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Oscillations and Physical Pendulum help

  1. Jul 23, 2008 #1
    1) SOLVED. In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the following expression, where x is in centimeters and t is in seconds. x = (7.00 cm) cos(5t + π/8). (a) at t=0, find the position (b) velocity (c) acceleration

    For part a, I tried plugging in 0 for t, giving 7cos(0 + pi/8), which equals 6.999 or 7? That's the only thing I know to do, and this is not the correct answer

    For b and c, I tried plugging in 0 to the equations v = -wAsin(wt + [tex]\phi[/tex]) and a = -w2Acos(wt + [tex]\phi[/tex]), but once again no luck.

    2) A very light rigid rod with a length of 1.81 m extends straight out from one end of a meter stick. The stick is suspended from a pivot at the far end of the rod and is set into oscillation. Ip = ICM + MD2 (a) Determine the period of oscillation. [Suggestion: Use the parallel-axis theorem equation given. Where D is the distance from the center-of-mass axis to the parallel axis and M is the total mass of the object.]

    [tex]T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{I}{mgd}}[/tex] is what I know I need to be using, but I'm not sure what to plug in for I. I tried plugging in [tex]\frac{1}{3}ML^{2}[/tex] for I, giving me [tex]2\pi\sqrt{\frac{\frac{1}{3}ML^{2}}{mgd}}[/tex]
    The mass cancelled out and I changed d to L, which cancelled out and reduced L2 to L. When I plugged in the numbers I didn't get the right answer. Is my value for d wrong or is it the I as a whole?

    Any help is appreciated.
    Last edited: Jul 23, 2008
  2. jcsd
  3. Jul 23, 2008 #2


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    Check to see if your calculator is in radian or degree mode! pi/8= .392... degrees would be very close to 0 so cos(pi/8) would be very close to 1. cosine of pi/8 radians however, is not 1 so 7 cos(pi/8) is not 7.

    Again, put your calculator into radian mode.

  4. Jul 23, 2008 #3
    You were right I had my calculator in degrees, thanks! #1 is down, now all I need help on is #2!
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