Question

A clock pendulum made of invar has a period of $0.5\,{\text{sec}}$ at ${\text{20}}^\circ {\text{C}}$. If the clock is used in a climate where average temperature is ${\text{30}}^\circ {\text{C}}$, approximately. How much fast or slow will the clock run in ${10^6}\,{\text{sec}}$. (${\alpha _{{\text{linear}}}} = 1 \times {10^{ - 6}}/^\circ {\text{C}}$)

Answer 1

Use the formula for time period of the simple pendulum. This formula gives the relation between the time period of the pendulum, length of the pendulum and acceleration due to gravity. From this equation, determine the variation in the time period of the pendulum. Thus, determine the time by which the clock slows down in ${10^6}\,{\text{sec}}$.

Formulae used:
The time period $t$ of a pendulum is
$t = 2\pi \sqrt {\dfrac{L}{g}}$ …… (1)
Here, $L$ is the length of the pendulum and $g$ is the acceleration due to gravity.
The expression for linear thermal expansion is
$\Delta L = \alpha L\Delta T$ …… (2)
Here, $\Delta L$ is the change in length of the material, $\alpha$ is the linear thermal expansion coefficient, $L$ is the original length of the material and $\Delta T$ is the change in temperature of the material.

Complete step by step answer:
We have given that the time period of the clock pendulum is $0.5\,{\text{sec}}$ at the temperature ${\text{20}}^\circ {\text{C}}$. Then the clock is used in the climate having temperature ${\text{30}}^\circ {\text{C}}$. The change in the temperature of the climate is ${\text{10}}^\circ {\text{C}}$.
$\Delta T = {\text{10}}^\circ {\text{C}}$
Rearrange equation (2) for $\dfrac{{\Delta L}}{L}$.
$\dfrac{{\Delta L}}{L} = \alpha \Delta T$

Substitute $1 \times {10^{ - 6}}/^\circ {\text{C}}$ for $\alpha$ and ${\text{10}}^\circ {\text{C}}$ for $\Delta T$ in the above equation.
$\dfrac{{\Delta L}}{{{L_0}}} = \left( {1 \times {{10}^{ - 6}}/^\circ {\text{C}}} \right)\left( {{\text{10}}^\circ {\text{C}}} \right)$
$\Rightarrow \dfrac{{\Delta L}}{L} = {10^{ - 5}}$
When the temperature of the climate changes, there is a small change in the time period and length of the pendulum.

From equation (1), the small variation in the time period of the pendulum is given by
$\dfrac{{\Delta t}}{t} = \dfrac{1}{2}\dfrac{{\Delta L}}{L}$

Substitute ${10^{ - 5}}$ for $\dfrac{{\Delta L}}{L}$ and $0.5\,{\text{sec}}$ for $t$ in the above equation.
$\Delta t = \dfrac{1}{2}\left( {{{10}^{ - 5}}} \right)\left( {0.5\,{\text{sec}}} \right)$
$\Rightarrow \Delta t = 2.5 \times {10^{ - 6}}\,{\text{sec}}$
Therefore, the clock has slowed down by $2.5 \times {10^{ - 6}}\,{\text{sec}}$.

The time difference by which the clock has slowed down in ${10^6}\,{\text{sec}}$ is calculated by
${\text{Time difference}} = \dfrac{{\Delta t}}{t}\left( {{{10}^6}\,{\text{s}}} \right)$

Substitute $2.5 \times {10^{ - 6}}\,{\text{sec}}$ for $\Delta t$ and $0.5\,{\text{sec}}$ for $t$ in the above equation.
${\text{Time difference}} = \dfrac{{2.5 \times {{10}^{ - 6}}\,{\text{sec}}}}{{0.5\,{\text{sec}}}}\left( {{{10}^6}\,{\text{s}}} \right)$
$\therefore {\text{Time difference}} = 5\,{\text{s}}$

Therefore, the clock will be slowed down by $5\,{\text{s}}$.

Note: We have not taken the variation in acceleration due to gravity while deriving the formula for variation in time period of the clock pendulum because the acceleration due to gravity on which time period of the pendulum depend does not change with the change in temperature of the climate.