- #1

NP04

- 23

- 1

- 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

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.

_{o}ΔTΔL = (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.

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