บทที่ 8 · การเคลื่อนที่แบบฮาร์มอนิกอย่างง่าย
Problem 8.49 Challenge
หัวข้อ: ลูกตุ้มอย่างง่าย (Simple Pendulum)
\(E=\frac{1}{2}mgd\theta _{0}^{2}\;\blacksquare\)
ดูวิธีทำ
a) จาก \(Στ = Iα\)
\(-mgdsin\theta = I\frac{d^{2}\theta}{dt^{2}}\)
\(\frac{d^{2}\theta}{dt^{2}}+\frac{mgd}{I}\sin \theta = 0\)
\(\frac{d^{2}\theta}{dt^{2}}+\frac{mgd}{I}\theta = 0; sinθ \approx θ\) เมื่อ \(θ\) น้อยมากๆ
\(\frac{d^{2}\theta}{dt^{2}}+ \omega _{0}^{2}\theta = 0; \omega _{0} =\sqrt{\frac{mgd}{I}}\)
\(\therefore θ\) จะเคลื่อนที่แบบฮาร์มอนิกอย่างง่าย
b) จาก \(ω_{0} = \sqrt{\frac{mgd}{I}}\)
\(2\pi f =\sqrt{\frac{mgd}{I}}\)
\(f =\frac{1}{2\pi}\sqrt{\frac{mgd}{I}}\;\blacksquare\)
c) จาก \(ω_{0} = \sqrt{\frac{mgd}{I}}\)
\(\frac{2\pi}{T}=\sqrt{\frac{mgd}{I}}\)
\(T = 2\pi \sqrt{\frac{I}{mgd}}\;\blacksquare\)
d) จาก \(\frac{d^{2}\theta}{dt^{2}}+ \omega _{0}^{2}\theta = 0; \omega _{0} =\sqrt{\frac{mgd}{I}}\)
\(\therefore \theta (t) = \theta _{0}\cos (\omega _{0}t + φ)\)
จาก \(\theta (0) = \theta _{0}\cos (\omega _{0}(0) + φ)\)
\(\theta _{0} = \theta _{0}\cos (φ)\)
\(\cos (φ) = 1\)
\(φ = 0 rad\)
\(\therefore \theta = \theta _{0}\cos (\sqrt{\frac{mgd}{I}}t) \;\blacksquare\)
e) จาก \(ω = \frac{d\theta}{dt}\)
\(\omega =\frac{d}{dt}(\theta _{0}\cos (\sqrt{\frac{mgd}{I}}t))\)
\(\omega = -\sqrt{\frac{mgd}{I}}\theta _{0}\sin (\sqrt{\frac{mgd}{I}}t) \;\blacksquare\)
f) จาก \(α = \frac{d\omega}{dt}\)
\(\alpha =\frac{d}{dt}(-\sqrt{\frac{mgd}{I}}\theta _{0}\sin (\sqrt{\frac{mgd}{I}}t))\)
\(\alpha = -\frac{mgd}{I}\theta _{0}\cos (\sqrt{\frac{mgd}{I}}t) \;\blacksquare\)
g) จาก \(ω^{2} + ω_{0}^{2}θ^{2} = (-ω_{0}θ_{0}sin(ω_{0}t))^{2} + ω_{0}^{2}(θ_{0}cos(ω_{0}t))^{2}\)
\(\omega ^{2} + \omega _{0}^{2}\theta ^{2} = \omega _{0}^{2}\theta _{0}^{2}\sin ^{2}(\omega _{0}t) + \omega _{0}^{2}\theta _{0}^{2}\cos ^{2}(\omega _{0}t)\)
\(\omega ^{2} + \omega _{0}^{2}\theta ^{2} = \omega _{0}^{2}\theta _{0}^{2}(\sin ^{2}(\omega _{0}t + φ) + \cos ^{2}(\omega _{0}t + φ))\)
\(\omega ^{2} + \omega _{0}^{2}\theta ^{2} = \omega _{0}^{2}\theta _{0}^{2}\)
\(\omega = \omega _{0}\sqrt{\theta _{0}^{2} - \theta ^{2}}\)
\(\omega =\sqrt{\frac{mgd}{I}(\theta _{0}^{2} - \theta ^{2})}\;\blacksquare\)
h) จาก \(E = K + U\)
\(E=\frac{1}{2}I\omega ^{2}+mgy\)
\(E=\frac{1}{2}I(\frac{mgd}{I}(\theta _{0}^{2} - \theta ^{2})) + mgd(1 - \cos \theta )\)
จาก \(\sin ^{2}(\theta ) =\frac{1 - \cos 2\theta}{2}\)
จะได้\(E=\frac{1}{2}mgd\theta _{0}^{2}-\frac{1}{2}mgd\theta ^{2}+mgd(2\sin ^{2}(\frac{\theta}{2}))\)
จาก \(sinθ \approx θ\) เมื่อ \(θ\) น้อยมากๆ
จะได้\(E=\frac{1}{2}mgd\theta _{0}^{2}-\frac{1}{2}mgd\theta ^{2}+\frac{1}{2}mgd\theta ^{2}\)
\(E=\frac{1}{2}mgd\theta _{0}^{2}\;\blacksquare\)