Thermal expansion and keeping time

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Homework Help Overview

The discussion revolves around a problem related to thermal expansion and its effect on the timekeeping of an Invar pendulum clock. The original poster seeks assistance in understanding how changes in temperature affect the clock's accuracy, specifically how much time it will gain or lose when the temperature increases from 20 degrees Celsius to 30 degrees Celsius.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the pendulum's period and its length, noting that an increase in temperature leads to thermal expansion, which in turn affects the period of the pendulum. There are inquiries about how to calculate the new length of the pendulum and the implications for timekeeping. Some participants express confusion regarding the units of the coefficient of thermal expansion.

Discussion Status

The conversation includes various interpretations of the problem and attempts to clarify the mathematical relationships involved. Some participants have provided guidance on how to approach the calculations, while others are still grappling with the implications of the thermal expansion coefficient and its application to the problem.

Contextual Notes

There is a noted uncertainty regarding the units of the coefficient of thermal expansion, with participants questioning its expression and how it relates to the calculations needed for the problem. Additionally, the original poster is struggling with how to express the new length of the pendulum in terms of the original length and the change due to thermal expansion.

brad sue
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HI, this problem is supposed to be easy but I don't 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 celsius, how much time will the clockgain or lose per day when it is at room temperature of 30 degree celsius?


can I have some suggestion ?
Thank you
 
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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.
 
Gamma said:
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.


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:
L=dL*(142857.14)

then
L+dL=(1.000007)*L
I still have L as unknow and I don't know how to get rid of it.

Can you help me on that?
thank you
B
 
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?
 
T = 2 pi sqrt(L/g)

T= 1 sec. You can find L
 
Gamma said:
T = 2 pi sqrt(L/g)

T= 1 sec. You can find L

Thanks a lot
 
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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