Solution Found!
a) If you have not already done so, do .38(a). (b) If the
Chapter 4, Problem 4.39(choose chapter or problem)
a) If you have not already done so, do 4.38(a). (b) If the amplitude eD is small then so is A = sin el3/2. 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, r = To = 2701g. (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/-/1 A2u2 ti 1 + 1A2u2 and show that, in this approximation, (4.103) gives r = ro[l sin2(43/2)]. What percentage correction does the second term represent for an amplitude of 45? (The exact answer for elp = 45 is 1.040 ro to four significant figures.)
Questions & Answers
QUESTION:
a) If you have not already done so, do 4.38(a). (b) If the amplitude eD is small then so is A = sin el3/2. 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, r = To = 2701g. (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/-/1 A2u2 ti 1 + 1A2u2 and show that, in this approximation, (4.103) gives r = ro[l sin2(43/2)]. What percentage correction does the second term represent for an amplitude of 45? (The exact answer for elp = 45 is 1.040 ro to four significant figures.)
ANSWER: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