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Simple Harmonic Motion and free-fall accelerations

  1. Apr 19, 2005 #1
    A "seconds" pendulum is one that goes through its equilibrium position once each second. (The period of the pendulum is 2.000 s.) The length of a seconds pendulum is 0.9927 m at Tokyo and 0.9942 m at Cambridge, England. What is the ratio of the free-fall accelerations at these two locations?
    Cambridge / Tokyo) = 1 +?

    How do I work this problem? Any help would be appreciated.
     
  2. jcsd
  3. Apr 19, 2005 #2
    As you observe, in Cambridge the value of g is higher than in Tokyo (g_C/g_T = 1 + ?).

    Knowing that the periods are equal, then:

    [tex]\frac{T_C}{T_T} = 1[/tex]

    or


    [tex]\frac{2\pi \sqrt{\frac{l_C}{g_C}}}{2\pi \sqrt{\frac{l_T}{g_T}}} = 1[/tex]

    You can go on to find g_C/g_T in terms of the ratio l_C/l_T.
     
  4. Apr 19, 2005 #3
    Are you familiar with this formula?

    [tex]T = 2\pi \sqrt\frac{L}{g}[/tex]

    [tex]T[/tex] is the peroid, so set it equal to two and solve for the value [tex]g[/tex] with your given [tex]L[/tex] value.
     
  5. Apr 19, 2005 #4
    It is neater to express the ratio of free fall accelerations in terms of the ratio of the 1-sec pendulum lengths, than finding the free fall acceleration in each location and then computing the ratio.

    Indeed the conclusion will be that:

    [tex]\frac{g_C}{g_T} = \frac{l_C}{l_T} = \frac{0.9942}{0.9927} = 1 + 1.51 * 10^{-3}[/tex]
     
  6. Apr 19, 2005 #5
    Yes, yours is a much better way to compute it. I dislike dealing with ratios throught out a whole problem so I attempted it a different way.
     
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