Problem 8.56 Challenge

แทนค่า \(x(t) = Acos(ωt + φ)\)

ในสมการ \(\frac{d^{2}x}{dt^{2}}+ 2\gamma \frac{dx}{dt}+ \omega _{0}^{2}x =\frac{F_{0}}{m}sin(ωt)\) จะได้

\(-\omega ^{2}Acos(\omega t + φ) - 2\gamma \omega Asin(\omega t + φ) + \omega _{0}^{2}Acos(\omega t + φ) =\frac{F_{0}}{m}\sin (\omega t)\)

\(-2\gamma \omega Asin(\omega t + φ) - (\omega ^{2} - \omega _{0}^{2})Acos(\omega t + φ) =\frac{F_{0}}{m}\sin (\omega t)\)

\(-2\gamma \omega Acosφ\sin (\omega t) - 2\gamma \omega Asinφ\cos (\omega t)\)

\(- (\omega ^{2} - \omega _{0}^{2})Acosφ\cos (\omega t) + (\omega ^{2} - \omega _{0}^{2})Asinφ\sin (\omega t) =\frac{F_{0}}{m}\sin (\omega t)\)

\(A[((\omega ^{2} - \omega _{0}^{2})\sin φ - 2\gamma \omega \cos φ)\sin (\omega t)\)

\(- (2\gamma \omega \sin φ + (\omega ^{2} - \omega _{0}^{2})\cos φ)\cos (\omega t)] =\frac{F_{0}}{m}\sin (\omega t)\)

สมการจะเป็นจริงเมื่อ

\(2\gamma \omega \sin φ + (\omega ^{2} - \omega _{0}^{2})\cos φ = 0\)

\(\tan φ =\frac{-(\omega ^{2} - \omega _{0}^{2})}{2\gamma \omega}\)

มองเป็นรูปสามเหลี่ยมมุมฉากได้ ดังนี้

\(2\gamma \omega\)

\(φ\)

\(-(\omega ^{2} - \omega _{0}^{2})\)

\(\sqrt{(\omega ^{2} - \omega _{0}^{2})^{2} + 4\gamma ^{2}\omega ^{2}}\)

จะได้ \(sinφ = \frac{-(\omega ^{2} - \omega _{0}^{2})}{(\omega ^{2} - \omega _{0}^{2})^{2} + 4\gamma ^{2}\omega ^{2}}\)

และ \(cosφ = \frac{2\gamma \omega}{(\omega ^{2} - \omega _{0}^{2})^{2} + 4\gamma ^{2}\omega ^{2}}\)

ตอนนี้สมการลดรูปลงมา

\(A((\omega ^{2} - \omega _{0}^{2})\sin φ - 2\gamma \omega \cos φ)\sin (\omega t) =\frac{F_{0}}{m}\sin (\omega t)\)

\(A((\omega ^{2} - \omega _{0}^{2})\sin φ - 2\gamma \omega \cos φ) =\frac{F_{0}}{m}\)

\(A((\omega ^{2} - \omega _{0}^{2})(\frac{-(\omega ^{2} - \omega _{0}^{2})}{(\omega ^{2} - \omega _{0}^{2})^{2} + 4\gamma ^{2}\omega ^{2}})\)

\(- 2\gamma \omega (\frac{2\gamma \omega}{(\omega ^{2} - \omega _{0}^{2})^{2} + 4\gamma ^{2}\omega ^{2}})) =\frac{F_{0}}{m}\)

\(A(-\frac{(\omega ^{2} - \omega _{0}^{2})^{2}}{(\omega ^{2} - \omega _{0}^{2})^{2} + 4\gamma ^{2}\omega ^{2}}-\frac{4\gamma ^{2}\omega ^{2}}{(\omega ^{2} - \omega _{0}^{2})^{2} + 4\gamma ^{2}\omega ^{2}}) =\frac{F_{0}}{m}\)

\(-A\frac{(\omega ^{2} - \omega _{0}^{2})^{2} + 4\gamma ^{2}\omega ^{2}}{(\omega ^{2} - \omega _{0}^{2})^{2} + 4\gamma ^{2}\omega ^{2}}=\frac{F_{0}}{m}\)

\(-A\sqrt{(\omega ^{2} - \omega _{0}^{2})^{2} + 4\gamma ^{2}\omega ^{2}}=\frac{F_{0}}{m}\)

จะได้ \(A = -\frac{F_{0}}{m(\omega ^{2} - \omega _{0}^{2})^{2} + 4\gamma ^{2}\omega ^{2}}\)

เนื่องจาก A คือ แอมพลิจูดจึงสนใจแต่ขนาด

\(\therefore A =\frac{F_{0}}{m(\omega ^{2} - \omega _{0}^{2})^{2} + 4\gamma ^{2}\omega ^{2}}\;\blacksquare\)