Pendulum Clocks & Thermal Expansion

1. The Physics of a Pendulum Clock

The timekeeping of a pendulum clock relies on the period of a simple pendulum. For small oscillations, the time period, $T$, is given by:
$T=2\pi \sqrt{\frac{L}{g}}$
where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity.

The critical component here is the pendulum's rod, which is typically made of metal. As we've learned, the length of a metal rod changes with temperature.

• When the temperature increases, the rod's length $L$ increases.
• When the temperature decreases, the rod's length $L$ decreases.

This change in length directly affects the time period $T$, causing the clock to run either too slow or too fast.

2. Deriving the Change in Time Period

Let's quantify how a change in temperature, $\Delta \theta$, affects the time period. Suppose a clock is calibrated to be accurate at a temperature ${\theta }_0$, where its pendulum has a length $L_0$ and a period $T_0$.

If the temperature changes to ${\theta }_0+\Delta \theta$, the new length $L$ of the pendulum becomes:
$L=L_0(1+\alpha \Delta \theta )$
where $\alpha$ is the coefficient of linear expansion.

The new time period, $T$, will be:
$T=2\pi \sqrt{\frac{L}{g}}=2\pi \sqrt{\frac{L_0(1+\alpha \Delta \theta )}{g}}$

To see how this relates to the original period $T_0=2\pi \sqrt{L_0/g}$, we can write:
$T=\left(2\pi \sqrt{\frac{L_0}{g}}\right)(1+\alpha \Delta \theta {)}^{1/2}=T_0(1+\alpha \Delta \theta {)}^{1/2}$

This derivation is explained very clearly in the following video.

As the video explains, since the term $\alpha \Delta \theta$ is much smaller than 1, we can use the binomial approximation $(1+x{)}^n\approx 1+nx$. Here, $x=\alpha \Delta \theta$ and $n=1/2$.

Applying this, we get:
$T\approx T_0\left(1+\frac{1}{2}\alpha \Delta \theta \right)$

The change in the time period for one oscillation is $\Delta T_{period}=T-T_0$.
$\Delta T_{period}\approx \frac{1}{2}{T}_0\alpha \Delta \theta$

The fractional change in time period, which is a more useful quantity, is:
$\frac{\Delta T_{period}}{T_0}=\frac{T-T_0}{T_0}\approx \frac{1}{2}\alpha \Delta \theta$

3. Calculating Time Gained or Lost

The fractional change $\frac{\Delta T_{period}}{T_0}$ tells us the time the clock gains or loses per unit time. If a clock runs for a total duration of $t_{total}$ (e.g., one day = 86400 seconds), the total time gained or lost ($\Delta t$) is:

$\Delta t=\left(\frac{\Delta T_{period}}{T_0}\right)\times t_{total}=\left(\frac{1}{2}\alpha \Delta \theta \right)\times t_{total}$

Important Interpretation:

• Summer ($\Delta \theta >0$): Temperature increases, $L$ increases, period $T$ increases. Each swing takes longer, so the clock runs slow and loses time.
• Winter ($\Delta \theta <0$): Temperature decreases, $L$ decreases, period $T$ decreases. Each swing is faster, so the clock runs fast and gains time.

Let's see this in action with a typical problem.

Example 1: Direct Calculation

Problem: A second's pendulum clock (which has a period of 2 seconds) has a steel wire. The clock is calibrated at 20°C. How much time does the clock lose or gain in one week when the temperature is increased to 30°C? (${\alpha }_{steel}=1.2\times 10^{-5}\text{ /°C}$).

This problem is solved in the resource below. Let's analyze the steps.

Solution Analysis:

1. Identify Knowns:

   • Calibration temperature = 20°C, New temperature = 30°C.
   • $\Delta \theta =30°C-20°C=10°C$.
   • $\alpha =1.2\times 10^{-5}\text{ /°C}$.
   • Total time $t_{total}=1\text{ week}=7\times 24\times 3600\text{ seconds}$.

2. Determine Slow/Fast: Since $\Delta \theta$ is positive, the clock will run slow and lose time.

3. Calculate Fractional Change:
   $\frac{\Delta T_{period}}{T_0}=\frac{1}{2}\alpha \Delta \theta =\frac{1}{2}(1.2\times 10^{-5})(10)=6\times 10^{-5}$
   This means the clock loses $6\times 10^{-5}$ seconds for every second that passes.

4. Calculate Total Time Lost:
   $\Delta t=\left(6\times 10^{-5}\right)\times (7\times 24\times 3600)\approx 36.28\text{ s}$
   So, the clock loses about 36.3 seconds in one week. The calculation in the PDF takes a slightly different route by first finding the new period, but the result is the same. Our method is often more direct.

4. Advanced Problem: Finding the Correct Temperature

JEE Advanced problems often require you to work backwards or solve a system of equations. Here's a classic example.

Problem (JEE-Main 2016): A pendulum clock loses 12s a day if the temperature is 40°C and gains 4s a day if the temperature is 20°C. Find the temperature at which the clock will show correct time, and the coefficient of linear expansion ($\alpha$) of the metal of the pendulum shaft.

This problem is available in the same PDF resource (LINK, section index 3). Let's set it up.

Analysis:
Let ${\theta }_c$ be the correct calibration temperature. Let $t_{day}=86400$ seconds.

1. Case 1: Temperature = 40°C

   • The clock loses 12s. This means it's running slow.
   • The temperature change is $\Delta {\theta }_1=40-{\theta }_c$.
   • Our equation is: $12=\frac{1}{2}\alpha (40-{\theta }_c)\times t_{day}$. (Equation 1)

2. Case 2: Temperature = 20°C

   • The clock gains 4s. This means it's running fast.
   • The temperature change is $\Delta {\theta }_2=20-{\theta }_c$. This value must be negative, so ${\theta }_c>20$.
   • The time gained is the absolute value of $\Delta t$.
   • Our equation is: $4=\left|\frac{1}{2}\alpha (20-{\theta }_c)\times t_{day}\right|=\frac{1}{2}\alpha ({\theta }_c-20)\times t_{day}$. (Equation 2)

Now you have a system of two linear equations with two variables (${\theta }_c$ and $\alpha$). This is a perfect setup for you to solve.

This type of problem demonstrates how a fundamental physics principle can be turned into an algebraic challenge, a common pattern in competitive exams.

Conclusion

In this lesson, we explored a practical and interesting consequence of linear thermal expansion. We saw how even small changes in temperature can affect the precision of a pendulum clock over time.

Key Takeaways:

• The time period of a pendulum increases with temperature (clock runs slow, loses time) and decreases with temperature (clock runs fast, gains time).
• The fundamental formula for the total time gained or lost, $\Delta t$, over a duration $t_{total}$ is:
  $\Delta t=\left(\frac{1}{2}\alpha \Delta \theta \right)\times t_{total}$
• Many problems, especially at the JEE Advanced level, can be solved by setting up a system of equations for different temperature conditions.
• To minimize this effect, high-precision pendulums are often made of materials with very low coefficients of thermal expansion, such as Invar.

Preview of the Next Lesson:
This lesson concludes our module on Thermal Expansion. We have explored its various manifestations, from expanding liquids and bimetallic strips to thermal stress and pendulum clocks.

Next, we will begin a new module, Steady-State Heat Conduction. We will shift our focus from the effects of temperature to the process of heat transfer itself. Our first topic will be Fourier's Law of Heat Conduction, the fundamental principle governing how heat flows through materials.