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Simple Harmonic Oscillator Problem

  1. Jan 10, 2013 #1
    1. The problem statement, all variables and given/known data

    The position of a mass that is oscillating on a Slinky (which acts as a simple harmonic oscillator) is given by 18.5 cm cos[ 18.0 s-1t]. What is the speed of the mass when t = 0.360 s?
    2. Relevant equations
    x(t)=Acos(ωt+θ)
    v(t)=-Aωsin(ωt+θ)


    3. The attempt at a solution

    I used the formula, v(t)=-Aωsin(ωt+θ) because you basically have everything you need such as:
    A=18.5cm
    ω=18.0s^-1
    t=0.360s

    What I get is:
    v(0.360)=-(18.5cm)(18.0)sin(18.0*0.360)
    to get -65.1cm/s
    which isn't the right answer.
    Please tell me where I went wrong!
     
  2. jcsd
  3. Jan 11, 2013 #2

    Borek

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    Speed is never negative.
     
  4. Jan 11, 2013 #3

    rude man

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    The highlighted formula makes little sense to me. First, the argument of the cos must be dimensionless but here it looks like it's time. 18.0 s must be the phase term but what's with the s? Is 18.0 in radians, deg or ??? The "1" in front of t must be √(k/m), k = spring const. & m = mass, aka ω. The phase is due to the fact that this mass had initial velocity and displacement. Anyway, no way do I see that ω = 18.0.

    Weird! I guess you could go

    x = 18.5cos(18 - t) cm
    x' = -18.5sin(18 - t) cm/s since I guess ω = 1 rad/s;
    so x'(t=0.36) = -18.5sin(18 - 0.36) = -18.5sin(17.64) cm/s.

    BTW v can be negative. Speed can't.
     
  5. Jan 12, 2013 #4

    ehild

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    v=-65.1 cm/s is the velocity. The speed is magnitude of velocity.

    ehild
     
  6. Jan 12, 2013 #5

    Borek

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    I believe the formula is just misformatted and is intended to be [itex]\cos(18s^{-1}\times t)[/itex] - so there is no problem with the units.
     
  7. Jan 13, 2013 #6
    Thanks a lot! I got it!
     
    Last edited: Jan 13, 2013
  8. Jan 13, 2013 #7

    rude man

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    OK. What can be done to get the OPs to accurately state the problem, I wonder in my oft leisure moments ...
     
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