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Simple Harmonic Motion and phase constant

  1. Jul 25, 2015 #1
    A simple harmonic oscillator consists of a block of mass 45 g attached to a spring of spring constant 240 N/m, oscillating on a frictionless surface. If the block is displaced 3.5 cm from its equilibrium position and released so that its initial velocity is zero, what is the phase constant, φ , of its oscillations? Assume the block moves according to the equation x=xmaxcos(ωt+φ) .
    • A :

      45o

    • B :

      30o

    • C :

      0o

    • D :

      90o

    • E :

      The phase constant is arbitrary.
     
  2. jcsd
  3. Jul 25, 2015 #2

    jbriggs444

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    What do you think?
     
  4. Jul 25, 2015 #3
    Im really stuck on how to solve this problem. I know that the Xm is 0.035m. I also know that the w can be solved for using sqroot(k/m). But I'm stuck now on how to use this information to solve for the answer
     
  5. Jul 25, 2015 #4

    jbriggs444

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    You should not need to do any arithmetic to answer this problem. You are not asked for omega.

    Assume that "initial" means t = 0. What is the situation at t=0?
     
  6. Jul 25, 2015 #5
    since the initial velocity is zero, would u take the derivative of the function to get v=-wxmsin(wt+φ). Then plug in the values to get 0 = -240(0.035)sin[(240)(0)+φ). Then solve for φ and get zero. According to the assignment the correct answer is zero. Is this the correct way to solve it?
     
  7. Jul 25, 2015 #6

    jbriggs444

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    Yes, that is one correct way.

    Another approach would be to convince yourself that the starting position is a position of maximum displacement and that the cosine function is maximized when its argument is zero.
     
  8. Jul 25, 2015 #7
    ok thank you
     
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