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Homework Help: Resonance (Differential Equations Class)

  1. Oct 24, 2007 #1
    [SOLVED] Resonance (Differential Equations Class)

    1. The problem statement, all variables and given/known data
    A front-loading washing machine is mounted on a thick rubber pad that acts like a spring; the weight W = mg (with g = 9.8 m/s^2) of the machine depresses the pad exactly 0.38 cm. When its rotor spins at \omega radians per second, the rotor exerts a vertical force
    F_0 cos(omega t)
    Newtons on the machine. Neglecting friction, determine at what speed (in revolutions per minute) resonance vibrations will occur?

    2. Relevant equations

    3. The attempt at a solution

    I decided to just set it up like a force equation in physics.


    Now solve for [tex]\omega[/tex] which is [tex]\sqrt{\frac{k}{m}}[/tex]


    So omega is:


    Transform to rpms

    [tex]RPMS = \omega \frac{(60)}{2\pi}[/tex]

    Which to the nearest RPM is 528. But this is wrong. Any clues?
    Last edited: Oct 25, 2007
  2. jcsd
  3. Oct 24, 2007 #2
    Try to recalculate omega, or better give the values you used to find omega.
  4. Oct 24, 2007 #3
    Yes, I'm confused as to where you got these numbers. Doing the same calculations I don't get the same result...
  5. Oct 24, 2007 #4


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    Homework Helper

    My suspicion is insufficient precision, for one thing. Also, where did 35/2 come from?
  6. Oct 24, 2007 #5
    Okay, found my mistake and I'll also show more steps now too.

    So from the part:


    k/m is omega squared. G is given as 9.8 m/s^2 and x is given as .38 cm, or .0038 m.

    Solving for Omega you get:

    [tex]\omega=\sqrt{\frac{k}{m}} = \sqrt{\frac{g}{x}}= \sqrt{\frac{9.8}{.0038}}[/tex]

    This is where my mistake is, I mistakenly entered .0032 instead of .0038 in my calculator and got the previous result.
  7. Oct 24, 2007 #6
    BTW it was 485 rpm's
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