Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

The period \(T\) of a pendulum of length \(L\) is given by \(T=2 \pi \sqrt{L / g}\), where \(g\) is the acceleration of gravity. Show that \(d T / T=\frac{1}{2}[d L / L-d g / g]\), and use this result to estimate the maximum percentage error in \(T\) due to an error of \(0.5 \%\) in measuring \(L\) and \(0.3 \%\) in measuring \(g\).

Step-by-Step Solution

Answer

The maximum percentage error in period \(T\) is 0.1\%.

Step 1: Understand the Formula

We start with the formula for the period of a pendulum: \[ T = 2\pi \sqrt{ \frac{L}{g}} \]where \(L\) is the pendulum length, and \(g\) is the acceleration due to gravity.

Step 2: Calculate Error in T

We need to calculate \( \frac{dT}{T} \). Start by taking the natural log of both sides of \( T = 2\pi \sqrt{ \frac{L}{g} } \). \[ \ln T = \ln (2\pi) + \frac{1}{2} \ln L - \frac{1}{2} \ln g \]Now differentiate both sides with respect to \(L\) and \(g\).

Step 3: Derive Relative Change

Differentiating, we get:\[ \frac{dT}{T} = \frac{1}{2} \frac{dL}{L} - \frac{1}{2} \frac{dg}{g} \]This relation shows how the errors in \(L\) and \(g\) affect the error in \(T\).

Step 4: Calculate Maximum Percentage Error

Substitute \(\frac{dL}{L} = 0.005\) and \(\frac{dg}{g} = 0.003\) into the derived equation:\[ \frac{dT}{T} = \frac{1}{2}(0.005) - \frac{1}{2}(0.003) = 0.001 \]Multiply by 100 to convert to percentage:\[ 0.001 \times 100 = 0.1\% \]

Step 5: Conclusion

The maximum percentage error in the period \(T\) due to measurement errors is \(0.1\%\).

Key Concepts

Pendulum Period Calculation

In the world of pendulums, the period of oscillation is a crucial aspect to understand. A pendulum's period is the time it takes to complete one full swing back and forth. Mathematically, it's represented by the formula: \[ T = 2\pi \sqrt{ \frac{L}{g} } \]where:

• \(L\) is the length of the pendulum,
• \(g\) is the acceleration due to gravity (approximately 9.8 m/s² near the surface of the Earth), and
• \(T\) is the pendulum period.

This formula showcases a direct relationship between the pendulum length and its period, where the period increases with a longer pendulum. Understanding this relationship is essential in fields like clockmaking and physics education, where pendulums are frequently utilized.

Error Estimation in Measurements

When dealing with physical measurements, error estimation is an important concept. It helps predict how uncertainties in measurements affect the results. In the case of the pendulum period calculation, small errors in measuring the pendulum length \(L\) and gravitational acceleration \(g\) can lead to errors in the calculated period \(T\). This can be crucial, especially in experiments.If there's a 0.5% error in measuring \(L\) and a 0.3% error in measuring \(g\), how do these affect the period \(T\)? To find out, we use the expression for the relative change in period:\[ \frac{dT}{T} = \frac{1}{2} \left( \frac{dL}{L} - \frac{dg}{g} \right) \]This tells us how the percentage errors in the parameters translate into the percentage error in \(T\). Estimating this error is vital to ensure accurate results in experiments and applications utilizing pendulums.

Differentiation and Relative Change

Differentiation is a fundamental tool in calculus used to understand how a function changes. Relative change focuses on how a dependent variable, like the pendulum period \(T\), varies in relation to changes in independent variables, such as the pendulum length \(L\) and gravitational acceleration \(g\).In our context, to determine the effect of measurement errors on \(T\), we use differentiation. Begin by taking the logarithm of the formula for the period \(T = 2\pi \sqrt{ \frac{L}{g} } \), and then differentiate:\[ \ln T = \ln (2\pi) + \frac{1}{2} \ln L - \frac{1}{2} \ln g \]Differentiating both sides, we get the expressed formula:\[ \frac{dT}{T} = \frac{1}{2} \frac{dL}{L} - \frac{1}{2} \frac{dg}{g} \]This technique allows us to express the changes in \(T\) in terms of relative changes in \(L\) and \(g\), providing a method to estimate errors in real-world applications.

Natural Logarithm Differentiation

The natural logarithm, denoted by \(\ln\), is a powerful concept in calculus and is used here to transform the pendulum period equation. Differentiating with the natural logarithm can simplify complex expressions, especially those involving products or quotients like our pendulum period formula. Taking the natural log of \(T = 2\pi \sqrt{ \frac{L}{g} }\) helps to simplify the multiplication and division into additive components:\[ \ln T = \ln (2\pi) + \frac{1}{2} \ln L - \frac{1}{2} \ln g \]Once simplified, this expression can be differentiated easily. The differentiation results in a relatively simple linear combination of derivatives:\[ \frac{dT}{T} = \frac{1}{2} \frac{dL}{L} - \frac{1}{2} \frac{dg}{g} \]Using this method, you can elegantly estimate changes in functions due to small changes in variables, which is particularly useful in pendulum period calculations and other scientific measurements.