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SHM and angles

  1. Nov 8, 2015 #1
    1. The problem statement, all variables and given/known data
    A 100g mass on a 1.0m long string is pulled 8.0 degrees to one side and released. How long does it take for the pendulum to reach 4.0 degrees on the opposite side?

    2. Relevant equations
    ##T = 2\pi \sqrt\frac{L}{g}##
    ##x(t) = A\cos\omega t##

    3. The attempt at a solution
    From the simple pendulum we get ##\omega = \sqrt\frac{g}{L}## which leads to:

    ##x(t) = A\cos\sqrt g t##

    ##A = 8.0## and ##x(t) = 4.0## can be substituted directly since it results in the cosine of an angle then:

    ##\cos\sqrt g t = -0.50##

    ##t\sqrt g = \frac{4\pi}{3}##

    ##t = 1.3s##

    Which happens to be twice the actual answer. The only possibility for error I see is ##\arccos(-0.5)##.
    What I don't see is why. My reasoning was that since we're looking at 4.0 degrees on the opposite side, the angle must be the second point where cosine is negative, ie. ##\frac{4\pi}{3}##. What am I missing here?
     
  2. jcsd
  3. Nov 8, 2015 #2

    PeroK

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    You need to be careful working in degrees.

    I think you've identified the error. How did you evaluate the inverse cosine?
     
  4. Nov 8, 2015 #3
    Because in this case it makes no difference since the division results in a dimensionless figure.
     
  5. Nov 8, 2015 #4

    PeroK

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    Yes, but I'd be careful. It's an easy way to go wrong. If you draw a graph of cos, you'll see where you've gone wrong.
     
  6. Nov 8, 2015 #5
    There are two possibilities for ##-0.5##, either ##\frac{2\pi}{3}## or ##\frac{4\pi}{3}##. I took the first figure to equate to 4.0 degrees on the side on which the swing started.
     
  7. Nov 8, 2015 #6

    PeroK

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    Wouldn't that be where ##cos = 0.5##?
     
  8. Nov 8, 2015 #7
    I'm sure that's not the case.
     
  9. Nov 8, 2015 #8

    PeroK

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    Are you sure you're sure?

    Can you say where the pendulum is when?

    ##cos(\omega t) = 1##
    ##cos(\omega t) = 0.5##
    ##cos(\omega t) = 0##
    ##cos(\omega t) = -0.5##
    ##cos(\omega t) = -1##

    (All for the first time)
     
  10. Nov 8, 2015 #9
    I just noticed there's a typo in my initial post. It's supposed to be ##x(t) = -4.0## That's where the negative sign comes from.
     
  11. Nov 8, 2015 #10
    I see what you mean here. The pendulum in question hasn't even completed one half of a full cycle. ##\frac{4\pi}{3}## is when it reaches 4.0 for the second time.
     
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