In this section real life problems provide practice solving equations with square roots.

  1. The period of the pendulum (T) is the time it takes the pendulum in seconds to move back and forth once as shown in the equation:

    where L is the length in meters and g is the acceleration due to gravity (approximately 9.8 m/sec2.)

  2. Answer the questions below.


Source: Flickr

  1. In 1851, French physicist Léon Foucault hung a 67 m pendulum from the dome of the Panthéon in Paris. Foucault's pendulum was the first simple proof that the Earth rotated in an easy-to-see experiment. What was the period of Foucault’s pendulum?
    Interactive popup. Assistance may be required.

    Check Your Answer

    T = 2π
    T = 2(π)√6.84
    T = 2π(2.615)
    T = 16.4 sec
    The period of Foucault’s pendulum is 16.4 sec. Close Pop Up
  2. If Foucault had done his experiment on the moon where gravity is 1/6th that of Earth’s, what would have been the period of his pendulum?
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    Check Your Answer

    g = 9.8 ÷ 6 = 1.633
    T = 2π
    T = 2π41.1
    T = 2π (6.41)
    T = 40.2 sec
    If the pendulum is on the moon the period is 40.2 sec. Close Pop Up
  3. If Foucault had wanted the period of his pendulum to be 10 seconds, what length would he have needed to make his pendulum? (Assume he’s still on Earth.)
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    Check Your Answer

    10 = 2π
    10 2π = 2π 2π
    10 2π =
    100 4π2 (9.8) = L 9.8 (9.8)
    980 4π2 = L
    L = 24.8 meters
    The length of the pendulum would have to be 24.8 meters to have a period of 10 sec. Close Pop Up
  4. If Foucault has a new pendulum of length x, it would have a period of 2π. If he then halves the length, the new period is 5 seconds shorter. What was the original length, x, of the pendulum?
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    Check Your Answer

    Graph - x= 72.3 meters Close Pop Up
  5. Video segment. Assistance may be required. Watch the video if you need to see a solution to this problem.

  6. Explain in your own words how you would solve the following equation for d using graphs on your graphing calculator.