Thermal expansion of a pendulum

AI Thread Summary
The discussion centers on calculating the maximum temperature variation a steel pendulum can withstand without delaying more than one second per day. The user initially calculated that the pendulum could be heated by up to 23°C, but later realized a mistake in applying the coefficient of thermal expansion. After correcting the coefficient to 11×10^-6°C^-1, the user found that the maximum allowable temperature increase is actually around 2.3°C. This correction aligns with the professor's feedback, confirming the accuracy of the revised calculation. The user expresses relief at identifying and rectifying the error.
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Hi,
I've done the problem but I'm unsure of my answer. I would be glad if you could check it out.

Homework Statement


A steal pendulum is considered a clock at a certain temperature.
What is the maximum variation of temperature we can submit to the pendulum if it cannot delay more than one second by day?



Homework Equations

Coefficient of dilation of steal : 11\times 10 ^{-6}°C^{-1}.



The Attempt at a Solution


First I notice that a period of the pendulum corresponds to a second. Then I calculated the number of seconds in day to be 86400.
Say it delays 1 second in a day and I want to calculate its period. We have that 86400T=86401T'. Replacing T with 2\pi \sqrt {\frac{g}{l}} then I get that l'=0.9999768523l where l is the length of the pendulum and l' the length of the heated pendulum.
Now I want to find the length it cannot overpass : l(1-0.9999768523)=0.00002314774628l.
Looking at the coefficient of dilation of steal, if I heat the pendulum by 1°C, it will grow 0.0000011l. From it, I just look and see that I can heat the pendulum up to 23°C more than it is.
To my intuition it looks too much. What do you say?
 
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LaTeX Code: 0.0000011l
Is this correct?
 
According to my assignments yes. However I just checked it up on wikipedia thanks to you. (the page is http://en.wikipedia.org/wiki/Coefficient_of_dilatation) and they give a range of 33.0 ~ 39.0 \times 10 ^{-6}. This reduce the temperature I got by 3. So it's around 7°C which makes more sense to me.
EDIT: Ah no! Sorry, it is what my assignment says, around 11\times 10 ^{-6}°C^{-1} since we're talking about the coefficient of linear thermal expansion...
 
Does the equation \delta l=L \alpha T not apply here? Well that is what I thought to use first when you found the extension.
 
Ah I made a little mistake! The coefficient of dilation of steel is 11 \times 10 ^{-6}°C^{-1}.
Quoting myself :
Looking at the coefficient of dilation of steal, if I heat the pendulum by 1°C, it will grow 0.0000011l .
should be
Looking at the coefficient of dilation of steal, if I heat the pendulum by 1°C, it will grow 0.000011l
. With this, the answer becomes 2.3 °C which is the right answer. (I asked the professor).
I'm glad I found my mistake.
 
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