Thermal expansion of pendulum clock · BPhO 2008 #11

Physics · thermodynamics

Problem

When a metal rod is heated, it expands uniformly with temperature. The coefficient of linear thermal expansivity, α (alpha), is equal to the fractional increase in length per unit temperature rise. If a rod of length ℓ expands by an amount Δℓ when the temperature rises by Δθ in °C, α is given by, α=ΔℓℓΔθ11\alpha = \frac{\Delta \ell}{\ell \Delta \theta} \frac{1}{1} a) What are the units of α? [1] A pendulum clock has a metal pendulum. The period of oscillation, T, of the pendulum is given by, T=2πℓgT = 2\pi \sqrt{\frac{\ell}{g}} where ℓ is the length of the pendulum and g is the acceleration due to gravity (g = 9.8 N/kg). The period of the pendulum is exactly 1 second when the room temperature is such that the clock gives the correct time. On days when the room temperature is 15.0 °C the clock runs 5 s fast per day. When the room temperature is 30.0 °C, the clock runs 10 s slow per day. b) When the clock gives the correct time, how many oscillations will occur in a day? [1] c) For the two temperatures quoted, write down the number of oscillations that would occur in one day. [2] d) Calculate the periods of the pendulum, T₁₅ and T₃₀, at the two temperatures. [2] e) Calculate the corresponding values of lengths, ℓ₁₅ and ℓ₃₀. [2] f) Calculate the value of α for the metal of the pendulum. [3]

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