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Chapter 6: Problem 250

Suppose that a pendulum is to have a period of 2 seconds and a maximum angle of \(\theta_{\max }=\frac{\pi}{6}\). Use \(T \approx 2 \pi \sqrt{\frac{L}{g}}\left(1+\frac{k^{2}}{4}\right)\) to approximate the desired length of the pendulum. What length is predicted by the small angle estimate \(T \approx 2 \pi \sqrt{\frac{L}{g}} ?\)

Short Answer

Expert verified
The small angle estimate gives \( L \approx 0.9937 \) m; the accurate formula gives \( L \approx 0.981 \) m.

Step by step solution

01

Understanding the Given Formulas

Firstly, we are given two formulas for the period of a pendulum. The first formula provided is: \[ T \approx 2\pi \sqrt{\frac{L}{g}} \left(1 + \frac{k^2}{4} \right) \]where \( k = \sin\left( \frac{\theta_{\max}}{2} \right) \). The second formula is the small angle approximation:\[ T \approx 2\pi \sqrt{\frac{L}{g}} \]We need to find the length \( L \) that satisfies these equations given that \( T = 2 \) seconds and \( \theta_{\max} = \frac{\pi}{6} \).
02

Calculate the Small Angle Estimate

The small angle approximation formula assumes \( T = 2\pi \sqrt{\frac{L}{g}} \), and we know \( T = 2 \) seconds and \( g \approx 9.81 \text{ m/s}^2 \). Solving for \( L \), we equate:\[ 2 = 2\pi \sqrt{\frac{L}{9.81}} \]Solving for \( L \):\[ 1 = \pi \sqrt{\frac{L}{9.81}} \]\[ \sqrt{\frac{L}{9.81}} = \frac{1}{\pi} \]\[ \frac{L}{9.81} = \frac{1}{\pi^2} \]\[ L = \frac{9.81}{\pi^2} \]
03

Compute the Value of L for Small Angle Approximation

Calculate \( L \) with the given value:\[ L = \frac{9.81}{\pi^2} \approx 0.9937 \text{ meters} \]
04

Calculate L Using the Modified Formula

For the more accurate formula, \( T \approx 2\pi \sqrt{\frac{L}{g}} \left(1 + \frac{k^2}{4}\right) \), calculate \( k \):\[ k = \sin\left(\frac{\pi}{6}\div 2\right) = \sin\left(\frac{\pi}{12}\right) \]Using a calculator, approximate \( \sin\left(\frac{\pi}{12}\right) \approx 0.2588 \). So, \( k^2 \approx 0.067 \).Solve for \( L \):\[ 2 = 2\pi \sqrt{\frac{L}{9.81}} \left(1 + \frac{0.067}{4}\right) \]Calculate the correction factor:\[ 1 + \frac{0.067}{4} = 1.01675 \]
05

Solve for L with the Correction Factor

Substitute the correction factor into the equation and solve for \( L \):\[ 1 = \pi \sqrt{\frac{L}{9.81}} \times 1.01675 \]\[ \sqrt{\frac{L}{9.81}} = \frac{1}{1.01675 \pi} \]\[ \frac{L}{9.81} = \left(\frac{1}{1.01675 \pi}\right)^2 \]\[ L \approx 0.981 \text{ meters} \]
06

Comparison of Lengths

The length predicted by the small angle approximation is approximately \( 0.9937 \) meters, while the more accurate formula predicts a length of approximately \( 0.981 \) meters.

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Key Concepts

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

Small Angle Approximation
The small angle approximation is a crucial concept in physics, especially when analyzing pendulum motion. This approximation simplifies complex equations by assuming that for small angles, the sine of the angle is approximately equal to the angle itself in radians. This assumption makes calculations more straightforward because
  • The sine function is nonlinear, but for small angles, it acts nearly linear.
  • It helps in simplifying mathematical expressions involving trigonometric functions.
In the context of pendulums, the small angle approximation comes into play when the amplitude of the pendulum's swing is small, making it easier to predict periods or lengths. Typically, angles less than about 15 degrees are considered suitable for this approximation. This means that the pendulum does not need to swing far from a vertical position, which keeps the math both simple and effective.
Pendulum Length
Pendulum length is an essential factor in determining the period of a pendulum's swing. In physics, this length refers to the distance from the pivot point to the center of mass of the pendulum bob. In our pendulum exercises:
  • The length greatly affects the time it takes for one complete oscillation or period, as described by the formula:
\[ T \approx 2\pi \sqrt{\frac{L}{g}} \]
  • Here, \( L \) is the length and \( g \) is the acceleration due to gravity.
  • The relationship is such that a longer pendulum will have a longer period and vice versa.
Finding the right pendulum length for a desired period involves rearranging the formula to solve for \( L \). This calculation is vital for designing clocks and other mechanisms where precise timing is necessary.
Sine Function
The sine function is an elemental trigonometric function that often appears in physics problems involving periodic motion, such as that of a pendulum. It relates the angle of a right triangle to the ratio of the length of the opposite side over the hypotenuse. This function is pertinent in pendulum calculations since
  • It describes the geometric nature of pendulum swings.
  • It's used to calculate correction factors when angles aren't small enough to rely solely on the approximation.
For example, in sinusoidal function corrections, the pendulum's maximum angle (\( \theta_{\max} \)) led us to determine \( k = \sin\left(\frac{\theta_{\max}}{2}\right) \), a crucial step for more precise period calculations.
Sinusoidal Function Corrections
Sinusoidal function corrections become necessary when the small angle approximation is insufficiently accurate. In our problem, we extended the standard formula with a correction factor:
  • This adjusted calculation better accounts for larger angles.
  • It serves to correct the period prediction when the simple model underestimates or overestimates period due to higher amplitudes.
The correction uses a term derived from the sine function: \( 1 + \frac{k^2}{4} \), where \( k = \sin\left(\frac{\theta_{\max}}{2}\right) \).
  • The calculation finds \( k \) and computes the correction factor.
  • Then, substitute this factor back into the period length formula to account for the additional nuance in the pendulum motion.
These corrections help refine the predicted length and make the model's results much more reliable.

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Most popular questions from this chapter

An equivalent formula for the period of a pendulum with amplitude \(\theta_{\max } \quad\) is \(T\left(\theta_{\max }\right)=2 \sqrt{2} \sqrt{\frac{L}{g}} \int_{0}^{\theta_{\max }} \frac{d \theta}{\sqrt{\cos \theta}-\cos \left(\theta_{\max }\right)}\) where \(L\) is the pendulum length and \(g\) is the gravitational acceleration constant. When \(\theta_{\max }=\frac{\pi}{3} \quad\) we \(\frac{1}{\sqrt{\cos t-1 / 2}} \approx \sqrt{2}\left(1+\frac{t^{2}}{2}+\frac{t^{4}}{3}+\frac{181 t^{6}}{720}\right)\). Integrate this approximation to estimate \(T\left(\frac{\pi}{3}\right)\) in terms of \(L\) and \(g\). Assuming \(g=9.806\) meters per second squared, find an approximate length \(L\) such that \(T\left(\frac{\pi}{3}\right)=2\) seconds.

Compute at least the first three nonzero terms (not necessarily a quadratic polynomial) of the Maclaurin series of \(f\). $$ f(x)=\sin \left(x+\frac{\pi}{4}\right)=\sin x \cos \left(\frac{\pi}{4}\right)+\cos x \sin \left(\frac{\pi}{4}\right) $$

Use the substitution \((b+x)^{r}=(b+a)^{r}\left(1+\frac{x-a}{b+a}\right)^{r} \quad\) in the binomial expansion to find the Taylor series of each function with the given center. $$ \begin{array}{l} \text { 181. } \quad \sqrt{2 x-x^{2}} \quad \text { at } \quad a=1 \quad \text { (Hint: } \\ \left.2 x-x^{2}=1-(x-1)^{2}\right) \end{array} $$

Use appropriate substitutions to write down the Maclaurin series for the given binomial. $$ (1-x)^{1.01} $$

Use the fact that if \(q(x)=\sum_{n=1}^{\infty} a_{n}(x-c)^{n} \quad\) converges in an interval containing \(c,\) then \(\lim _{x \rightarrow c} q(x)=a_{0}\) to evaluate each limit using Taylor series. $$ \lim _{x \rightarrow 0} \frac{\cos x-1}{x^{2}} $$
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