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Motion of Pendulum Exercise

  1. Dec 1, 2007 #1
    1. The problem statement, all variables and given/known data

    An aluminum clock pendulum having a period of 1.00 s keeps perfect time at 20 degrees celcius. (A) When placed in a room at a temperature of -5.0 Celcius, will it gain or lose time? (B)How much time will it gain or lose every hour.

    2. Relevant equations

    I am pretty sure that this has to due with thermal expansion and motion of a pendulum so:

    (A) Delta L = (alpha) (Lo) (delta T)
    (B) T = 2pi *sqrt(L/G)

    3. The attempt at a solution

    I am thinking that the pendulum itself would shrink, therefore making it gain speed, but I just don't know how to set it up correctly??? Any help guys and gals?

  2. jcsd
  3. Dec 1, 2007 #2
    What is the length of the pendulum once it has changed due to temperature?
  4. Dec 1, 2007 #3
    If I assume length = 1 m, then I come up with a Delta L of .0006 m
  5. Dec 1, 2007 #4
    You don't really need to make assumptions- just do it algebraically.
    Then what is the period of your new pendulum?
  6. Dec 1, 2007 #5
    ok, so :

    T = (2pi) *Sqrt (.9994/9.8)

    t = 2.006 s

    So it loses time?
  7. Dec 1, 2007 #6
    Well like I said, I'd do it algebraically (i.e. in terms of L). But that's the next step yes.
  8. Dec 1, 2007 #7
    How would you do it algebraically? Would you solve in terms of Lo ?
  9. Dec 1, 2007 #8
    Just replace 1-0.0006 with Lo-delta L.
  10. Dec 1, 2007 #9
    T = (2pi) *Sqrt (Lo-delta L /9.8) ???

    How do I find out how much time it loses? Or would it simply be .0006 sec/min
  11. Dec 2, 2007 #10
    Having read your question again... you know the original period of your pendulum! Divide T' (the period of your new pendulum) by T (the period of your old pendulum) and see what happens. Use no numbers (except 2pi) until you have a formula for T' in terms of T.
  12. Dec 3, 2007 #11
    So would it be :

    (2 Pi * sqrt ( l / g)) / 1 sec ?????
  13. Dec 3, 2007 #12
    Think about your equations for T and T'. The ratio T' to T(i.e. T' over 1...) will be equal to the ratio of the RHSs of the two equations.
  14. Dec 3, 2007 #13
    What are you referring to when you say RHSs ?
  15. Dec 3, 2007 #14
    Right Hand Side. Sorry-that's what too much maths does to you...
  16. Dec 3, 2007 #15
    I was getting really confused with the algebra, so I just solved for the original length, Lo. Doing this I found it to be .25 meters.

    After that I used the equation :

    Delta L = (alpha) (Lo) (delta T), to find the change in length of the aluminum.

    Delta L = (24 X 10 ^ -6) ( .25) (-25)

    Delta L = -.00015 m

    So, the new lenth was .24985 m

    I then placed that in the equation for a period and found it to be :

    T = 2 pi * sqrt (.24985)/(9.8)

    T = 1.003 seconds.

    Therefore, the clock slows by .003 seconds per period.

    .003 X 3600 s in one hour = Loses 10.8 seconds per hour

    Is this correct?
  17. Dec 3, 2007 #16
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