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

\[\Delta\]ΔT be  the change in temperature.
^​So, the time periods of pendulum at different values of g will be t₁ and t₂ , such that

\[t_1 = 2\pi\sqrt{\frac{l_1}{g_1}}\]

\[ t_2 = 2\pi\sqrt{\frac{l_2}{g_2}}\]

\[ = \frac{2\pi\sqrt{l_1 \left( 1 + \alphaΔT \right)}}{\sqrt{g_2}} \left( \because l_2 = l_1 \left( 1 + \alpha ∆ T \right) \right)\]

\[Given, t_1 = t_2 \]

\[ \Rightarrow \frac{2\pi\sqrt{l_1}}{\sqrt{g_1}} = \frac{2\pi\sqrt{l_1 \left( 1 + \alpha ∆ T \right)}}{\sqrt{g_2}}\]

\[ \Rightarrow \sqrt{\left( \frac{l_1}{g_1} \right)} = \frac{\sqrt{l_1 \left( 1 + \alpha ∆ T \right)}}{\sqrt{g_2}}\]

\[ \Rightarrow \frac{1}{9 . 8} = \frac{1 + 12 \times {10}^{- 6} \times ∆ T}{9 . 788}\]

\[ \Rightarrow \frac{9 . 788}{9 . 8} = 1 + 12 \times {10}^{- 6} \times ∆ T\]

\[ \Rightarrow \frac{9 . 788}{9 . 8} - 1 = 12 \times {10}^{- 6} \times ∆ T\]

\[ \Rightarrow ∆ T = \frac{- 0 . 00122}{12 \times {10}^{- 6}}\]

\[ \Rightarrow T_2 - 20 = - 102 . 4\]

\[ \Rightarrow T_2 = - 102 . 4 + 20\]

\[ = - 82° 4\]

⇒ T₂ ≈ -82° C

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.