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Determining the period of a periodic motion

  1. Aug 31, 2005 #1
    Hi. I am currently working on a problem involving a ball being dropped from a hieght of 4m making a perfectly elastic collision with the ground. Assuming no mechanical energy is lost due to air resistance, I must find the period of the motion.
    I used the equation
    V(final) = V(initial) - gt
    where V(initial)=0
    therefore V(final)= -9.8t

    and V^2(final) = V^2(initial) - 2gh
    h= -4m
    V(final)= -9.8t
    so i replaced these values in the equation, and i find t= 0.9035 seconds
    To find T, I simply multiplied 0.9035 sec by 2. However, my answer does not correspond that what it says in the book, and now i'm lost! Can someone help me by telling me if my method is right or wrong? I would really appreciate it!!
    Thank you!
  2. jcsd
  3. Aug 31, 2005 #2
    And also, why is this type of motion NOT Simple Harmonic?
  4. Aug 31, 2005 #3
    It is simple harmonic motion if the collisions are completely elastic and no air resistance/friction is present.

    It takes 1.807s for the ball to fall, hit the ground, and go back to its original height. What does the book say?
  5. Sep 1, 2005 #4


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    1) it would have been more straightforward to use
    x = x_0 + v_0 *t + .5 g t^2 => 4m = .5(9.8m/ss) t^2

    2) yes, the time Period T = 1.807s . Did they give frequency?

    3) "Harmonic" means location function x(t-t_0) can be written
    as a Sum of sine waves with (a) w_i = n * w_0
    "Simple Harmonic" means that there's only one term:
    x(t) = A sin(wt) (if you set your starting time right, so no phase).
    Do you think this ball's motion is Simple Harmonic?
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