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Simple Harmonic Motion & Pendulum Problem

  1. Feb 10, 2015 #1
    1. The problem statement, all variables and given/known data
    What length must the pendulum be changed to in order to show the correct time?
    L=0.5m
    After 12hours the clock is behind by 30minutes.

    2. Relevant equations
    w=sqrt(g/L)
    w=2πf

    3. The attempt at a solution
    I thought if I set the frequency equal to 1Hz and solved for length it would work.

    2πf=sqrt(g/L)
    4π^2=g/L <-got rid of f since its equal to 1
    L=g/(4π^2)
    This unfortunately doesn't give the correct answer, I also tried making a ratio involving the time but that failed miserably.

     
  2. jcsd
  3. Feb 10, 2015 #2

    ehild

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    Why should be the frequency of the pendulum 1 Hz?
     
  4. Feb 10, 2015 #3
    I thought that if frequency is 1Hz then it would be doing one cycle per second which would result in 1 tick of the clock per second but since its for the entire cycle and each cycle contains 2 ticks? So I should set frequency to 0.5Hz?
     
  5. Feb 10, 2015 #4

    ehild

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    The swing of the pendulum drives the hands of the clock through some rather complicated mechanical system, including cogwheels. http://visual.merriam-webster.com/s...easure-time/weight-driven-clock-mechanism.php. The time period of the bob is proportional to the period of the hands of the clock, but is not equal to it. In case of a 0.5 m pendulum, the time period is 1.41 s. But the transmission system makes the clock going too slow - by what percent? By what percent have you change the time period of the pendulum, and what does it mean on the length?
     
  6. Feb 11, 2015 #5
    tclock=41 400s
    t=43 200s
    tclock/t=23/24

    The clock is going (1/24)% slower than the actual time, to fix this I should multiply the period by (23/24)?
    T=1.41*(23/24)≈1.351

    Then w=sqrt(g/L) ->w=2pi/T
    (2pi/T)^2=g/L
    L=g/(2pi/T)^2=0.453m

    Thanks!!!
     
  7. Feb 11, 2015 #6

    ehild

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    The method is correct, but you but you made it very complicated, and rounded off too early.
    You know that Tcorrect/T = 23/24. Tcorrect/T = sqrt(Lcorrect/0.5). How do you get the correct length from here?
     
  8. Feb 11, 2015 #7
    Yea that is definitely an easier way to solve for Lcorrect but I'm curious as to how you found this relationship between the period and length.
     
  9. Feb 11, 2015 #8

    Nathanael

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    [itex]\frac{T_a}{T_b}=\frac{2\pi\sqrt{L_a/g}}{2\pi\sqrt{L_b/g}}=\sqrt{\frac{L_a}{L_b}}[/itex]
     
  10. Feb 11, 2015 #9

    ehild

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    You know how the formula ##ω=\sqrt{\frac{g}{L}}## was derived and you know the relation between angular frequency and time period: ##\omega=\frac{2\pi}{T}##.
    So ##\frac {2\pi}{T}=\sqrt{\frac{g}{L}}##,
    isolate T:
    ##{T}=2\pi\sqrt{\frac{L}{g}}##
     
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