a) If you have not already done so, do .38(a). (b) If the | StudySoup

Textbook Solutions for Classical Mechanics

Chapter 4 Problem 4.39

Question

a) If you have not already done so, do 4.38(a). (b) If the amplitude \(\Phi\) is small then so is \(A=\sin\Phi/2\). If the amplitude is very small, we can simply ignore the last square root in (4.103). Show that this gives the familiar result for the small-amplitude period, \(\tau=\tau_0=2\pi\sqrt{l/g}\). (c) If the amplitude is small but not very small, we can improve on the approximation of part (b). Use the binomial expansion to give the approximation \(1/\sqrt{1-A^2u^2}\approx1+\frac{1}{2}A^2u^2\) and show that, in this approximation, (4.103) gives

\(\tau=\tau_{\mathrm{o}}\left[1+\frac{1}{4}\sin^2(\Phi/2)\right]\).

What percentage correction does the second term represent for an amplitude of \(45^{\circ}\)? (The exact answer for \(\Phi=45^{\circ}\) is \(1.040\tau_{\mathrm{o}}\) to four significant figures.)

Solution

Step 1 of 7

 

(a) Using the trig identity \((1-\cos \phi)=2 \sin ^{2}(\phi / 2)\), we can write the PE as \(U(\phi)=2 m g l \sin ^{2}(\phi / 2)\).

The KE is \(\frac{1}{2} m l^{2} \dot{\phi}^{2}\) and the total energy is just \(E=U(\Phi)\), since \(T = 0\) at the endpoint \(\phi=\Phi\).

Thus we can solve the equation \(T=U(\Phi)-U(\phi)\) to give

 

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full solution

Title Classical Mechanics 0 
Author John R Taylor
ISBN 9781891389221

a) If you have not already done so, do .38(a). (b) If the

Chapter 4 textbook questions

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