Linear Thermal Expansion with respect to time

[NP04]

Homework Statement

The length of an aluminum pendulum in a certain clock is 0.2480 m on a day when the temperature is 23.30°C. This length was chosen so that the period of the pendulum is exactly 1.000s. The clock is then hung on a wall where the temperature is -5.00°C and set to the correct local time. Assuming the acceleration due to gravity is the same at both locations, by how much time is the clock incorrect after one day at this temperature?

Relevant Equations

ΔL= αLoΔT
α for aluminum = 23 * 10^-6

ΔL= αL_oΔT
ΔL = (23*10^-6)(0.2480 m)(28.30°C)
ΔL = 1.61 * 10^-4

Period is 1s, so each second the pendulum moves 1.61 * 10^-4 m
1.61 * 10^-4 m/s *(60s/1min)*(60min/1hr)*(24hr/1day) = 13.95 m/day

T = s (1:1 ratio)
13.95 seconds. But the answer is actually 69.3 s. Is the equation T = 2π√L/g necessary for this problem? I don't see how it could be used. Please help.

 

[PeroK]

Homework Statement:: The length of an aluminum pendulum in a certain clock is 0.2480 m on a day when the temperature is 23.30°C. This length was chosen so that the period of the pendulum is exactly 1.000s. The clock is then hung on a wall where the temperature is -5.00°C and set to the correct local time. Assuming the acceleration due to gravity is the same at both locations, by how much time is the clock incorrect after one day at this temperature?
Homework Equations:: ΔL= αLoΔT
α for aluminum = 23 * 10^-6

ΔL= αL_oΔT
ΔL = (23*10^-6)(0.2480 m)(28.30°C)
ΔL = 1.61 * 10^-4

Period is 1s, so each second the pendulum moves 1.61 * 10^-4 m
1.61 * 10^-4 m/s *(60s/1min)*(60min/1hr)*(24hr/1day) = 13.95 m/day

T = s (1:1 ratio)
13.95 seconds. But the answer is actually 69.3 s. Is the equation T = 2π√L/g necessary for this problem? I don't see how it could be used. Please help.

The idea is that if the length of the pendulum is different at different temperatures, then its period is different at different temperatures.

 

[Cutter Ketch]

Is the equation T = 2π√L/g necessary for this problem?

Yes. Note that it shows that the period is proportional to $\sqrt L$. You already figured out how much the length changed. How did the period change?

 

[haruspex]

Period is 1s, so each second the pendulum moves 1.61 * 10^-4 m

Huh? Moves that far which way in one second?
It gets shorter by that amount over the day, assuming it comes to ambient temperature by then.
What is unknown is how quickly it comes to ambient temperature. I think you will have to assume it is immediate.