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Simple Harmonic Oscillation frequency

  1. Aug 22, 2009 #1
    1. The problem statement, all variables and given/known data
    "A mass stands on a platform which executes simple harmonic oscillation in a vertical direction at a frequency of 5 Hz. Show that the mass loses contact with the platform when the displacement exceeds 10^-2 m."


    2. Relevant equations

    x(t) = a cos(wt - phi)

    frequecy = 1/T = w/2(pi)




    3. The attempt at a solution

    Ok, so I solved for w: 5 = w/(2 * pi) w = 31.41592 rad/s

    But the thing is, I don't understand the fundamentals of this question, how can a mass stay on a platform without gravity pushing it against the board? I.E. What am I solving for? I'm guessing there's a threshold somewhere to solve for but I'm just not seeing it. I.E. if the question specified a mass and gravity constant I could use that as the force holding the mass on the platform and when the platform's force exceeded the force of gravity the mass would lose contact - but I don't have that here.

    I'm utterly confused.
     
  2. jcsd
  3. Aug 22, 2009 #2

    nicksauce

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    The mass will presumably lose contact with the with the platform if the gravitational force exceeds the spring force. So your equation should be kx = mg. Now you need some way to relate k and m. How is the frequency of vibration related to these two quantities?
     
  4. Aug 23, 2009 #3
    well I have w = (k/m)^(1/2)

    Did you mean that the mass will leave the platform once the spring force exceeds gravity and not the other way around since gravity is what's holding the mass on the board?

    in any event, the equation kx = mg

    1/x = k/mg

    (1/x)^1/2 = (1/g)^1/2 * w ; w = 31.41

    so plugging 10^-2 in for x and 9.8 for g and 31.41 = w

    I get
    10 = 10

    So the value of 10^2 m does satisfy the equation, but that equation just set the two equal to each other not one over the other - how is having the spring force = weight show the mass loses contact?
     
  5. Aug 23, 2009 #4

    kuruman

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    The mass is not in contact with the spring therefore the spring cannot affect it directly. More correctly, the mass will lose contact when the normal force exerted by the platform is zero because, when that happens, it doesn't matter whether the platform is there or not. One then writes

    N - mg = ma
    N = mg + ma

    For a spring-mass system, a = -(k/m)x so that

    N = mg - kx

    Note that x is an algebraic quantity, negative when the mass is below the equilibrium point, which makes N non-zero below equilibrium. Therefore N can be zero only above the equilibrium point.
     
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