?The period of a pendulum with length \(L\) that makes a maximum angle \(\theta_{0}\)

Chapter 11, Problem 38

(choose chapter or problem)

The period of a pendulum with length \(L\) that makes a maximum angle \(\theta_{0}\) with the vertical is

                         

                                                 \(T=4 \sqrt{\frac{L}{g}} \int_{0}^{\pi / 2} \frac{d x}{\sqrt{1-k^{2} \sin ^{2} x}}\)

where \(k=\sin \left(\frac{1}{2} \theta_{0}\right)\) and \(g\) is the acceleration due to gravity. (In Exercise 7.7.42 we approximated this integral using Simpson's Rule.)

(a) Expand the integrand as a binomial series and use the result of Exercise 7.1.56 to show that

\(T=2 \pi \frac{L}{q}\left[1+\frac{1^{2}}{2^{2}} k^{2}+\frac{1^{2} 3^{2}}{2^{2} 4^{2}} k^{4}+\frac{1^{2} 3^{2} 5^{2}}{2^{2} 4^{2} 6^{2}} k^{6}+\ldots\right]\)

If \(\theta_{0}\) is not too large, the approximation \(T \approx 2 \pi \sqrt{L / g}\) obtained by using only the first term in the series, is often used. A better approximation is obtained by using two terms:

                               

                                                   \(T \approx 2 \pi \sqrt{\frac{\bar{L}}{g}}\left(1+\frac{1}{4} k^{2}\right)\)

(b) Notice that all the terms in the series after the first one have coefficients that are at most \(\frac{1}{4}\).Use this fact to compare this series with a geometric series and show that

               

                                                  \(2 \pi \sqrt{\frac{\bar{L}}{g}}\left(1+\frac{1}{4} k^{2}\right) \leq T \leq 2 \pi \sqrt{\frac{\bar{L}}{g} \frac{4-3 k^{2}}{4-4 k^{2}}}\)

(c) Use the inequalities in part (b) to estimate the period of a pendulum with \(L=1\) meter and \(\theta_{0}=10^{\circ}\). How does it compare with the estimate \(T \approx 2 \pi \sqrt{L / g}\)? What if \(\theta_{0}=42^{\circ}\)?

Equation Transcription:

Text Transcription:

L

theta_0

T = 4sqrt frac{L}{g} integral^0_pi/2 frac{dx}{sqrt 1-k^2 sin^2 x}

k = sin (frac{1}{2} theta_0)

g

T = 2pi L/q [1+frac{1^2}{2^2} k^2 + frac{^23^2}{2^2 4^2} k^4 + frac{1^2 3^2 5^2}{2^2 4^2 6^2} k^6+...]

theta_0

T approx 2pi sqrt L/g

T approx 2pi sqrt frac{L}{g}(1 + frac{1}{4} k^2)

frac{1}{4}

2pi sqrt frac{L}{g} (1+frac{1}{4} k^2) leq T leq 2pi sqrt frac{L}{g} frac{4-3k^2}{4-4k^2}

L = 1

theta_0 = 10 degree

T approx 2pi sqrt L/g

theta_0 = 42 degree