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Forced oscillations: solution of Differential Equation at resonance

  1. Mar 19, 2014 #1
    1. The problem statement, all variables and given/known data

    Following a worked example in my book, I have been trying to get a solution for the equation

    [itex]\frac{d^2u}{dt^2} + \frac{k}{m}u = Fcos\omega t[/itex]

    The book says that at resonance, i.e. when [itex]\omega_0[/itex] (the natural frequency) = [itex]\omega[/itex] (the forcing frequency), the term [itex]F cos\omega t[/itex] is a solution to the homogenous equation and the solution to the differential equation above is[tex]Acos\omega_0t +Bsin\omega_0t+\frac{F}{2m\omega_o^2}tsin\omega_0t[/tex]

    2. Relevant equations

    3. The attempt at a solution

    To get the full solution:

    Complementary Function:

    [tex]Acos\omega_0t+Bsin\omega_0t[/tex], where [tex]\omega_0=\sqrt{\frac{k}{m}}[/tex]

    To get the Particular Integral:
    Assume [itex]u= Ctcos\omega t +Dtsin\omega t[/itex]

    Then [tex]\frac{du}{dt}= C cos\omega t - Ct\omega sin \omega t + D sin\omega t +Dt \omega cos \omega t[/tex]

    [tex]\frac{du}{dt}= (Dt\omega +C ) cos\omega t + (D-Ct\omega) sin\omega t[/tex]

    And [tex]\frac{d^2u}{dt^2} = -C \omega sin \omega t -C\omega sin \omega t - C \omega^2 t cos \omega t + D\omega cos \omega t + D\omega cos \omega t - D \omega^2 t sin \omega t[/tex]

    [tex]\frac{d^2u}{dt^"}=(2D\omega -c \omega^2t)cos\omega t +(-2C\omega -D\omega^2t)sin \omega t [/tex]

    Back substituting these into the original differential equation:

    [tex](2D\omega -C\omega^2t + \frac{k Ct}{m}) cos\omega t + (\frac{KDt}{m}-2C\omega -D\omega^2t) sin \omega t = \frac{F}{m} cos\omega t[/tex]

    Equating coefficients:

    [tex](2D\omega -C\omega^2t + \frac{k Ct}{m}) =\frac{F}{m}[/tex]

    and [tex](\frac{kDt}{m} -2 C \omega - D\omega^2 t) = 0[/tex]

    After this, I have tried solving for D and C but it seems to end up in a pretty intractable mess.

    I know somehow D should be equal to [itex] \frac{F}{2m\omega_0^2} [/itex] (and I assume since we are talking about a situation in which [itex]\omega_0=\omega[/itex], D= [tex] \frac{F}{2m\omega_0^2} = \frac{F}{2m\omega} [/tex]) but can not see how to get there. Can anyone tell me if I have been going in the right direction so far?
    Last edited by a moderator: Mar 19, 2014
  2. jcsd
  3. Mar 19, 2014 #2


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    From the form of the particular solution, you're apparently assuming that ##\omega=\omega_0 = \sqrt{k/m}##, so the two equations reduce to
    2\omega D &= \frac Fm \\
    -2\omega C &= 0.
    \end{align*} The solution you're trying to obtain is not correct: ##\frac{F}{m\omega_0^2}t\sin\omega_0 t## does not have units of length.
  4. Mar 19, 2014 #3
    Thanks a lot Vela. I should have noticed that when equating coefficients.

    I don't quite understand this though:

    I can't see anything to do with units of length in the equation...
  5. Mar 19, 2014 #4


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    The denominator is ##m\omega^2 = k## where ##k## has units of force/length, so ##\frac{F}{m\omega^2}## will have units of length. That would be fine if the coefficient multiplied only ##\sin \omega t##, but it doesn't. It multiplies ##t \sin\omega t##. The factor of ##t## introduces a unit of time.
  6. Mar 19, 2014 #5
    Oh, I see... That makes sense, thanks!
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