A clock pendulum made of aluminum,which has a coefficient of linear expansion of 24 $\times$ 10^-6/K,has a period of exactly 1 s at 20ºC.Before a homeowner leaves town for one week,he turns the thermostat down to 10ºC.When he returns,the clock is A)fast by about 73 s. B)fast by about 7.6 min. C)exactly on time. D)slow by about 7.6 min. E)slow by about 73 s.

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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 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 of the pendulum, and $\Delta T$ is the change in temperature.In this problem, the coefficient of linear expansion is $24 	imes 10^{-6}/K$, the original length of the pendulum is not given, and the change in temperature is $-10°C$.The change in length is:$$\Delta L = (24 	imes 10^{-6}/K)(-10°C) = -2.4 	imes 10^{-4} m$$The new length of the pendulum is:$$L' = L + \Delta L = L - 2.4 	imes 10^{-4} m$$The new period of the pendulum is:$$T' = 2\pi\sqrt{\frac{L'}{g}} = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}}$$The change in period is:$$\Delta T = T' - T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L}{g}}$$$$\Delta T = 2\pi\sqrt{\frac{L - 2.4 	imes 10^{-4} m}{g}} - 2\pi\sqrt{\frac{L

A Clock Pendulum Made of Aluminum,which Has a Coefficient of Linear