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Harmonic motion

  1. Apr 2, 2005 #1
    A 2 kg mass is attached to a spring aand placed on a horizontal, smooth surface. A force of 20N is required to hold the mass at reast when it is pulled .200m from the equilibrium position. The mass is released and undergoes simple harmoic motion.
    a) what is the force constant?
    b) frequency?
    c) max speed and where does this occur?
    d) mass acceleration and where does this occur?
    e) total energy
    f) speed when displacement equals one third of the max value
    g) acceleration when displacement equals one third of the max value

    a)
    N/m=20/0.2m=100N/m

    b)
    [tex]T=2\pi\sqrt{\frac{m}{k}}[/tex]
    [tex]T=2\pi\sqrt{\frac{2kg}{100N/m}}[/tex]
    [tex]T=0.8886s[/tex]

    [tex]f=1/T=1.1125Hz[/tex]

    c)
    [tex]v_{max}=\sqrt{\frac{m}{k}}*A[/tex]
    [tex]v_{max}=\sqrt{\frac{2kg}{100N/m}}*0.2m[/tex]
    [tex]v_{max}=0.0283m/s@x=0m[/tex]

    d)
    [tex]a_{max}=\frac{mA}{k}}[/tex]

    [tex]a_{max}=\frac{2kg*0.2m}{100N/m}}[/tex]
    [tex]a_{max}=0.004m/s^2@x=+/-0.2m[/tex]

    e)
    [tex]E=0.5kA^2[/tex]
    [tex]E=0.5(100N/m)(0.2m)^2[/tex]
    [tex]E=2J[/tex]

    f)
    [tex]d_{max}/3=0.2/3=\frac{1}{15}m[/tex]
    [tex]v=\sqrt{\frac{k}{m} (A^2-x^2)}[/tex]
    [tex]v=\sqrt{\frac{100N/m}{2kg} ((0.2m)^2-(\frac{1}{15}m)^2)}[/tex]
    [tex]v=+/- \frac{4}{3}m/s[/tex]

    g)
    [tex]v=-\omega A sin \omega t[/tex]

    [tex]\frac{4}{3}m/s=-\sqrt{\frac{100N/m}{2kg}} *0.2m sin \sqrt{\frac{100N/m}{2kg}} t[/tex]

    [tex]\frac{4}{3}m/s=-1.414213562sin7.071067812t[/tex]

    [tex]t=-0.1740839504s[/tex]

    [tex]a=-\omega^2 A cos \omega t[/tex]

    [tex]a=- \frac{100N/m*0.2m}{2kg} cos 7.071067812*-0.1740839504s[/tex]

    [tex]a=+/- \frac{10}{3} m/s^2[/tex]

    did I do this right? especially letter (g) thanks
     
    Last edited: Apr 3, 2005
  2. jcsd
  3. Apr 3, 2005 #2
    could someone please check just letter f and g for me? thanks
     
  4. Apr 3, 2005 #3

    quasar987

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    The steps are fine. I didn't checked the calculations.
     
  5. Apr 3, 2005 #4
    I just didnt know if there was an easier way to do letter g. For letter f, there is an equation that has both velocity and distance, but for g, I dont know if there is an equation that has acceleration in terms of distance.

    this is why I'm hesitant on my way of answering letter g
     
  6. Apr 4, 2005 #5
    any help??
     
  7. Apr 4, 2005 #6

    quasar987

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    I don't know of such an equation, but it doesn't matter. As long as you have A way to do it.
     
  8. Apr 4, 2005 #7
    You used the correct formula on f. For g) a simpler method would be to use F=kx, so a=kx/m. Honestly can't be troubled to check your method.
     
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