Question

A pendulum clock gives the correct time at \[{\text{20}}^\circ {\text{C}}\] at a place where \[g = 9.8m/s\] The pendulum consists of a light steel rod connected to a heavy ball. When taken to a different place where \[g = 9.778m/s\]. What temperature would result in the correct time? Coefficient of linear expansion of steel \[12 \times {10^{ - 6}}{C^{ - 1}}\]

Answer 1

Hint: We should consider the equations of the time period of a simple pendulum while solving this question. The ratio between the two circumstances will give out one unknown variable which can be solved (in this case, temperature). Using equations of a simple pendulum substituted with the coefficient of linear expansion of the steel wire can result in a simple ratio.

Complete step by step answer:
A simple pendulum’s time period consisting of the length of its string and gravitational constant value at a certain place is given by the equation:

\[T = 2\pi \sqrt {\dfrac{l}{g}} \]
While solving this question, we should consider two instances. Instance 1 as the place where g = 9.800 m/s and temperature is 20℃ and instance 2 as the place where g = 9.788 m/s and temperature is to be found.
Now, the linear expansion of the steel rod is given by the equation:
\[\dfrac{{\Delta L}}{L} = \alpha \Delta T\]
If we take the differential equation of time period of the pendulum using logarithms on both sides, we get the following equation.
\[\dfrac{{\Delta T}}{T} = \dfrac{1}{2}\left( {\dfrac{{\Delta L}}{L} - \dfrac{{\Delta g}}{g}} \right)\]
Now substituting this on the equation of coefficient of linear expansion, we get:
\[\dfrac{{\Delta L}}{L} = \dfrac{{\Delta g}}{g}\] at \[\Delta T = 0\]
Then,
\[\alpha \Delta T = \dfrac{{\Delta g}}{g}\]
\[\left( {T - 20} \right) = \dfrac{{ - 0.012}}{{\left( {9.8 \times 12 \times {{10}^{ - 6}}} \right)}}\]
\[T - 20{\text{ }} = {\text{ }}102.04\;\] which gives the temperature as \[ - 82.04^\circ C\]
Therefore, the required temperature is $ 82.04^\circ C$.

Note:
The end result might seem very absurd compared to real life values. We can infer the fact that this is an impossible case for getting the right time and that is the reason why pendulums are not used in clocks in areas with such a low gravitational constant. The replacement of an analog pendulum clock paved way to proper time management across the world with standard time fixations at certain regions and they are required to follow one particular time for a small area of land. An example of this would be the entire country of India following the Indian Standard Time (IST which is GMT \[ + 05:30\] ).