Thermal expansion

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Question

A clock pendulum made of invar has a period of 2s at 20^oC. If the clock is used in a climate where average temperature is 40^oC, what correction (in seconds) may be necessary at the end of 10 days to the time given by clock? $(α_{invar} = 7 \times {10^{- 7}}^{} C^{- 1}, 1 day = 8 .64 \times 10^{4} s)$

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Solution

$\begin{array}{l} T = 2 π \sqrt{\frac{l}{g}} \Rightarrow \frac{ΔT}{T} = \frac{1}{2} \frac{Δl}{l} = \frac{1}{2} αΔθ \Rightarrow ΔT = \frac{T}{2} (αΔθ) \\ = (\frac{10 \times 8 .64 \times 10^{4}}{2}) (7 \times 10^{- 7}) (20) = 6 s \end{array}$

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Thermal expansion

Remember concepts with our Masterclasses.

A clock pendulum made of invar has a period of 2s at 20oC. If the clock is used in a climate where average temperature is 40oC, what correction (in seconds) may be necessary at the end of 10 days to the time given by clock? αinvar =7×10−7 C−1,1 day =8.64×104s

T=2πlg⇒ΔTT=12Δll=12αΔθ⇒ΔT=T2(αΔθ)=10×8.64×10427×10−7(20)=6s

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$\begin{array}{l} T = 2 π \sqrt{\frac{l}{g}} \Rightarrow \frac{ΔT}{T} = \frac{1}{2} \frac{Δl}{l} = \frac{1}{2} αΔθ \Rightarrow ΔT = \frac{T}{2} (αΔθ) \\ = (\frac{10 \times 8 .64 \times 10^{4}}{2}) (7 \times 10^{- 7}) (20) = 6 s \end{array}$