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Pendulum oscillation

  1. Aug 3, 2007 #1
    1. The problem statement, all variables and given/known data
    A student decides to use a simple pendulum to check the shutter time of her camera. A meter stick is laid horizontally, and a simple pendulum is set up so that, when it is hanging vertically, the mass falls on the 50.0 cm mark of the meter stick. The pendulum is set into oscillation such that at its extreme positions, the mass falls on the 40.0 cm and 60.0 cm graduations. The student has chosen the length of the pendulum to be 99.3 cm.

    She then takes a photography of the pendulum as it is oscillating. Her camera is set to a nominal 1/10 of second shutter time. By careful inspection of the photography, she observes that the mass moved directly from the 41.2 cm to 43.9 cm scale markings on the meter stick.

    How long does it take for the mass to move directly from the 41.2 cm to the 43.9 cm scale markings?


    2. Relevant equations
    omega = sq. root g / Length
    y = 50 cm +10 cm(sin omega t)


    3. The attempt at a solution
    I received the 2nd equation from an example my prof did in class, and I followed his steps precisely, but received an answer that was way too high--a difference of 9.5 seconds. This makes the 1/10 second marking look ridiculous. I tried a few other things, including using a cosine function in case I chose the sine function incorrectly, and I just end up with a negative value for time. I promise that I followed my prof's instructions to the letter--does anyone have any better ideas? Because clearly, my answer doesn't make sense. (And yes, my calculator is in radian mode and I double-checked my decimal points and conversions between cm and m).

    Thanks!
     
  2. jcsd
  3. Aug 3, 2007 #2
    You can use either sine or cosine, though you will find that cosine is more conventional (in this case it would require a phase angle). Now you said that you tried something, and you want a better idea, yet you never showed the work of what you tried. Did you simply put in the two distances to solve for time? Also, you mention that you are off by a factor of 10, did you convert your centimeters into meters?
     
  4. Aug 4, 2007 #3
    First, I solved for omega using the first formula. Then I inserted the values 41.2 and 43.9 (separate equations, of course) as "y", and solved for t from there. I originally had not done the conversions, but even after I did, I still came up with the same answer. My prof didn't actually show us how to use a phase angle. Were you able to come up with a reasonable answer? It's possible that I keep making the same mistake somewhere.
     
  5. Aug 4, 2007 #4

    dlgoff

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    You are starting at 50cm (when mass is hanging motionless) on the scale. So should 50 be in y = 50 cm +10 cm(sin omega t)? Wouldn't you want to use a delta y?
     
  6. Aug 5, 2007 #5
    I think you have the right idea for the most part. You have two equations, x2(t) and x1(t), and you want to find the time it took for the pendulum to travel between those two points (x2 - x1 = delta(x) = change in x).

    [tex]x_2(t) = .5 + .1sin(\sqrt{\frac{g}{.993}}t) = .412[/tex]
    [tex]x_1(t) = .5 + .1sin(\sqrt{\frac{g}{.993}}t) = .439[/tex]

    Should give you a reasonable answer for the time.

    Also: I converted the cm to meters to keep the gravitational constant in meters, but you could also just make g = 98cm/s^2, and keep the cm whole.
     
    Last edited: Aug 5, 2007
  7. Aug 5, 2007 #6
    That does look much better. Still can't figure out what I did wrong the first time. Oh well...thanks for the help!
     
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