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Change in Pendulum's Period with temperature.

  1. Feb 15, 2007 #1
    Hi! I have been racking my brain trying to solve the relationship between the effect of temperature increase and the change in period of a pendulum. I am suppose to prove a relationship, and am given that the relationship is

    delta(P) = 1/2*alpha*Pinitial

    where alpha is the coefficient of linear expansion with temperature.

    We are given that we are dealing with a simple pendulum, so the period is equal to 2*pi*sqrt(L/g)

    I know the pendulum length will change under change of temperature to alpha*L*delta(T), for a total new length of L+alpha*L*delta(T)

    the change in period then I would guess is:

    delta(P) = 2*pi*sqrt((L+alpha*L*delta(T)/g) - 2*pi*sqrt(L/g)

    I have tried to simplify this a dozen different ways but I never seem to get the required end formula. I think I can pull out Pinital by dividing everything by 2*pi*sqrt(L/g) to get

    delta(P) = Pinitial*(sqrt(1+alpha*delta(T))-1)

    but am stuck after that.

    thanks for any help!

    p.s. is there a summary list for the symbol set this forum uses to generate symbols like pi, square roots etc?
  2. jcsd
  3. Feb 15, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    That relationship isn't quite right. How can the change in period be independent of change in temperature? (A delta(T) is missing.)

    You've got it. Hint: Approximate the answer using a binomial expansion.

    The best way to write equations is using Latex: Introducing LaTeX Math Typesetting
  4. Feb 16, 2007 #3
    Great! Thanks, how do to anything with that sqrt term was driving me nuts!
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