Second-Order Linear ODE (7) - Forced Oscillation

12. Driving Force

 

$ my''+cy'+ky=r(t)=F_{0}cos\omega t $

$ r(t)=F_{0}cos\omega t $: Driving Force

 

$ y=y_{h}+y_{p} $

$ y_{p}=a\cos \omega t+b\sin \omega t $

$ y_{p}'=-\omega a\sin \omega t+\omega b\cos \omega t $

$ y_{p}''=-\omega ^{2}a\cos \omega t-\omega ^{2}b\sin \omega t $

$ \rightarrow m(-\omega ^{2}a\cos \omega t-\omega ^{2}b\sin \omega t)+c(-\omega a\sin \omega t+\omega b\cos \omega t)+k(a\cos \omega t+b\sin \omega t)=F_{0}cos\omega t $

$ [(k-m\omega ^{2})a+\omega cb]\cos \omega t+[-\omega ca+(k-m\omega ^{2})b]\sin \omega t=F_{0}cos\omega t $

$ \rightarrow (k-m\omega ^{2})a+\omega cb=F_{0} $: ㄱ

   $ -\omega ca+(k-m\omega ^{2})b=0 $: ㄴ

 

We want to know a and b.

ㄱ($ k-m\omega ^{2} $)+ㄴ($ -\omega c $) → $ (k-m\omega ^{2})^{2}a+\omega ^{2}c^{2}a=F_{0}(k-m\omega ^{2}) $

ㄱ($ \omega c $)+ㄴ($ k-m\omega ^{2} $) → $ \omega ^{2}c^{2}b+(k-m\omega ^{2})^{2}b=F_{0}\omega c $

$ \rightarrow a=F_{0}\frac{k-m\omega ^{2}}{(k-m\omega ^{2})^{2}+\omega ^{2}c^{2}} $

   $ b=F_{0}\frac{\omega c}{(k-m\omega ^{2})^{2}+\omega ^{2}c^{2}} $

($ \omega _{0}=\sqrt{\frac{k}{m}} \rightarrow k=m\omega _{0}^{2} $)

$ \Rightarrow a=F_{0}\frac{m(\omega _{0}^{2}-\omega ^{2})}{m^{2}(\omega _{0}^{2}-\omega ^{2})^{2}+\omega ^{2}c^{2}} $

   $ b=F_{0}\frac{\omega c}{m^{2}(\omega _{0}^{2}-\omega ^{2})^{2}+\omega ^{2}c^{2}} $

$ \therefore y=y_{h}+y_{p}=C\cos (\omega _{0}t-\delta )+(F_{0}\frac{m^{2}(\omega _{0}^{2}-\omega ^{2})}{m^{2}(\omega _{0}^{2}-\omega ^{2})^{2}+\omega ^{2}c^{2}}\cos \omega t+F_{0}\frac{\omega c}{m^{2}(\omega _{0}^{2}-\omega ^{2})^{2}+\omega ^{2}c^{2}}\sin \omega t) $

 

 

13. Forced Oscillation - Undamped

 

Small Damping & Short Time → Disregard Damping(c=0)

$ my''+ky=F_{0}cos\omega t $

$ \rightarrow a=\frac{F_{0}}{m(\omega _{0}^{2}-\omega ^{2})}, b=0 $

$ \therefore y=y_{h}+y_{p}=C\cos (\omega _{0}t-\delta )+\frac{F_{0}}{m(\omega _{0}^{2}-\omega ^{2})}\cos \omega t $

 

i) Resonance

$ \omega =\omega _{0} $

$ y=y_{h}+y_{p}=(A\cos \omega _{0}t+B\sin \omega _{0}t)+y_{p} $

By modification rule, $ y_{p}=\alpha t\cos \omega _{0}t+\beta t\sin \omega _{0}t $

$ \rightarrow \alpha =0, \beta =\frac{F_{0}}{2m\omega _{0}} $

$ \therefore y_{p}=\frac{F_{0}}{2m\omega _{0}}t\sin \omega _{0}t $: amplitude becomes larger and larger

 

ii) Beats

$ \omega \approx \omega _{0} $

$ y_{p}=\frac{F_{0}}{m(\omega _{0}^{2}-\omega ^{2})}(\cos \omega t-\cos \omega_{0} t) $: $ y_{h} $의 일부를 가지고 온다

$ \therefore y_{p}=\frac{2F_{0}}{m(\omega _{0}^{2}-\omega ^{2})}\sin (\frac{\omega _{0}+\omega }{2}t)\sin (\frac{\omega _{0}-\omega }{2}t) $

 

 

14. Forced Oscillation - Damped

 

$ my''+cy'+ky=F_{0}cos\omega t  $

$ y=y_{h}+y_{p} $: Transient Solution

$ y=y_{p} $: Steady-State Solution

 

Theorem) Steady-State Solution

After a sufficiently long time, $ y_{h} \rightarrow 0 $ so $ y \rightarrow y_{p} $

 

$ y_{p}=a\cos \omega t+b\sin \omega t=C^{*}\cos (\omega t-\eta ) $

$ C^{*}(\omega )=\sqrt{a^{2}+b^{2}}=\frac{F_{0}}{\sqrt{m^{2}(\omega _{0}^{2}-\omega ^{2})^{2}+\omega ^{2}c^{2}}} $

$ \tan \eta (\omega )=\frac{b}{a}=\frac{\omega c}{m(\omega _{0}^{2}-\omega ^{2})} $

 

Find maximum of $ C^{*}(\omega ) $

$ \frac{\mathrm{d} C^{*}}{\mathrm{d} \omega }=-\frac{1}{2}F_{0}\frac{2m(\omega _{0}^{2}-\omega ^{2})(-2\omega )+2\omega c^{2}}{(m^{2}(\omega _{0}^{2}-\omega ^{2})^{2}+\omega ^{2}c^{2})^{\frac{3}{2}}}=0 $

$ 2m(\omega _{0}^{2}-\omega ^{2})(-2\omega )+2\omega c^{2}=0 $

$ c^{2}=2m^{2}(\omega _{0}^{2}-\omega ^{2}) $

$ 2m^{2}\omega ^{2}=2m^{2}\omega _{0}^{2}-c^{2}=2mk-c^{2} $

→ If $ c^{2}>2mk $, $ C^{*}(\omega ) $ decreases as $ \omega $ increases.

    If $ c^{2}<2mk $, $ \omega _{max}^{2}=\omega _{0}^{2}-\frac{c^{2}}{2m^{2}} $

$ C^{*}(\omega _{max})=\frac{F_{0}}{\sqrt{m^{2}(\omega _{0}^{2}-\omega _{max}^{2})^{2}+\omega _{max}^{2}c^{2}}}=\frac{F_{0}}{\sqrt{\frac{c^{4}}{4m^{2}}+(\omega _{0}^{2}-\frac{c^{2}}{2m^{2}})c^{2}}}=\frac{F_{0}}{\sqrt{\frac{c^{4}+4m^{2}\omega _{0}^{2}c^{2}-2c^{4}}{4m^{2}}}}=\frac{2mF_{0}}{c\sqrt{{4m^{2}\omega _{0}^{2}-c^{2}}}} $