Change in Pendulum's Period with temperature.

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.
  • #1
rmorelan
8
0
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?
 
Physics news on Phys.org
  • #2
rmorelan said:
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"
 
Last edited by a moderator:
  • #3
Great! Thanks, how do to anything with that sqrt term was driving me nuts!
 

1. How does temperature affect the period of a pendulum?

The period of a pendulum is directly proportional to the square root of its length and inversely proportional to the square root of the acceleration due to gravity. Since temperature affects the length of the pendulum and the acceleration due to gravity, it can indirectly affect the period of a pendulum.

2. Does a change in temperature always result in a change in the period of a pendulum?

No, a change in temperature does not always result in a change in the period of a pendulum. It depends on the material and properties of the pendulum, as well as the magnitude of the change in temperature.

3. How does the material of the pendulum affect its period with temperature?

The material of the pendulum can affect its period with temperature in two ways. First, the coefficient of thermal expansion of the material can affect the length of the pendulum and therefore, its period. Second, the material's thermal conductivity can affect the heat transfer between the pendulum and its surroundings, potentially altering its period.

4. Is there a specific temperature range where the period of a pendulum is most affected?

Yes, the period of a pendulum is most affected by temperature changes when the pendulum is made of materials with high coefficients of thermal expansion and is exposed to large temperature variations. In general, the period may be affected more at extreme temperatures rather than at moderate temperatures.

5. Can the period of a pendulum be controlled to remain constant despite changes in temperature?

Yes, the period of a pendulum can be controlled to remain constant despite changes in temperature by using materials with low coefficients of thermal expansion, insulating the pendulum from temperature variations, and using temperature compensation mechanisms such as mercury-filled pendulums.

Similar threads

  • Introductory Physics Homework Help
Replies
14
Views
437
  • Introductory Physics Homework Help
Replies
20
Views
973
  • Introductory Physics Homework Help
Replies
12
Views
293
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
636
  • Introductory Physics Homework Help
Replies
23
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
26
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
853
  • Introductory Physics Homework Help
Replies
15
Views
1K
Back
Top