# Change in Pendulum's Period with temperature.

• rmorelan
In summary, the relationship between the effect of temperature increase and the change in period of a pendulum can be approximated by using a binomial expansion.f

#### rmorelan

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?

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

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

p.s. is there a summary list for the symbol set this forum uses to generate symbols like pi, square roots etc?
The best way to write equations is using Latex: https://www.physicsforums.com/showthread.php?t=8997"

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Great! Thanks, how do to anything with that sqrt term was driving me nuts!