as

    [tex]\frac{d^2x}{dt^2} + w_o^2 x = \frac{F_o}{m} cos(wt)[/tex]

    If [tex] w \not= w_o[/tex]

    Assuming (Particular solution)
    [tex]x_p = acos(wt) + bsin(wt)[/tex] Why we have assuming this ?

    How find [tex]a(w_o^2 - w^2)cos(wt) + b(w_o^2 - w^2)sin(wt) = \frac{F_o}{m} cos(wt)[/tex] ???

    And why find that: [tex]x_p[/tex] is [tex]x_p = \frac{F_o}{m(w_o^2 -w^2)}cos(wt)[/tex] ???
  2. jcsd
  3. Dec 20, 2009 #2


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    No, you can have beats with a friction force- but they die away quickly. In "real life" there is always friction, but you can hear "beats" if you play two guitar strings that are almost tuned to the same note.

    Defining [itex]w_0[/itex] to be a square root of k, yes.

    I suspect you will soon learn what to do in case those are equal!

    Because it works! You should have learned, long before this, that the solutions to "linear differential equations with constant coefficients" are only of a few limited kinds of functions:
    exponentials, sine and cosine, polynomials, and products of those.

    And the method of "undetermined coefficients" for finding "specific solutions" for such equations uses the fact that if the right hand side is one of those kinds of functions, the "specific solution" will also be of that kind, with certain adjustments that you may still be learning about.

    Note that this equation has an infinite number of solutions and we are looking for just one. So it doesn't hurt to look specifically for particular types of solutios.

    The equation you get from the above,
    [tex]a(w_o^2 - w^2)cos(wt) + b(w_o^2 - w^2)sin(wt) = \frac{F_o}{m} cos(wt)[/tex]
    must be true for all t. In particular, at t= 0, we must have
    [tex]a(w_0^2- w^2)(1)+ b(w_0^2- w^2)(0)= a(w_0^2- w^2)= \frac{F_0}{m}(1)[/quote]
    which says that
    [tex]a= \frac{F_0}{m(w_0^2- w^2)}[/tex]

    and taking [itex]t= \pi/2[/itex], we must have
    [tex]a(w_0^2- w^2)(0)+ b(w_0^2- w^2)(1)= b(w_0^2- w^2)= \frac{F_0}{m}(0)= 0[/tex]
    so that b= 0.
  4. Dec 20, 2009 #3
    Why we start this equation [tex]a(w_o^2 - w^2)cos(wt) + b(w_o^2 - w^2)sin(wt) = \frac{F_o}{m} cos(wt)[/tex] ?
    Where does it come from?
  5. Dec 20, 2009 #4


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    You don't seem to be grasping how the general-and-particular solution method works.

    For the general solution, you pretend the RHS is 0.

    For the particular solution, you make an intelligent guess (that's all it is! :wink:), which in this case is the given xp.

    You then put that xp into the original equation, differentiate where appropriate, and that gives you …

    that equation! :rolleyes:.
  6. Dec 20, 2009 #5
    Ok. Thanks.

    The solution we have is:

    [tex] x(t) = \frac{F_o}{m(w_o^2 - w^2)} (cos(wt) - cos(w_ot))[/tex]

    From this equation how find the solution:
    [tex]x(t) = [ \frac{2F_o}{m(w_o^2-w^2)} sin(\frac{(w_o-w)t}{2}) ] sin(\frac{(w_o+w)t}{2}) [/tex] ???

    I think when [tex]w[/tex] is near [tex]w_o[/tex] produces beat, movement with varying amplitude
  7. Dec 20, 2009 #6


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    (type "\left(" and "\right)" for big brackets, and have an omega: ω :wink:)

    That uses one of the trigonometric identities which you need to learn …

    cosA - cosB = 2sin((A+B)/2)sin((A-B)/2)​

    Try putting ω = ω0 + λ, where λ is small. :wink:
  8. Dec 20, 2009 #7
    Thanks. Beat it is.

    About Resonance, need [tex]w = w_o[/tex]

    [tex]x(t) = Acos(w_o t) + Bsin(w_o t)[/tex]

    Where the particular solution

    [tex]x_p(t) = t(Acos(w_ot)+Bsin(w_ot))[/tex]

    How I find the solution: => [tex]x_p(t) = \frac{F_o}{2mw_o}tsin(w_ot)[/tex] ???
  9. Dec 20, 2009 #8


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    Same as before … you put that xp into the original equation, and differentiate where appropriate. :smile:
  10. Dec 20, 2009 #9
    I do not know that equation we put XP on the original equation, I could not solve
  11. Dec 20, 2009 #10


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    Now you're worrying me, did you not understand how the particular solution was found in the original case?

    In this new case, put the new xp into md2x/dt2 + kx = F0cosωt, to find what A and B are.
  12. Dec 20, 2009 #11
    Derivative great. I do after the with a program :smile:

    The general solution is

    [tex]x(t) = Acos(w_ot) + Bsin(w_ot) + \frac{F_o}{2mw_o}t sin(w_ot)[/tex]

    Is corret ?

    A and B determine to initial conditions again ?
  13. Dec 20, 2009 #12


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  14. Dec 20, 2009 #13
    Thank you
  15. Dec 21, 2009 #14
    Someone has a graphic example of the system to beat and resonance?
  16. Dec 21, 2009 #15
    This system [tex]w_o[/tex] is the frequency of oscillation of the system and [tex]w[/tex] is frequency of oscillation of the external force ????
  17. Dec 21, 2009 #16
    Actually there's a general technicall method called the variation of parameters which produces the general solution given any arbitrary "input" function. So it's not just a guess.

    In specific cases (where the derivative of the input has its own form) it's much more easy to guess the solution.
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