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Homework Help: Find position after t for simple harmonic motion

  1. Jun 9, 2009 #1
    A 28.0 kg block at rest on a horizontal frictionless air track is connected to the wall via a spring. The equilibrium position of the mass is defined to be at x=0. Somebody pushes the mass to the position x= 0.350 m, then lets go. The mass undergoes simple harmonic motion with a period of 4.70 s. What is the position of the mass 3.854 s after the mass is released?

    T=2pi*sqrt m/k
    KE1+PE1=KE2+PE2 from this you get mv^2=KA^2
    x=Asin *omega*t
    frequency =1/T

    I used the first equation get k=3.92
    then I use that k in the second equation, but I dont know how to get v to then get A. Any help? Would I then take that A and plug it into the third equation to get my X?
  2. jcsd
  3. Jun 10, 2009 #2


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    In the given problem the amplitude A is given. ( 0.35 m). Find x.
  4. Jun 10, 2009 #3
    So I plugged in A (.35m) into the equation,
    x=.35sin(1.34*3.854) and got x=.032 which is still incorrect? Any help on where I am going wrong?
    The second part of this problem calls for the maximum acceleration- would I use the equation a=omega^2cos omega *t? I don't think that I will get this problem correct if I got the last one incorrect.
  5. Jun 10, 2009 #4
    Your formula for posistion does not work at t=0! You need a phase angle in it, or use cosine instead of sine.
  6. Jun 10, 2009 #5
    I figured this problem out. Thank you!
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