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Harmonic oscillator problem

  1. Sep 23, 2014 #1
    I have the following homework problem that I am having trouble with. Any guidance would be appreciated. Thank you in advance.

    Consider an object hanging on a spring, immersed in a cup of water. The water
    exerts a linear viscous force -bv on the object, where v is the speed of the object
    relative to the liquid.
    a. The object is displaced vertically from its equilibrium, and is then
    released. The cup remains at rest. Find the equation of motion for the
    position of the object.
    b. Now, the cup is moved up and down. Assuming the motion of the cup is
    simple harmonic motion, find the equation of motion for the position of
    the object hanging from the spring. Give your answer in terms of the mass
    of the object, the spring constant, the damping constant b, the amplitude
    through which the cup is moved, and the frequency with which the cup is
    moved. Remember that the viscous force depends on the speed of the
    object relative to the liquid.
    c. What is the steady-state amplitude of oscillation for the object?
    d. Now, in addition to moving the cup up and down, we also drive the object
    by moving the suspension end of the spring (the opposite end from the one
    that the object is attached to). This drive has the same frequency as the
    drive that is moving the cup, but it has a different phase. Find the
    equation of motion for the object in this case.
    e. If the suspension end of the spring is driven with a particular amplitude
    and phase, the object will be motionless in steady state. Find the
    amplitude and phase for which this occurs.


    1. The problem statement, all variables and given/known data

    At equilibrium, when the object is at rest in the water the forces acting on it are
    mg down
    ky and pVg up, with pVg being a bouyant force that is always acting up.
    mg-ky-pVg=0 -> mg=pVg+ky

    When it starts to move
    mg-k(y+x)-bx'+pVg=mx'' -> x''+γx'+ω^2x=2pVg

    I am not confident in the effect that the bouyant force has on the mass. Half of the time it is acting as a damping force, half the time as a restoring force, which makes it seem like it would cancel itself out.

    So if the effect of the water is just incorporated into the values of the damping force, γ, then the equation of motion would be the same as it is in air, x''+γx'+ω^2x=0.

    please let me know if I am missing something. Thanks again!

    Phil
     
  2. jcsd
  3. Sep 24, 2014 #2

    Simon Bridge

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    Hormones going up and down aye? I sympathize... have you tried cold showers or playing a sport? ;)

    ... how does the buoyant force act as a damping force? Doesn't it always act upwards just like gravity always acts downwards?
     
  4. Sep 24, 2014 #3

    Orodruin

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    To expand on what Simon said. Here your buoyant force is all of a sudden acting in the same direction as gravity wheras in your previous work, it was acting against... So, yes, it will cancel out if you do things correctly.

    The big difference to the case of just air is the value of the damping.
     
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