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

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