A clock pendulum made of aluminum, which has a coefficient of linear expansion of 24 × 10^-⁶/K, has a period of exactly 1 s at 20^oC. Before a homeowner leaves town for one week, he turns the thermostat down to 10^oC. When he returns, the clock is

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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 changes, 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, and $\Delta T$ is the change in temperature.In this case, the coefficient of linear expansion is $24 	imes 10^{-6}/K$, the original length is not given, and the change in temperature is $-10^oC$.The change in length is:$$\Delta L = (24 	imes 10^{-6}/K) 	imes L 	imes (-10^oC) = -2.4 	imes 10^{-4}L$$The new length of the pendulum is:$$L' = L + \Delta L = L - 2.4 	imes 10^{-4}L = 0.99976L$$The new period of the pendulum is:$$T' = 2\pi\sqrt{\frac{L'}{g}} = 2\pi\sqrt{\frac{0.99976L}{g}} = 0.99988T$$The change in period is:$$\Delta T = T' - T = 0.99988T - T = -1.2 	imes 10^{-3}T$$The change in period is about -1.2 ms. Over one week, this will add up to about -73 s.Therefore, the clock will be slow by about 73 s.

A Clock Pendulum Made of Aluminum, Which Has a Coefficient