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Thermal expansion and keeping time

  1. Jan 18, 2006 #1
    HI, this problem is supposed to be easy but I dont get it:

    The coefficient of thermal expansion for Invar is 0.70*10-6K-1.
    ( Invar is a steel alloy). Given that an Invar pendulum clock keeps perfect time at a room temperature of 20 degree celcius, how much time will the clockgain or lose per day when it is at room temperature of 30 degree celcius?

    can I have some suggestion ?
    Thank you
  2. jcsd
  3. Jan 19, 2006 #2
    Pendulum period T is a measure of time assuming that the period remains constant. Say one swing of the pendulum is equal to one second. Period of a pendulum is given by T = 2 pi sqrt(L/g)

    If the room temp. increased, then the string length is going to increase due to thermal expansion, and as a result, T will increase (say new T is T')

    That means, it will take a little longer than one second for your pendulum to make one full swing. This means your pendulum clock is said to be slow. Slow by T' - T seconds each second. Or you loose T'-T seconds each second. So in 24 hours how many seconds would you have lost?

    (T'-T) 24*60*60.

    T' = 2pi sqrt(L'/g). L' can be found from the given info.
  4. Jan 19, 2006 #3

    Thank you Gamma,
    I understand your explanation.
    But I have some issues to find the value of L'= L+dL.

    With the data I have I found:

    I still have L as unknow and I dont know how to get rid of it.

    Can you help me on that?
    thank you
  5. Jan 19, 2006 #4
    Normally thermal expansion is in units length per units temperature.

    what is this?: 0.70*10-6K-1.

    do you mean .7*10-6 m/K?
  6. Jan 19, 2006 #5
    T = 2 pi sqrt(L/g)

    T= 1 sec. You can find L
  7. Jan 20, 2006 #6
    Thanks a lot
  8. Jan 20, 2006 #7


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    Homework Helper

    Homer - thermal expansion is a fractional amount per degree :
    dL/L = coefficient * dT , so the coefficient only has units 1/K .
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