Problem 8.53 Challenge

จาก \(x = A(t)cos(ωt + φ)\)

\(_{-\gamma t}\)

\(x=Ae\cos (\omega t+φ)\)

\(_{-\gamma t-\gamma t}\)

\(\frac{dx}{dt}=-\omega Ae\sin (\omega t+φ)-\gamma Ae\cos (\omega t+φ)\)

กำหนดให้ \(θ = ωt + φ\)

\(_{-\gamma t-\gamma t}\)

จะได้ \(\frac{dx}{dt}=-\omega Ae\sin \theta -\gamma Ae\cos \theta\)

\(_{-\gamma t-\gamma t}\)

\(v=-\omega Ae\sin \theta \frac{\cos \theta}{\cos \theta}-\gamma Ae\cos \theta ;v=\frac{dx}{dt}\)

\(_{-\gamma t}\)

\(v=-(\omega \tan \theta +\gamma )Ae\cos \theta ; \tan \theta =\frac{\sin \theta}{\cos \theta}\)

\(_{-\gamma t}\)

\(v = -(\omega \tan \theta + \gamma )x;x=Ae\cos \theta\)

จาก \(E = K + U\)

\(E=\frac{1}{2}mv^{2}+\frac{1}{2}kx^{2}\)

\(E=\frac{1}{2}m(-(\omega \tan \theta +\gamma )x)^{2}+\frac{1}{2}kx^{2}\)

\(E=\frac{1}{2}m(\omega \tan \theta +\gamma )^{2}x^{2}+\frac{1}{2}kx^{2}\)

\(E=\frac{1}{2}x^{2}(m(\omega \tan \theta +\gamma )^{2}+k)\)

\(E=\frac{1}{2}kx^{2}(\frac{m}{k}(\omega \tan \theta + \gamma )^{2} + 1)\)

\(E=\frac{1}{2}kx^{2}(\frac{1}{\omega _{0}^{2}}(\omega \tan \theta + \gamma )^{2} + 1); \omega _{0} =\sqrt{\frac{k}{m}}\)

\(E=\frac{1}{2}kx^{2}(1+(\frac{\gamma  + \omega \tan \theta}{\omega _{0}})^{2})\)

\(_{-\gamma t}\)

\(E=\frac{1}{2}k(Ae\cos \theta )^{2}(1+(\frac{\gamma  + \omega \tan \theta}{\omega _{0}})^{2})\)

\(_{-2\gamma t}\)

\(E=\frac{1}{2}kA^{2}e(1 + (\frac{\gamma  + \omega \tan \theta}{\omega _{0}})^{2})\cos ^{2}\theta\)

\(_{-2\gamma t}\)

\(E=\frac{1}{2}kA^{2}e(1 + (\frac{\gamma  + \omega \tan (\omega t + φ)}{\omega _{0}})^{2})\cos ^{2}(\omega t + φ)\)

เพิ่มเติม : ในกรณีไม่มีความหน่วง \((γ = 0)\)

\(_{-2(0)t}\)

จะได้\(E=\frac{1}{2}kA^{2}e(1 + (\frac{(0) + \omega _{0}\tan (\omega t + φ)}{\omega _{0}})^{2})\cos ^{2}(\omega _{0}t + φ)\)

\(E=\frac{1}{2}kA^{2}(1+\tan ^{2}(\omega _{0}t+φ))\cos ^{2}(\omega _{0}t+φ)\)

\(E=\frac{1}{2}kA^{2}\sec ^{2}(\omega _{0}t+φ)\cos ^{2}(\omega _{0}t+φ)\)

\(E=\frac{1}{2}kA^{2}\) ซึ่งเป็นพลังงานในกรณีของ SHM นั่นเอง