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Simple harmonic motion derivative of position function

  1. Apr 18, 2014 #1
    1. The problem statement, all variables and given/known data

    The function
    x = (7.4 m) cos[(5πrad/s)t + π/5 rad]
    gives the simple harmonic motion of a body. At t = 6.2 s, what are the (a) displacement, (b) velocity, (c) acceleration, and (d) phase of the motion? Also, what are the (e) frequency and (f) period of the motion?

    2. Relevant equations

    3. The attempt at a solution

    Hi! I understand that you're supposed to plug in your t into the position function, and then take the derivative and continue plugging in. I think I'm messing up the math somehow because I keep getting the wrong answer...

    (7.4)cos(98.0177)= -0.1395
  2. jcsd
  3. Apr 18, 2014 #2
    So I went ahead and tried to derive the position function, and I feel like I did it correctly. However, I'm still getting incorrect answers when plugging in my t value.

    v(t)= -7.4sin(5[itex]\pi[/itex]t+[itex]\frac{\pi}{5}[/itex])*5[itex]\pi[/itex]
    v(6.2)= -115 m/s

    a(t)= -7.4cos(5[itex]\pi[/itex]t+[itex]\frac{\pi}{5}[/itex])*5[itex]\pi[/itex]*5[itex]\pi[/itex]
    a(6.2)= -255 m/s2
  4. Apr 18, 2014 #3


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    Homework Helper

    For a) I'm getting -5.987 m.

    For b) your derivative looks fine to me. I'm getting 68.324 m/s for the velocity though.

    c) I'm getting -1825.880 m/s^2.

    Is your calculator in degree mode perhaps?
  5. Apr 18, 2014 #4
    Haha ohh, thanks so much! That was the problem :)
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