Timekeeping Changes with Temperature: Pendulum Clock at 1°C

AI Thread Summary
The discussion focuses on calculating the time gained by an aluminum pendulum clock when the temperature drops from 20.0°C to 1°C. The original period of the pendulum is 1.20 seconds, and the user calculated the new length and period after accounting for thermal expansion. They found a difference in the period of approximately 0.000273631 seconds per oscillation. However, the user is unclear about how to convert this difference into a time gain over one hour. The key takeaway is that the final step involves multiplying the period difference by the number of oscillations in one hour to determine the total time gained.
starfish794
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An aluminum clock pendulum having a period of 1.20 s keeps perfect time at 20.0°C. When placed in a room at a temperature of 1°C, how much time will it gain every hour?

I used T=2pi*squareroot(x/9.8) and solved for x to be .357823847 as the length of the pendulum. Then I plugged that into delta L = (alpha)(Lo)(delta T) and found the change in L to be 0.000163168. I subtracted 0.000163168 from .357823847. Then I used T=2pi*squareroot(L/g) to find the period with the new length and got 1.199726369 so that the difference between the original period and the new period would be .000273631.

I don't understand why this is the wrong answer. Did I do it correctly and just make a math error?
 
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What is your answer to the question? You have computed the difference in time for 1 oscillation of the pendulum. Now you need to find the error over 1 hour.
 

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