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Simple harmonic motion (finding omega)

  1. Jul 2, 2010 #1
    1. The problem statement, all variables and given/known data
    A particle executes simple harmonic motion with an amplitude of 2.51cm. At what positive displacement from the midpoint of its motion does its speed equal one half of its max speed?


    2. Relevant equations
    A=2.51 cm
    x(t) = Acos(wt)
    v(t) = -Awsin(wt)
    vmax = |-Aw|


    3. The attempt at a solution
    I just want to know how to find w so I can plug it into the vmax equation.
     
  2. jcsd
  3. Jul 2, 2010 #2
    You actually don't need to plug it in. If [tex]v_{max} = Aw[/tex] then [tex]\frac{v_{max}}{2} = \frac{Aw}{2}[/tex]

    Just use the energy equation

    [tex]\frac{1}{2}kA^2 - \frac{1}{2}kx^2 = \frac{1}{2}mv^2[/tex]

    where [tex]v= \frac{Aw}{2}[/tex], and remember to sub [tex]\frac{k}{m}[/tex] for [tex]w^2[/tex].
     
    Last edited: Jul 2, 2010
  4. Jul 2, 2010 #3
    I don't know what k is though. How do I find k?
     
  5. Jul 2, 2010 #4
    Divide both sides by k!
     
  6. Jul 2, 2010 #5
    O ok haha sorry I'm really tired. Thanks for the help I solved it!
     
  7. Jul 2, 2010 #6
    You don't need the velocity or the angular frequency in order to answer the question. You're only asked at what displacement the velocity has half its maximum value.

    [tex]x=A\cos{(\omega t)}[/tex]

    [tex]v=-\omega A \sin{(\omega t)}[/tex]

    [tex]v=-\omega A \sqrt{(1-\cos ^2 {(\omega t))}}[/tex]

    [tex]v = -\omega \sqrt{(A^2-x^2)}[/tex]

    [tex]|v|=\tfrac{1}{2} v_{max}=\tfrac{1}{2}\omega A[/tex]

    [tex]\tfrac{1}{2}\omega A = \omega \sqrt{(A^2-x^2)}[/tex]

    You don't always need all the constants involved in a problem in order to solve for a particular value. :)
     
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