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Chapter 3: Problem 31

Writing \(g\) for the acceleration due to gravity, the period, \(T,\) of a pendulum of length \(l\) is given by $$ T=2 \pi \sqrt{\frac{l}{g}} $$ (a) Show that if the length of the pendulum changes by \(\Delta l\), the change in the period, \(\Delta T\), is given by $$ \Delta T \approx \frac{T}{2 l} \Delta l $$ (b) If the length of the pendulum increases by \(2 \%,\) by what percent does the period change?

Short Answer

Expert verified
(a) Use differentiation to find \( \Delta T = \frac{T}{2l} \Delta l \). (b) The period increases by 1%.

Step by step solution

01

Differentiate the Period Formula

Start by differentiating the given period formula \( T = 2 \pi \sqrt{\frac{l}{g}} \) with respect to \( l \). We'll use implicit differentiation here. The derivative of \( T \) with respect to \( l \) is:\[ \frac{dT}{dl} = \frac{d}{dl} \left(2\pi \sqrt{\frac{l}{g}}\right) = 2\pi \times \frac{1}{2} \left( \frac{1}{\sqrt{\frac{l}{g}}} \right) \times \left( \frac{1}{g} \right) = \frac{\pi}{\sqrt{lg}} \times \frac{1}{g} = \frac{\pi}{g \sqrt{\frac{l}{g}}} \]
02

Approximate Change in Period

To find the approximate change in the period \( \Delta T \) when the length changes by \( \Delta l \), multiply the derivative by \( \Delta l \):\[ \Delta T \approx \frac{dT}{dl} \Delta l = \frac{\pi}{g \sqrt{\frac{l}{g}}} \Delta l \]Next, substituting the expression for \( T \): \[ T = 2\pi \sqrt{\frac{l}{g}} = \frac{4\pi^{2} l}{gT} \]Hence:\[ \Delta T \approx \frac{T}{2l} \Delta l \]
03

Calculate Percent Change in Length

If the length of the pendulum increases by \( 2\% \), then \( \Delta l = 0.02l \).
04

Calculate Percent Change in Period

Substitute \( \Delta l = 0.02l \) into the approximation for \( \Delta T \):\[ \Delta T \approx \frac{T}{2l} \times 0.02l = \frac{T}{2} \times 0.02 = 0.01T \] The period \( T \) changes by \( 1\% \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pendulum Period Formula
The period of a pendulum is the time it takes for one complete cycle of swinging back and forth. This is crucial for understanding how potential changes in the length of the pendulum affect its motion. The formula given by \[ T = 2 \pi \sqrt{\frac{l}{g}} \]where \( T \) is the period, \( l \) is the length of the pendulum, and \( g \) is the acceleration due to gravity, illustrates this relationship. This formula indicates:
  • The period \( T \) is directly proportional to the square root of the pendulum's length \( l \).
  • The period \( T \) is inversely proportional to the square root of gravity \( g \).
If the length \( l \) increases, the period \( T \) also increases, meaning the pendulum takes longer to complete a swing. Understanding this formula allows us to explore how changes in length can impact the pendulum's motion.
Implicit Differentiation
Implicit differentiation is a technique used to differentiate equations that are not explicitly solved for a variable. It is particularly useful when dealing with functions composed of other functions.In the context of the pendulum formula, we implicitly differentiate the formula \( T = 2\pi \sqrt{\frac{l}{g}} \) with respect to \( l \) to find how changes in length affect the period.The steps involve:
  • Applying the chain rule to handle the square root and fraction.
  • Relating the rate of change of period with respect to the change in length (\( \frac{dT}{dl} \)).
This differentiation helps determine the approximate change in the pendulum period when there is a small change in length, guiding us further into the calculation.
Approximation of Change
When a variable undergoes a small change, often linear approximations make it easier to predict their effect on related quantities. In the pendulum problem, we approximate the change in period \( \Delta T \) using \[ \Delta T \approx \frac{T}{2l} \Delta l \]This formula is derived from the differentiation process and helps in estimating minor impacts on period based on small changes in length. It considers:
  • How the derivative \( \frac{dT}{dl} \) relates changes in \( T \) to changes in \( l \).
  • Linearity of small changes assuming other variables in the formula remain constant.
Such approximation techniques are widespread in calculus as they simplify complex real-world problems into manageable computations.
Percentage Change Calculation
After establishing the formula for the approximate change, we can further analyze by calculating percentage changes. For the pendulum, if the length \( l \) changes by a certain percent, how does that affect the period \( T \)?Assuming a 2% increase in length:
  • The change in length \( \Delta l = 0.02l \).
  • We substitute into our earlier found formula, yielding \( \Delta T \approx \frac{T}{2} \times 0.02 = 0.01T \).
This calculation shows the period \( T \) changes by approximately 1%.Understanding percentage changes relative to length and their subsequent effects on period is fundamental for practical implementations, such as pendulum-based clocks where precision in timing is paramount.

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Most popular questions from this chapter

The following problem is a generalization of Problem \(31 .\) Show that any function of the form $$y=A e^{x}+B e^{-x}, \quad A>0, B>0$$ can be written, for some \(K\) and \(c,\) in the form $$y=K \cosh (x-c)$$ What does this tell you about the graph of \(y=A e^{x}+\) \(B e^{-x} ?\)

Let \(P=f(t)\) give the US population \(^{18}\) in millions in year \(t\) (a) What does the statement \(f(2014)=319\) tell you about the US population? (b) Find and interpret \(f^{-1}(319) .\) Give units. (c) What does the statement \(f^{\prime}(2014)=2.44\) tell you about the population? Give units. (d) Evaluate and interpret \(\left(f^{-1}\right)^{\prime}(319) .\) Give units.

A rubber duck bounces up and down in a pool after a stone is dropped into the water. The height of the duck, in inches, above the equilibrium position of the water is given as a function of time \(t,\) in seconds, by $$d(t)=e^{-t}(\cos t+\sin t)$$ (a) Find and interpret the practical meaning of the derivative \(d^{\prime}(t)\) (b) Determine when \(d^{\prime}(t)=0\) for \(t \geq 0 .\) What can you say about the duck when \(d^{\prime}(t)=0 ?\) (c) Determine \(\lim _{t \rightarrow \infty} d(t)\) and explain why this limit makes sense in practical terms.

Let \(f\) be a differentiable function and let \(L\) be the linear function \(L(x)=f(a)+k(x-a)\) for some constant a. Decide whether the following statements are true or false for all constants \(k\). Explain your answer. (a) \(L\) is the local linearization for \(f\) near \(x=a\) (b) If \(\lim _{x \rightarrow a}(f(x)-L(x))=0,\) then \(L\) is the local linearization for \(f\) near \(x=a\)

give an example of: Two different functions that have the same linear approximation near \(x=0\)
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