An invar pendulum clock keeps a period of 0.5 s at 20 degC. The clock is then run where the average temperature is 30 degC. Over an interval of 10⁶ s, by approximately how much, and in which sense, does the clock run wrong? (alpha_invar = 1 * 10⁻⁶ /degC)

0.5 s fast
0.5 s slow
5 s fast
5 s slow

Correct answer: 5 s slow

Solution

Heating lengthens the pendulum, increasing the period, so the clock runs slow by (1/2)*alpha*deltaT*t = (1/2)(1e-6)(10)(1e6) = 5 s.

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