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Length of a String - Pendulum

  1. Jan 12, 2009 #1
    1. The problem statement, all variables and given/known data
    A 125 g ball is tied to a string. It is pulled to an angle of 4.80 degrees and released to swing as a pendulum. 12 oscillations take 9.00 s.

    Please help, I'm not sure which formulas to use
  2. jcsd
  3. Jan 12, 2009 #2

    Doc Al

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    Staff: Mentor

    What do you know about pendulums? What's the formula for the period of a pendulum?
  4. Jan 12, 2009 #3
    Period (T) = 1/f
    T = 2π √(l/g)
  5. Jan 12, 2009 #4

    Doc Al

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    That's true, but that's true in general, not just for pendulums. What formula allows you to calculate the period of a pendulum from its physical properties, such as its length?

    That's the one.

    Hint: Use the data provided to find the period. Then solve for the unknown length.
  6. Jan 12, 2009 #5
    I did that, but I wasn't sure if I got the correct answer.

    T = 12/9 = 1.33333

    1.3333(2π)= √(L/9.8)
    squared the left side and multiplied it by 9.8 and..

    I eventually got 687.07m
  7. Jan 12, 2009 #6

    Doc Al

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    Careful. The period is the time for one oscillation, so it should be total time divided by the number of oscillations. (You have them switched.)
  8. Jan 12, 2009 #7
    ah, I see.


    I haven't had any practice in physics for a long time, so I'm trying to re-learn everything.
  9. Jan 12, 2009 #8
    I just realized that I didn't use the mass or degree of angle in solving the question.

    Is there a way to answer this question using another equation?
  10. Jan 12, 2009 #9

    Doc Al

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    Perhaps the period does not depend on either of those variables. :wink:

    Are there are more parts to this question?
  11. Jan 12, 2009 #10
    no, this was the entire question.

    I guess the mass and angle is there to play mind tricks.
  12. Jan 12, 2009 #11
    Another question:

    Astronauts on the first trip to Mars take along a pendulum that has a period on earth of 1.50 s. The period on Mars turns out to be 2.45 s.

    My answer:

    I used the equation T=2pi √ (L/g)
    I first plugged in the 1.5s = 2pi √ (L/9.8) and found L = .5585

    Then I plugged in 2.45s = 2pi √(.5585/g) to solve for g of mars.

    Now I got 3.6733 m/s2 = g for mars.

    Can someone correct me if I'm wrong or if there is an easier way to solve this problem?
  13. Jan 13, 2009 #12

    Doc Al

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    Yours is a perfectly fine way to solve this problem.

    Here's how I would solve it:

    [tex]T_e = 2\pi \sqrt{L/g_e}[/tex]

    [tex]T_m = 2\pi \sqrt{L/g_m}[/tex]

    Divide the two and square:
    [tex](T_e/T_m)^2 = g_m/g_e[/tex]

    [tex]g_m = g_e(T_e/T_m)^2[/tex]
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