Thermal expansion and keeping time

AI Thread Summary
The discussion revolves around calculating the time gained or lost by an Invar pendulum clock due to thermal expansion when the temperature increases from 20°C to 30°C. The coefficient of thermal expansion for Invar is given as 0.70*10^-6 K^-1, indicating that the pendulum's length will increase, resulting in a longer period for each swing. This leads to the clock running slow, as each swing takes more time than one second. Participants discuss the formula for the pendulum's period and how to derive the new length after thermal expansion, but there is confusion regarding the calculations involving the unknown length L. Ultimately, the conversation emphasizes understanding the relationship between temperature change and pendulum length in the context of timekeeping.
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 celcius, how much time will the clockgain or lose per day when it is at room temperature of 30 degree celcius?


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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