[SHIVOK SP212] March 10, 2016 CH 31 Electromagnetic Oscillations and Alternating Current I. Review A. LastSemesteryoulearnedinMechanics(CH‐15) 1. LinearSimpleHarmonicOscillator(fig15‐15) 2. UsingF==()Newton’sLaw 3. Also…F=Hooke’sLaw 4. Thus:=()k= 5. Sow=(angularfrequency) 6. Rememberthat: a) B. velocity b) PotentialEnergy c) KineticEnergy d) period AboutamonthagoinChapter25youlearned: 1. EnergystoredinanElectricField(CH‐25) Page1 [SHIVOK SP212] March 10, 2016 2. orPotentialEnergyinanElectricField C. LastweekinChapter30youlearned: 1. EnergystoredinaMagneticField(CH‐30) 2. orPotentialEnergyinaMagnetic Field II. ElectroMagneticOscillations A. Inductor/Capacitor(LC)combinedCKTOverview(fig31‐1) Fig 31‐ 1 Page2 [SHIVOK SP212] March 10, 2016 1. =0. Lookat(a):AssumeCapacitorischarged,nocurrent.UE=MaxandUB 2. In(b),Capacitorstarts_______________________________,andcurrentis ___________________…thusInductor’smagneticfieldstartsto_________________. 3. In(c),Capacitorisfully__________________;Inductor’scurrentisat _______________.UE=0andUB=Max. 4. In(d),Capacitorstarts_____________________again,butthistimein __________________polarity.Inductor’scurrentis_______________________________. 5. In(e),Capacitorreaches________________________charge.Againnotice polarity;nocurrent. 6. In(f),Capacitorstarts___________________________withcurrentflowin _______________________directionfrom(b),andcurrentis________________________…thus Inductor’smagneticfieldstartsto___________________________________________ direction. 7. In(g),Capacitoris____________________________________;Inductor’scurrentis atmax.UE=0andUB=Max. 8. B. Finallyin(h),Capacitorstartschargingagainascurrentdecreases. UsingMechanicalAnalog 1. LookbackatA6inyouroutline;Ifwereplacexwithqthen: Page3 [SHIVOK SP212] March 10, 2016 1. 2. ThespringwhichhasPotentialEnergyisreplacedwithaCapacitor 3. Themassbecomestheinductor L= 4. Thuswherew= 5. a) thenw= b) c) Let’sPROVEwithCalculus Block‐SpringExample a) (1) (2) U=UB+US= = ()=0Totalenergyremains constantwithtime. (a) =0,butrememberthat (b) …So (c) =0(block‐springOscillations) ;thus=0 (3) SolutiontothisDiffEQis: (a) x=(displacement) Page4 [SHIVOK SP212] March 10, 2016 C. ForOurLCOscillator 1. U=UB+UE=or= a) = ()=0Totalenergy (1) b) Thus 0 But,rememberthat: and = (1) (a) Therefore:=0 (LCOscillations) c) NoticethisequationisofthesametypeofDIFFEQasbefore (1) Thereforesolutionis:q= (charge) (2) = Thus (current) (3) SO ori= Page5 [SHIVOK SP212] March 10, 2016 d) AngularFrequency(w) (1) Since then (2) Usingthestarequationandsubstitutingfromabove: (a) =0 (b) =1 (c) w=sameaswesaid before.NOTE:w= ,wherefis frequencyinHz. 2. ElectricalandMagneticEnergyOscillations a) b) andsubstitutingw= /√ Page6 [SHIVOK SP212] March 10, 2016 D. ExampleLCProblems 1. Asshownbelow,aseriesLCcircuitcontainsa100mHInductoranda 36mFCapacitorthatwaschargedbya12Vbattery.Whatisthefrequencyin Hz.oftheelectromagneticoscillationsinthecircuit(neglectresistanceofthe circuitwires)? Page7 [SHIVOK SP212] March 10, 2016 2. Exampletwo:AnLCcircuitproduceselectromagneticoscillationswitha frequencyof 1.50 x10 6 Hz. Ifthecapacitanceis2.00nF,theinductanceinthecircuit isclosestto: A. 1.98 x106 H B. 4.35 x106 H C. 3.47 x104 H D. 5.63 x106 H E. 2.22 x104 H Show all work: a) Answeris.Ifyouchosethenyouforgot ______________________________________________________________________________. Page8 [SHIVOK SP212] March 10, 2016 3. Example Three: In the circuit below, R = 2.0 Ω, C = 1.4 F, L = 0.50 H and E = 6.0 V. The switch has been in position 1 for a long time (t = ∞) and the capacitor has no charge. The switch is then quickly thrown to position 2. After the switch is thrown to position 2, the maximum charge on the capacitor is closest to: A. 1.7 C. B. 2.5 C. C. 2.9 C. D. 3.4 C. E. 3.8 C. Show all work: Page9 [SHIVOK SP212] March 10, 2016 E. RealWorldLCOscillatorsareDAMPED 1. RememberdampingisFrictionintheBlock‐Springexample.Its electricalanalogisResistance. a) Note:EveniftheCircuitonlyhasanInductorandaCapacitorin theCircuit,thewirehasresistance.ResistanceremovesEnergyinthe formofheat. 2. IfwestartwithU=UB+UE=andrememberfrom CH‐26thattheRATEofheattransferis: a) = b) Then =but (1) Therefore:=0(LRCOscillations) c) Thereforesolutionis:q=Q (Eq) (1) / and = cos(w’t+f)(charge) w’=andw= /√ F. ExampleLRCProblem 1. a) Answeris Page 10