Question

A pendulum clock shows correct time at 20°C at a place where g = 9.800 m s^–2. The pendulum consists of a light steel rod connected to a heavy ball. It is taken to a different place where g = 9.788 m s^–1. At what temperature will the clock show correct time? Coefficient of linear expansion of steel = 12 × 10^–6 °C^–1.

Answer 1

Given:
The temperature at which the pendulum shows the correct time, T₁ = 20 °C
Coefficient of linear expansion of steel, $\alpha$ = 12 × 10^–6 °C^–1
Let T₂ be the temperature at which the value of g is 9.788 ms^–2 and $\Delta$T be the change in temperature.
^​So, the time periods of pendulum at different values of g will be t₁ and t₂ , such that
$$\begin{gathered} t_1=2\mathrm{\pi }\sqrt{\frac{l_{\mathit{1}}}{g_1}} \\ {t}_2=2\mathrm{\pi }\sqrt{\frac{l_2}{g_2}} \\ =\frac{2\mathrm{\pi }\sqrt{l_1\left(1+\alpha \Delta T\right)}}{\sqrt{g_2}}\left(\because l_2=l_1\left(1+\alpha ∆T\right)\right) \\ \mathrm{Given},t_1=t_2 \\ \Rightarrow \frac{2\pi \sqrt{l_1}}{\sqrt{g_1}}=\frac{2\pi \sqrt{l_1\left(1+\alpha ∆T\right)}}{\sqrt{g_2}} \\ \Rightarrow \sqrt{\left(\frac{l_1}{g_1}\right)}=\frac{\sqrt{l_1\left(1+\alpha ∆T\right)}}{\sqrt{g_2}} \\ \Rightarrow \frac{1}{9.8}=\frac{1+12\times {10}^{-6}\times ∆T}{9.788} \\ \Rightarrow \frac{9.788}{9.8}=1+12\times {10}^{-6}\times ∆T \\ \Rightarrow \frac{9.788}{9.8}-1=12\times {10}^{-6}\times ∆T \\ \Rightarrow ∆T=\frac{-0.00122}{12\times {10}^{-6}} \\ \Rightarrow T_2-20=-102.4 \\ \Rightarrow T_2=-102.4+20 \\ =-82.4 \\ \Rightarrow T_2\approx -82°\mathrm{C} \end{gathered}$$
Therefore, for a pendulum clock to give correct time, the temperature at which the value of g is 9.788 ms^–2 should be $-$ 82 ^oC.