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Quick Question - Simple Harmonic Motion - (I think I just need a formula)

  1. Mar 15, 2008 #1
    1. The problem statement, all variables and given/known data

    If a particle undergoes Simple Harmonic Motion with amplitude 0.34 m, what is the total distance it travels in one period?

    2. Relevant equations

    x=Asin(wt+theta)

    3. The attempt at a solution

    I know A, (.34), yet it seems like this problem doesn't give me nearly enough information to find anything else! I think I just need a different formula...or perhaps knowledge of another concept. We just started this unit so it might be just something really simple that I'm missing. Thank you!
     
  2. jcsd
  3. Mar 15, 2008 #2
    The amplitude is the maximum distance the spring is compressed or stretched from its equilibrium position. In an ideal spring, which is one that represents Simple Harmonic Motion, if one were to stretch (or compress) a spring to a certain amplitude and let go, the spring would oscillate back and forth, with its maximum distance from the equilibrium position being the amplitude. One period is basically the time it takes for the spring to make one full oscillation, which is basically when the spring makes a full trip back and forth...using that, how much would you think the spring travels in one period?
     
  4. Mar 15, 2008 #3
    Oh!! Okay, I didn't even realize thats what amplitude meant! Thanks. In an ideal spring, that means it would just perfectly double the distance (of going back and forth once). =) Thanks!!

    So that gives me the correct answer of 1.36m. Thank You!!
     
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