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Math Help - adjusting a clock

  1. #1
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    adjusting a clock

    A pendulum clock can be approximated as a simple pendulum of length 1.14 m and keeps accurate time at a location where g = 9.90 m/s2. In a location where g = 9.75 m/s2, what must be the new length of the pendulum, such that the clock continues to keep accurate time (that is, its period remains the same)?
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by lovinhockey26 View Post
    A pendulum clock can be approximated as a simple pendulum of length 1.14 m and keeps accurate time at a location where g = 9.90 m/s2. In a location where g = 9.75 m/s2, what must be the new length of the pendulum, such that the clock continues to keep accurate time (that is, its period remains the same)?
    if i remember it right, T= 2\pi \sqrt{\frac{L}{g}} where L is the length of the pendulum, g is the gravity and T is the period.

    thus, if you want to solve for the length of the pendulum that will yield the same period at different gravity,

    2\pi \sqrt{\frac{L_1}{g_1}} = 2\pi \sqrt{\frac{L_2}{g_2}}
    or \sqrt{\frac{L_1}{g_1}} = \sqrt{\frac{L_2}{g_2}}
    or just \frac{L_1}{g_1} = \frac{L_2}{g_2}.. and just plug in the values correspondingly..
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