Problem 68
Question
The time required for one complete oscillation of a pendulum is called its period. If \(L\) is the length of the pendulum and the oscillation is small, then the period is given by \(P=2 \pi \sqrt{L / g},\) where \(g\) is the constant acceleration due to gravity. Use differentials to show that the percentage error in \(P\) is approximately half the percentage error in \(L\).
Step-by-Step Solution
Verified Answer
The percentage error in \( P \) is half the percentage error in \( L \).
1Step 1: Understand the given formula
The formula for the period of a pendulum is given as \( P = 2 \pi \sqrt{ \frac{L}{g} } \), where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.
2Step 2: Differentiate the equation
To find how a small change in \( L \) (\( dL \)) affects \( P \), differentiate the period formula with respect to \( L \). The differential \( dP \) is given by: \ \[ dP = \frac{d}{dL}(2 \pi \sqrt{ \frac{L}{g} }) \, dL \] \ Compute this derivative using the chain rule.
3Step 3: Apply the chain rule to differentiate
Apply the chain rule: \ \[ \frac{d}{dL}(\sqrt{L/g}) = \frac{1}{2} \cdot (L/g)^{-1/2} \cdot \frac{1}{g} \] \ Therefore, \ \[ \frac{d}{dL} \left( 2 \pi \sqrt{\frac{L}{g}} \right) = 2 \pi \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{L/g}} \cdot \frac{1}{g} = \frac{\pi}{\sqrt{Lg}} \]
4Step 4: Express the differential relationship
Substitute back to get \( dP \) in terms of \( dL \): \ \[ dP = \frac{\pi}{\sqrt{Lg}} \, dL \] \ Relate \( dP \) and \( dL \) to \( P \) and \( L \): \ \[ dP = 2 \pi \cdot \frac{1}{2} \cdot \frac{dL}{2\sqrt{Lg}} \] \ Simplifying: \ \[ dP = \frac{\pi}{\sqrt{Lg}} \, dL \]
5Step 5: Solve for the percentage error
Prove if the percentage error in \( P \) is half of that in \( L \): \ \[ \frac{dP}{P} = \frac{\pi}{\sqrt{Lg}} \cdot \frac{dL}{2\pi}\sqrt{L/g} \] \ Simplify the terms: \ \[ \frac{dP}{P} = \frac{1}{2} \cdot \frac{dL}{L} \] \ This shows that the percentage error in \( P \) is indeed half of the percentage error in \( L \).
6Step 6: Conclusion of the derivation
The differentiation implies that the relative increment (or percentage error) in the period \( P \) is half the relative increment (or percentage error) in the length \( L \). We have thus shown mathematically using differentials that small errors in \( L \) translate to smaller errors in \( P \).
Key Concepts
Error AnalysisPendulum PeriodChain Rule Differentiation
Error Analysis
Error analysis is a powerful tool in mathematics and science that allows us to understand how small changes in measurements affect derived quantities. In the pendulum problem, we are analyzing the small change in the period of a pendulum in response to a change in its length. The idea is to use differentials, which are a way to approximate small changes by considering infinitesimally small increments.
Imagine you have a pendulum with a certain length, and there's a small error or variation in measuring this length. By using differentials, we can also predict how this error affects the period, the time it takes for a pendulum to complete one swing back and forth.
In general, to analyze the percentage error, we compare the change in the derived quantity to the original quantity. For the pendulum period, the step-by-step solution shows that the percentage change in the period is half of that in the length. This analysis provides a clearer understanding of how sensitive the period is to changes in the length.
Imagine you have a pendulum with a certain length, and there's a small error or variation in measuring this length. By using differentials, we can also predict how this error affects the period, the time it takes for a pendulum to complete one swing back and forth.
In general, to analyze the percentage error, we compare the change in the derived quantity to the original quantity. For the pendulum period, the step-by-step solution shows that the percentage change in the period is half of that in the length. This analysis provides a clearer understanding of how sensitive the period is to changes in the length.
Pendulum Period
The period of a pendulum, denoted as \( P \), is the time taken for one complete swing back and forth. When the pendulum undergoes small oscillations, its period is calculated by the formula \( P = 2\pi \sqrt{\frac{L}{g}} \), where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, m/s^2 \) on Earth).
This formula reveals that the period depends on the square root of the length. Therefore, even a small error in measuring the length can impact the calculation of the period. However, as derived in the solution, the actual sensitivity of the period \( P \) to changes in \( L \) is minimal, because it varies with the square root.
This aspect of pendulums is one of the reasons they have historically been used in clocks, as changes in their length will not proportionally affect the timekeeping by a large amount. It all boils down to the most important property of pendulums: longer pendulums swing more slowly, proportionally to the square of the length.
This formula reveals that the period depends on the square root of the length. Therefore, even a small error in measuring the length can impact the calculation of the period. However, as derived in the solution, the actual sensitivity of the period \( P \) to changes in \( L \) is minimal, because it varies with the square root.
This aspect of pendulums is one of the reasons they have historically been used in clocks, as changes in their length will not proportionally affect the timekeeping by a large amount. It all boils down to the most important property of pendulums: longer pendulums swing more slowly, proportionally to the square of the length.
Chain Rule Differentiation
The chain rule is a fundamental technique in calculus used to differentiate composite functions. In the context of the pendulum period, it helps us find the derivative of the function \( P = 2\pi \sqrt{\frac{L}{g}} \) with respect to \( L \).
Using the chain rule, we can break down the differentiation into more manageable parts.
Using the chain rule, we can break down the differentiation into more manageable parts.
- We start by noting the function is composed of an outer function, the multiple of \(2\pi\), and an inner function, \( \sqrt{\frac{L}{g}} \).
- Applying the chain rule, we differentiate the outer function while keeping the inner function the same, multiplies by the derivative of the inner function.
- For \( \sqrt{\frac{L}{g}} \), using power rule and properties of fractions, the derivative is managed stepwise to find its contribution to changes in \( P \).
Other exercises in this chapter
Problem 67
(a) Show that \(f(x)=x^{3}-3 x^{2}+2 x\) is not one-to-one on \((-\infty,+\infty)\) (b) Find the largest value of \(k\) such that \(f\) is one-to-one on the int
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(a) Explain why L'Hôpital's rule does not apply to the $$\lim _{x \rightarrow 0} \frac{x^{2} \sin (1 / x)}{\sin x}$$ (b) Find the limit.
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(a) Show that the function \(f(x)=x^{4}-2 x^{3}\) is not one-toone on \((-\infty,+\infty)\) (b) Find the smallest value of \(k\) such that \(f\) is one-to-one o
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Find \(\lim _{x \rightarrow 0^{+}} \frac{x \sin (1 / x)}{\sin x}\) if it exists.
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