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Pendulum on earth and mars

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  1. Sep 11, 2014 #1
    1. The problem statement, all variables and given/known data
    A pendulum has a period of 4.50s on Earth. What is its period on Mars, where the acceleration of gravity is about 0.37 that on Earth? (2 sig figs)


    2. Relevant equations
    T= 2Pi(L/g)^.5


    3. The attempt at a solution

    first must solve for the length on earth

    L = g(T/2Pi)^2 = (9.8m/s^2)(4.5s/2Pi)^2 =489.7m

    now that i know the length i can take the same pendulum to mars and solve for T on mars

    T= 2Pi(L/.37g)^.5 = 2Pi(489.7m/(.37*9.8m/s^2))^.5 = 73 s

    mastering physics says I am wrong. What did I miss?
     
  2. jcsd
  3. Sep 11, 2014 #2
    Your length is wrong. Check the units.

    However you don't need the length at all. It's enough to calculate the ratio of the two periods, knowing the ratio of the two values of gravity.
     
  4. Sep 11, 2014 #3
    g is m/s^2 T is s and since its being squared it becomes s^2 which cancels with the 1/s^2 in g and leaves m. unless i am mistaken?
     
  5. Sep 11, 2014 #4

    HallsofIvy

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    It isn't necessary to solve for the length. Saying that T= 2Pi(L/g)^.5 means that the period is inversely proportional to the square root of g. It g on Mars is 0.37 g on earth then the period, on Mars, of the same pendulum is [tex]\frac{1}{\sqrt{0.37}}[/tex] times the period on earth.
     
  6. Sep 11, 2014 #5
    I thought you may have that value in cm, it would be reasonable.

    Then check your calculations. 4.5 divided by 2pi is less than one. When you square it is still less than one.
    How can you get over 400 when multiplied by 9.8?
     
  7. Sep 12, 2014 #6
    It was a calculator mistake on my part, sorry guys. In your way of doing it can you ignore 2 , Pi and L because they don't change?
     
  8. Sep 12, 2014 #7

    ehild

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    If you write the formula both for TMoon and TEarth and divide them, everything cancels but the g-s.

    ehild
     
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