A pendulum clock has a pendulum shaft made of aluminum which has a coefficient of linear expansion of 24.0 × 10^-⁶ K^-1. At 22°C the period of swing of the pendulum is 1.00 s. If the temperature is reduced to 5°C, how much time does the clock gain or lose in one day?

Question

Quizplus · Accepted Answer

The period of a pendulum is given by:$$T = 2\pi\sqrt{\frac{L}{g}}$$where L is the length of the pendulum and g is the acceleration due to gravity.If the temperature of the pendulum is changed, the length of the pendulum will change due to thermal expansion. The change in length is given by:$$\Delta L = \alpha L \Delta T$$where $\alpha$ is the coefficient of linear expansion, L is the original length of the pendulum, and $\Delta T$ is the change in temperature.In this problem, the coefficient of linear expansion is $24.0 	imes 10^{-6} K^{-1}$, the original length of the pendulum is not given, the original temperature is $22°C$, and the final temperature is $5°C$. The change in temperature is:$$\Delta T = 5°C - 22°C = -17°C$$The change in length is:$$\Delta L = (24.0 	imes 10^{-6} K^{-1}) (L) (-17°C) = -4.08 	imes 10^{-4} L$$The new length of the pendulum is:$$L' = L + \Delta L = L - 4.08 	imes 10^{-4} L = 0.9996 L$$The new period of the pendulum is:$$T' = 2\pi\sqrt{\frac{L'}{g}} = 2\pi\sqrt{\frac{0.9996 L}{g}} = 0.9998 T$$The change in period is:$$\Delta T = T' - T = 0.9998 T - T = -0.0002 T$$The change in period per day is:$$\Delta T_{day} = \Delta T 	imes 86400 s/day = -0.0002 T 	imes 86400 s/day = -18 s/day$$Therefore, the clock loses 18 s per day.

A Pendulum Clock Has a Pendulum Shaft Made of Aluminum