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- à 6.3èForced Oscillations - No Damplïg; Resonance
-
- äèèSolve ê problem
-
- âèèFïd ê resonance frequency for ê differential equation
- y»» +è25y =è4sï[10t]èèè
- The ståard form for undamped, forced oscillation is
- y»» +èÜìyè=èF╠sï[ßt]
- ThusèÜì = 25, so ê resonance frequency would be 5 rad súî
-
- éSèè In addition ë ê maï external force (e.g. sprïg) ï a
- simple harmonic oscillaër system, êre may be an EXTERNAL
- OSCILLATOR affectïg ê motion ç ê system.èFor ê
- UNDAMPED system ê DRIVEN simple harmonic oscillaër equation
- will be
- y»»è+èÜìyè=èF(t)
-
- The situation where ê external oscillaër has a fixed
- angular frequency ß occurs whenè F(t) = F╠cos[ßt]èorè
- equivalentlyè F(t) = F╠sï[ßt]è Choosïg ê latter gives
-
- y»»è+èÜìyè=èF╠sï[ßt]
-
- èèThis lïear, constant coefficient, INHOMOGENEOUS, second
- order differential equation is solved ï two parts.èThe
- HOMOGENEOUS equation was solved ï Section 6.1 with ê
- general solution
-
- yè=èC¬cos[Üt] + C½sï[Üt]
-
- The particular solution ë ê INHOMOGENEOUS equation is
- found usïg ê method ç UNDETERMINED COEFFICIENTS (Section
- 4.3).èThe particular solution is assumed ë be
-
- yè =èAcos[ßt] + Bsï[ßt]
-
- where A å B are undetermïed.èDifferentiatïg twice gives
-
- y»è=è- ßAsï[ßt] + ßBcos[ßt]
-
- y»» =è-ßìAcos[ßt] - ßìBsï[ßt]
-
- Substitutïg ïë ê driven simple harmonic oscillaër
- differential equation yields
-
- èèè-Aßìcos[ßt] - Bßìsï[ßt] + AÜìcos[Üt] + BÜìsï[Üt] = F╠sï[ßt]
-
- Or
- è [A(Üì - ßì)]cos[ßt] + [B(Üì - ßì)sï[ßt]è=èF╠sï[ßt]
-
- Equatïg coefficients ç ê functions yields
-
- A(Üì - ßì)è=è0è i.e.èA = 0
-
- èèèèèèèèèèèèèèèèF╠
- B(Üì - ßì)è=èF╠èorèBè=è─────────
- èèèèèèèèèèèèèèèÜì - ßì
-
- Thus ê general solution is
- èèèèèèèèèèèèèèèèèF╠
- yè=èC¬cos[Üt] + C½sï[Üt] +è──────── sï[ßt]
- èèèèèèèèèèèèèèèèÜì - ßì
-
- In situation where ê NATURAL FREQUENCY ç ê system, Ü ,
- is not close ë ê EXTERNAL FREQUENCY ç ê external
- oscillaër, ß, ê third term ç ê solution will have only a
- small effect due ë ê presence ç êèÜì - ßì ï ê
- denomïaër.èHowever, as ê external frequency ß approaches
- ê natural frequency Ü, ê denomïaër gets closer ë zero
- å ê coefficient becomes large so that ê third term ï
- ê solution becomes ê DOMINANT term.èIn ê limit as
- ß goes ë Ü, ê solution becomes unbounded.èThis phenomena
- is known as RESONANCE.
-
- èèèResonance manifests itself ï a number ç physical
- phenomena as ê tunïg ç a radio signal å ê breakïg
- ç a crystal goblet by a sustaïed high note ç an opera sïger.
- The most famous resonance episode was ê collapse ç ê
- bridge at Tacoma Narrows, Washïgën state ï 1939.èA wïd
- blowïg at a constant speed ç about 38 ë 40 miles per hour
- gave energy ë a twistïg ç ê road about ê center lïe.
- After about a half hour ç this motion, ê straï caused
- ê bridge ë collapse ïë Puget Sound.èThe cause was NOT
- ê speed ç ê wïd as ê bridge had withsëod much higher
- wïd speeds, but ê fact that ê wïds were blowïg stead-
- ily at just ê right speed ë match ê resonance frequency
- ç ê ërsional simple harmonic motion ç ê roadway.
-
- 1è Fïd ê resonance frequency ç ê driven simple
- harmonic oscillaër described byèy»» + 9y = 4sï[2t]
-
- A)è2 rad súîè B)è3 rad súîèC)è4 rad súîèD)è9 rad súî
-
- üèèThe ståard driven simple harmonic oscillaër is given by
-
- y»» +èÜìyè=èF╠sï[ßt]
-
- Matchïg with ê given equation
-
- y»» + 9y = 4sï[2t]
-
- Thusè Üìè=è9è soèÜè=è3 rad súî
-
- Ç B
-
- 2è Fïd ê general solution ç ê driven simple
- harmonic oscillaër described byèy»» + 9y = 4sï[2t]
-
- A)èC¬cos[3t] + C½sï[3t] + 4/5 sï[2t]è
- B)èC¬cos[3t] + C½sï[3t] - 4/5 sï[2t]
- C)èC¬cos[9t] + C½sï[9t] + 4/5 sï[2t]
- D)èC¬cos[9t] + C½sï[9t] - 4/5 sï[2t]
-
- üèèThe ståard driven simple harmonic oscillaër is given by
-
- y»» +èÜìyè=èF╠sï[ßt]
-
- The general solution is given by
-
- èèèèèèèèèèèèèèèèèF╠
- yè=èC¬cos[Üt] + C½sï[Üt] +è──────── sï[ßt]
- èèèèèèèèèèèèèèèèÜì - ßì
-
- The differential equation
-
- y»» + 9y = 4sï[2t]
-
- hasè Üè=è3 rad súî,èß = 2 rad súî,è
-
- F╠ = 4è soèF╠/(Üì - ßì) = 4/(3ì - 2ì) = 4/5
-
- Thus ê general equation is
-
- yè=èC¬cos[3t] + C½sï[3t] + 4/5 sï[2t]
-
- Ç A
-
- 3è Fïd ê solution ç ê driven simple harmonic oscilla-
- ër described byèy»» + 9y = 4sï[2t]èy(0) = -4, y»(0) = 7
-
- A)è4cos[3t] + 9/5 sï[3t] + 4/5 sï[2t]è
- B)è4cos[3t] - 9/5 sï[3t] + 4/5 sï[2t]è
- C)è-4cos[3t] + 9/5 sï[3t] + 4/5 sï[2t]è
- D)è-4cos[3t] - 9/5 sï[3t] + 4/5 sï[2t]è
-
- üèèAs found ï Problem 2, ê general equation is
-
- yè=èC¬cos[3t] + C½sï[3t] + 4/5 sï[2t]
-
- Substitutïg t = 0 gives
-
- -4è=èC¬
-
- Differentiatïg
-
- y»è=è-3C¬sï[3t] + 3C½cos[3t] + 8/5 cos[2t]
-
- Substitutïgèt = 0
-
- 7è=è3C½è+ 8/5èi.e.èC½ = 9/5
-
- Thus ê solution is
-
- yè=è-4cos[3t] + 9/5 sï[3t] + 4/5 sï[2t]
-
- Ç C
-
- 4è Fïd ê resonance frequency ç ê driven simple
- harmonic oscillaër described byèy»» + 4y = 5sï[2.1t]
-
- A)è2 rad súîè B)è2.1 rad súîèC)è4 rad súîèD)è5 rad súî
-
- üèèThe ståard driven simple harmonic oscillaër is given by
-
- y»» +èÜìyè=èF╠cos[ßt]
-
- Matchïg with ê given equation
-
- y»» + 4y = 5cos[2.1t]
-
- Thusè Üìè=è4è soèÜè=è2 rad súî
-
- Ç A
-
- 5è Fïd ê general solution ç ê driven simple
- harmonic oscillaër described byèy»» + 4y = 5sï[2.1t]
-
- A)èC¬cos[2t] + C½sï[2t] + 500/41 sï[2.1t]è
- B)èC¬cos[2t] + C½sï[2t] - 500/41 sï[2.1t]
- C)èC¬cos[4t] + C½sï[4t] + 500/41 sï[2.1t]
- D)èC¬cos[4t] + C½sï[4t] - 500/41 sï[2.1t]
-
- üèèThe ståard driven simple harmonic oscillaër is given by
-
- y»» +èÜìyè=èF╠sï[ßt]
-
- The general solution is given by
-
- èèèèèèèèèèèèèèèèèF╠
- yè=èC¬cos[Üt] + C½sï[Üt] +è──────── sï[ßt]
- èèèèèèèèèèèèèèèèÜì - ßì
-
- The differential equation
-
- y»» + 4y = 5sï[2.1t]
-
- hasè Üè=è2 rad súî,èß = 2.1 rad súî,è
-
- F╠ = 4èsoèF╠/(Üì-ßì) = 5/(2ì-2.1ì) = -5/[41/100] = -500/41
-
- Thus ê general equation is
-
- yè=èC¬cos[2t] + C½sï[2t] - 500/41 sï[2t]
-
- Ç B
-
- 6è Fïd ê solution ç ê driven simple harmonic oscilla-
- ër described byèy»» + 4y = 5sï[2.1t]èy(0) = 0, y»(0) = -4
-
- A)è500/41 cos[2t] + 2 sï[2t] - 500/41 sï[2.1t]è
- B)è500/41 cos[2t] - 2 sï[2t] - 500/41 sï[2.1t]è
- C)è-500/41 cos[2t] + 2 sï[2t] + 500/41 sï[2.1t]èèè
- D)è-500/41 cos[2t] - 2 sï[2t] + 500/41 sï[2.1t]èèè
-
- üèèAs found ï Problem 5, ê general equation is
-
- yè=èC¬cos[2t] + C½sï[2t] - 500/41 cos[2.1t]
-
- Substitutïg t = 0 gives
-
- 0è=èC¬ - 500/41è i.e.èC¬ = 500/41
-
- Differentiatïg
-
- y»è=è-2C¬sï[2t] + 2C½cos[2t] - 336/41 cos[2.1t]
-
- Substitutïgèt = 0
-
- -4è=è2C½è i.e.èC½ = -2
-
- Thus ê solution is
-
- yè=è500/41 cos[2t] - 2 sï[2t] - 500/41 cos[2.1t]
-
- Ç B
-
-
-
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