Pendulum period

  1. An aluminum clock pendulum having a period of 1.20 s keeps perfect time at 20.0°C. When placed in a room at a temperature of 1°C, how much time will it gain every hour?

    I used T=2pi*squareroot(x/9.8) and solved for x to be .357823847 as the length of the pendulum. Then I plugged that into delta L = (alpha)(Lo)(delta T) and found the change in L to be 0.000163168. I subtracted 0.000163168 from .357823847. Then I used T=2pi*squareroot(L/g) to find the period with the new length and got 1.199726369 so that the difference between the original period and the new period would be .000273631.

    I don't understand why this is the wrong answer. Did I do it correctly and just make a math error?
  2. jcsd
  3. Integral

    Integral 7,345
    Staff Emeritus
    Science Advisor
    Gold Member

    What is your answer to the question? You have computed the difference in time for 1 oscillation of the pendulum. Now you need to find the error over 1 hour.
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