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Urgent help with Period of a pendulum!

  1. Nov 12, 2012 #1
    #7. Number7.png

    This is what i tried to do... but I got it wrong :( Please help

    T ≈ 2π√(L/g)

    1.31 = 2π√(L/9.81)
    1.31/2π = √(L/9.81)
    2.057 = √(L/9.81)
    2.057² = L/9.81
    4.2343 = L/9.81
    4.2343(9.81) = L
    41.54/1000 = L
    0.0415 = L

    "INCORRECT: Your response differs from the correct answer by more than 100%. "
     
  2. jcsd
  3. Nov 12, 2012 #2
    Wow where is everybody?
     
  4. Nov 12, 2012 #3
    Hello?
     
  5. Nov 12, 2012 #4

    SammyS

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    One clue is the mention of the moment of inertia.

    Another is that it's a physical pendulum.

    Also,
    Don't bump your thread until it's been here for 24 hours.​
     
  6. Nov 12, 2012 #5
    I =T2MgD/4π2
    I = [(1.312)(0.246)(9.81)(18)]/39.478
    I = 74.545/39.478
    I = 1.8882

    "INCORRECT"

    -___-
    AKFJDJFDAKLNDJK
     
  7. Nov 12, 2012 #6

    SammyS

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    Convert centimeters to meters.
     
  8. Nov 12, 2012 #7
    Oh...

    0.0188

    "CORRECT"

    -.- The day I graduate from Webassign...........

    Thank you...

    What about this last one? I have no idea where to start/what equation to use.

    #3. A 40.0-g object is attached to a horizontal spring with a force constant of 10.0 N/m and released from rest with an amplitude of 20.0 cm. What is the velocity of the object when it is halfway to the equilibrium position if the surface is frictionless?
    m/s
     
  9. Nov 12, 2012 #8

    SammyS

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    Describe the motion of the object after being released.
     
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