Problem Statement
A pendulum clock keeps correct time at 20°C. The pendulum is made of steel with $\alpha = 1.2 \times 10^{-5}$ °C$^{-1}$. How much time does the clock lose per day when the temperature rises to 40°C?
Given Information
- All quantities, constants, and constraints stated in the problem above
- Physical constants used as needed (see Concepts section)
Physical Concepts & Formulas
This problem draws on fundamental physical principles. The key is to identify which conservation law or field equation governs the system, then apply it systematically. Dimensional analysis can always be used to verify that the final answer has the correct units. Working from first principles — rather than memorising formulas — builds deeper understanding and allows tackling novel problems.
- Identify the relevant physical law (Newton’s laws, conservation of energy/momentum, Maxwell’s equations, etc.)
- State the mathematical form of that law as it applies here
- Check dimensions at every step: both sides of an equation must have the same units
Step-by-Step Solution
Problem Statement
A pendulum clock keeps correct time at 20°C. The pendulum is made of steel with $\alpha = 1.2 \times 10^{-5}$ °C$^{-1}$. How much time does the clock lose per day when the temperature rises to 40°C?
Given Information
- $T_1 = 20°C$, $T_2 = 40°C$, $\Delta T = 20°C$
- $\alpha = 1.2 \times 10^{-5}$ °C$^{-1}$
- Total time in a day: $t = 86400$ s
Physical Concepts & Formulas
Period of a pendulum: $T = 2\pi\sqrt{L/g}$. If $L$ increases, $T$ increases (clock runs slow).
Fractional change in period:
$$\frac{\Delta T_{period}}{T_{period}} = \frac{1}{2}\alpha \Delta T$$
Time lost per day:
$$\Delta t = \frac{1}{2}\alpha \Delta T \times 86400 \text{ s}$$
Step-by-Step Solution
Step 1: Fractional increase in period.
$$\frac{\Delta T_p}{T_p} = \frac{1}{2} \times 1.2 \times 10^{-5} \times 20 = 1.2 \times 10^{-4}$$
Step 2: Compute time lost per day.
$$\Delta t = 1.2 \times 10^{-4} \times 86400$$
Step 3: Result.
$$\Delta t = 10.368 \text{ s} \approx 10.4 \text{ s per day}$$
Worked Calculation
$$\Delta t = \frac{1}{2} \times 1.2 \times 10^{-5} \times 20 \times 86400 = 1.2 \times 10^{-4} \times 86400 \approx 10.4 \text{ s}$$
Answer
The clock loses $\approx 10.4$ seconds per day.
Physical Interpretation
A longer pendulum swings more slowly, causing the clock to run slow. Precision pendulum clocks use Invar (iron-nickel alloy, $\alpha \approx 10^{-6}$ °C$^{-1}$) or temperature-compensated pendulums to minimize this effect. Even a 10-second loss per day adds up to ~1 hour per year!
Worked Calculation
Substituting all given numerical values with their units into the derived formula:
$$\text{Numerical result} = \text{given expression substituted with values}$$
Answer
$$\boxed{\Delta t = \frac{1}{2} \times 1.2 \times 10^{-5} \times 20 \times 86400 = 1.2 \times 10^{-4} \times 86400 \approx 10.4 \text{ s}}$$
Physical Interpretation
The answer should be checked for dimensional consistency and physical reasonableness: is the magnitude in the expected range for this type of problem? Does the answer change in the correct direction when parameters are varied (e.g., increasing mass should increase momentum, increasing distance should decrease field strength)? These sanity checks are as important as the calculation itself.
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