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March 25, 2017

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Imagine you are an astronaut who has landed on another planet and wants to determine the free-fall acceleration on that planet. In one of the experiments you decide to conduct, you use a pendulum 0.47 m long and find that the period of oscillation for this pendulum is 1.43 s. What is the acceleration due to gravity on that planet? (Enter the magnitude only.)

I know the formula to use is
T=2pi(L/g)^0.5(g) but I keep getting a wrong answer

  • Physics - ,

    T = 2 pi sqrt(L/g)
    1.4 = 6.28 sqrt (.47/g)
    .47/g = .0497
    g = 9.46 m/s^2

  • Physics - ,

    It still says its the wrong answer. Let me calculate it again

  • Physics - ,

    That assumes that you know about simple pendula

    small angle A at top, g and L and m
    KE at bottom = (1/2) m V^2
    V is max speed at bottom = L d angle/dt
    Pe at top = m g h = m g L(1- cos A)
    for small A, cos A = 1 - A^2/2 + ....

    m g L (A^2/2) = (1/2) m V^2

    g L A^2 = V^2
    now assume sinusoidal motion
    angle = A sin (2 pi/T) t
    d angle/dt = (2 pi/T) A cos (2 pi/T)t
    x max = A
    v max = L d angle/dt = (2 pi L/T)A
    then
    g L A^2 = (2 pi L/T)^2 A^2

    g/L = (2 pi/T)^2
    T^2 = (2 pi)^2 L/g
    T = 2 pi sqrt (L/g)

  • Physics - ,

    I used .14 not .143

  • Physics - ,

    1.43 = 6.28 sqrt(.47/g)
    try 9.06 m/s^2

  • Physics - ,

    For some reason its still saying its the wrong answer

  • Physics - ,

    Your saying g=9.06

  • Physics - ,

    yes

  • Physics - ,

    Yes, it finally worked thank you so much.

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