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Given a harmonic oscillator with mass m, and spring constant

  1. Mar 29, 2015 #1
    1. The problem statement, all variables and given/known data
    Given a harmonic oscillator with mass m, and spring constant k, is subject to damping force F= cdx/dt and driven by an external force of the form F[ext]= FoSin(wt).

    A) Find the steady state solution.
    B) Find the amplitude and the phase.

    2. Relevant equations
    F=-kx

    the steady state is usually in the form of X(t)= Acos(wt+Φ)

    3. The attempt at a solution\
    So i came up with this equation for the Fnet force.

    F[net]= -kx+c(dx/dt)+FoSin(wt)
     
  2. jcsd
  3. Mar 29, 2015 #2

    rude man

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    What should be the sign of your c dx/dt term? Think about it.
     
  4. Mar 29, 2015 #3
    The sign of my dx/dt term should be X(dot)
    or are you saying it should be negative instead of positive?

    Once I have that should I move all the signs to the other side
     
  5. Mar 29, 2015 #4

    BvU

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    Hello star,

    Rudy wants you to think about the sign: a + or a -
    In other words: if dx/dt > 0, which way does a damping force point ?

    Usually we take the damping coefficient (in your case c) as a positive value and therefore we need a + or - sign in the equation of motion to let the force point the correct way.
     
  6. Mar 29, 2015 #5
    When dx/dt >0 then the force has displacement to the right

    Well I rearranged the equation of motion into
    X(doubledot)=-(k/m)-cx(dot)+Fosin(wt)

    which then turns into X(doubledot)= -Wo^2X-2γx(dot)+(Fo/m)Sin(wt)

    Then I balanced the equation by bringing everything to the otherside
    and using known equation for the angular frequence Wo=√(k/m) and C = 2γm
    we receive

    X(doubledot)= -Wo^2X-2γx(dot)+(Fo/m)Sin(wt)
    X(doubledot)+Wo^2X+2γx(dot)=(Fo/m)Sin(wt)

    Then in order to find the steady state solution we must assume the gamma is equal to 0 but why do we consider gamma as equal to zero?
     
  7. Mar 29, 2015 #6

    BvU

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    $$\ddot x + 2\gamma\;\dot x + \omega_0^2\; x = {F_0\over m} \sin(\omega t)$$ looks good to me. But the
    does not. As you say: "why consider ##\gamma = 0 ## ?". What is your perception of this steady state solution you are looking for ?

    is indeed correct. What it tells you (should tell you) is that after an initial reponse that dampens out (-- thanks to the non-zero gamma! --), the oscillator will oscillate with the period of the driving force and with a certain amplitude; there will also be (or may be) a phase difference between the driving force and the oscillator.

    That the given form is actually a solution can be shown by substituting it in the equation. That will also help you on your way to find A and ##\phi##.
     
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