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User: As a text script expert, please help me to write a short text script with the topic "Taking into consideration (4) and (5) we can of three forces, i.e. the restoring force of a spring F (t)= 0 res easilydeducethatthephase-lagangleϕ(ω )dependsonthe −kx(t), where k is the stiffness of the spring, the damping A amountofdampinginthesystemandtakesonarangeofval- force F (t)=−bx˙(t), where b>0 is the damping constant, d ues, i.e. ϕ(ω )∈(0,π/2), where the limits of the range are and the external driving force F ext(t)=F 0cos(ωt), where ω A √ istheangularfrequencyofthedrivingforce.A(ω)= F 0 (3) ϕ(ω A)≈ϕ(ω V)onlyifγ ≪ω 0.(4) =−F 0ωA(ω)cos(ωt)sin(ωt−ϕ), (8) 0 5202 naJ 8 ]hp-ssalc.scisyhp[ 1v79740.1052:viXra Drivendampedharmonicoscillatorrevisited: energyresonance 2 whilethesteady-stateinstantaneouspowerlossisgivenby istheamplitudeofthepotentialenergy,and P (ω,t)=F (t)v (t) mω2A2(ω) out d s E (ω)= (14)Denoting the time average over one period by brackets, i.e. ω2/ω2, for driving frequencies ω < ω we have E (ω) > ⟨f(t)⟩=T−1(cid:82) tt+T f(t′)dt′,whereT =2π/ω istheperiodof E K0 (ω),i.e.thetotalenergy(12)isnotco0 nstant,butraP theros- thedrivingforce,weget⟨cos(ωt)sin(ωt−ϕ)⟩=−sin(ϕ)/2 cillates with peak values equal to E (ω).Since dE(ω,t)/dt =P (ω,t)+ in F ωA(ω)sin(ϕ) ⟨P in(ω)⟩= 0 =mγω2A2(ω) (10) P out(ω,t),anddE(ω,t)/dt=0forω =ω 0,theinstantaneous 2 input power and power loss are in balance, i.e. P (ω ,t)= out 0 isthetime-averagedsteady-stateinputpower,and −P in(ω 0,t) at the velocity resonance.Thus, for ω =ω 0 the systembehavesasanundampedfreeoscillatorthatoscillates ⟨P (ω)⟩=−mγω2A2(ω) (11) withtotalenergyE(ω ,t)=mω2A2(ω )/2.Wenoteherethat out 0 0 0 it is stated in textbook3 that the input power and power loss is the time-averaged steady-state power loss.For driving frequencies ω >ω we have E (ω)<E (ω), i.e. the total energy (12) 2γω 0 P K sin(ϕ)= (cid:113) , oscillateswithpeakvaluesequaltoE K(ω).The resonant frequencies ω and ω are alsoThe time-averaged input power (10) resonantfrequenciesoftheamplitudesofpotentialandkinetic is equal in magnitude to the time-averaged power loss (11) energy,i.e.of(13)and(14),respectively.Theresonantampli- for any driving frequency ω and damping coefficient γ, i.e. tudeofpotentialenergyasafunctionofγ is ⟨P (ω)⟩=−⟨P (ω)⟩forallω >0andγ >0.Wenotehere thao tut the expressioin n for ⟨P out(ω)⟩ in some physics textbooks3 E (ω )= F 02 , (15) ismissingtheminussignand,inaddition,itisstatedseveral P A 8mγ2(1−(γ/ω 0)2) times that ⟨P (ω)⟩=⟨P (ω)⟩, which could potentially be in out andtheresonantamplitudeofkineticenergyasafunctionof confusing to students.We see that 0 level textbooks3–6, is that the frequency ω V also maximizes E P(ω A)>E K(ω V) is valid, and due to the discussion in the thepeakvalueofinstantaneousinputpower(8).ENERGY RESONANCE energygivenby 1 (cid:90) t+T E (ω)+E (ω) Thetotaloscillatorsteady-stateinstantaneousenergyis ⟨E(ω)⟩= E(t′)dt′= P K .(18) T t 2 E(ω,t)=E P(ω)cos2(ωt−φ)+E K(ω)sin2(ωt−φ), (12) Wewill referto(18) simplyasthe averageenergy.mω2A2(ω) onance of the average energy is rarely discussed in the liter- E P(ω)= 0 (13) ature, and even then only in the weak damping limit, i.e. for 2 Drivendampedharmonicoscillatorrevisited: energyresonance 3 γ ≪ω .3,6 Thus, to the best of our knowledge, an exact ex- ω is(21) condition d⟨E(ω)⟩ ∝ d(ω 02+ω2)A2(ω) =0,andweget ω ⟨E⟩ 2 2(cid:112) 1−(γ/ω 0)2−1 dω dω (cid:114) TheMaclaurinseriesofthisexpressionismissingthefirstfive (cid:113) ω ⟨E⟩= 2ω 0 ω 02−γ2−ω 02 (19) powersinγ/ω 0andisgivenby 1 (cid:18) γ (cid:19)6 19 (cid:18) γ (cid:19)8 fortheresonantfrequencyoftheaverageenergy.Thevalueof R≈ + +..., (22) 16 ω 128 ω theaverageenergyatresonanceisgivenby 0 0 whichshowsthatR≲1%forγ ≲0.65ω .⟨E(ω )⟩≈E (ω )≈Using the condition 2ω 0(cid:113) ω 02−γ2−ω 02 >0, we determine E K(ω V) = F 02/(8mγ2), when γ ≪ ω 0.to occur between ω and ω , i.e. ω <ω <ω , for anyResonantfrequenciesωA (solidbluecurve), ωV (solidred ciesω ={ω A,ω ⟨E⟩,ω V}andγ =0.5ω 0.and(γ,ωAV)=( 2ω0/2,ω0/2) √respectively,whilethedashedblack curveendsatpoint(γ,ω √⟨E⟩)=( 3ω0/2,0).TheapproximationωAV ⟨P in(ω ⟨E⟩)⟩−⟨P in(ω 0)⟩⟨E(ω ⟨E⟩)⟩−⟨E(ω 0)⟩ ·100% by ω AV =(ω A+ω V)/2.The filled red cir- anddrivingfrequenciesωA(solidbluecurve),ω ⟨E⟩(solidredcurve) cles designate the values of A(ω ⟨E⟩),V(ω ⟨E⟩) and a(ω ⟨E⟩), which andωV (dashedblackhorizontalline).TheunitsofenergyareE 0= correspondtotheresonantaverageenergy(20).⟨P in(ω ⟨E⟩)⟩T(ω ⟨E⟩)−⟨P in(ω 0)⟩T(ω 0) incomparisontostandardtextbooks, ourresultscanalsoen-In tally obtained (ω,A(ω)) data, students could calculate the termsofinputenergypercycle,drivingthesystemwithω (ω,ω2A2(ω)),(ω,ω2A2(ω))and(ω,ω2A(ω))dataandplotandtodeterminethevaluesofA(ω), V(ω), and a(ω) that correspond to maximal steady-state av- where a(ω) = ω2A(ω) is the acceleration amplitude.Drivendampedharmonicoscillatorrevisited: energyresonance Driven damped harmonic oscillator revisited: energy resonance K. Lelas1 and N. Poljak2 1)FacultyofTextileTechnology,UniversityofZagreb,Croatia 2)DepartmentofPhysics,FacultyofScience,UniversityofZagreb,Croatia (*Electronicmail:<PRESIDIO_ANONYMIZED_EMAIL_ADDRESS>) (*Electronicmail:<PRESIDIO_ANONYMIZED_EMAIL_ADDRESS>) (Dated:10January2025) Wederivetheexactexpressionfortheresonantfrequencyofthetime-averagedsteady-stateenergyandshowthatthis frequencyisexcellentlyapproximatedbythearithmeticmeanoftheamplitudeandvelocityresonantfrequencies.The state are analyzed, but energy resonance is not discussed at driving frequency that maximizes the steady-state displace- all,whiletime-averagedenergyresonanceisconsideredrarely mentamplitude(3)is4 andonlyintheweakdampinglimit.Standard physics textbooks for undergraduate studies3–6 thoroughly discuss how the steady-state behavior of the We see that ω V does not depend on damping, and thus, con- damped driven harmonic oscillator is determined by the par- trary to the amplitude resonance, velocity resonance effects ticularsolutionofnonhomogeneousdifferentialequation(1), are present for any value of the damping coefficient γ.We i.e.thesteady-statesolutionofequation(1)isgivenby should note that in this type of resonance the phase-lag an- gle is ϕ(ω ) = π/2 for any damping coefficient γ, sinceThus, only in the weak damping limit are both resonant frequencies and where phase-lag angles approximately equal, i.e. ω ≈ ω and(4) =−F 0ωA(ω)cos(ωt)sin(ωt−ϕ), (8) 0 5202 naJ 8 ]hp-ssalc.scisyhp[ 1v79740.1052:viXra Drivendampedharmonicoscillatorrevisited: energyresonance 2 whilethesteady-stateinstantaneouspowerlossisgivenby istheamplitudeofthepotentialenergy,and P (ω,t)=F (t)v (t) mω2A2(ω) out d s E (ω)= (14)Denoting the time average over one period by brackets, i.e. ω2/ω2, for driving frequencies ω < ω we have E (ω) > ⟨f(t)⟩=T−1(cid:82) tt+T f(t′)dt′,whereT =2π/ω istheperiodof E K0 (ω),i.e.thetotalenergy(12)isnotco0 nstant,butraP theros- thedrivingforce,weget⟨cos(ωt)sin(ωt−ϕ)⟩=−sin(ϕ)/2 cillates with peak values equal to E (ω).Since dE(ω,t)/dt =P (ω,t)+ in F ωA(ω)sin(ϕ) ⟨P in(ω)⟩= 0 =mγω2A2(ω) (10) P out(ω,t),anddE(ω,t)/dt=0forω =ω 0,theinstantaneous 2 input power and power loss are in balance, i.e. P (ω ,t)= out 0 isthetime-averagedsteady-stateinputpower,and −P in(ω 0,t) at the velocity resonance.Thus, for ω =ω 0 the systembehavesasanundampedfreeoscillatorthatoscillates ⟨P (ω)⟩=−mγω2A2(ω) (11) withtotalenergyE(ω ,t)=mω2A2(ω )/2.Wenoteherethat out 0 0 0 it is stated in textbook3 that the input power and power loss is the time-averaged steady-state power loss.For driving frequencies ω >ω we have E (ω)<E (ω), i.e. the total energy (12) 2γω 0 P K sin(ϕ)= (cid:113) , oscillateswithpeakvaluesequaltoE K(ω).The time-averaged input power (10) resonantfrequenciesoftheamplitudesofpotentialandkinetic is equal in magnitude to the time-averaged power loss (11) energy,i.e.of(13)and(14),respectively.Theresonantampli- for any driving frequency ω and damping coefficient γ, i.e. tudeofpotentialenergyasafunctionofγ is ⟨P (ω)⟩=−⟨P (ω)⟩forallω >0andγ >0.we can refer to ω =ω as the resonant frequency of input total instantaneous energy is not taken anywhere else in themω2A2(ω) onance of the average energy is rarely discussed in the liter- E P(ω)= 0 (13) ature, and even then only in the weak damping limit, i.e. for 2 Drivendampedharmonicoscillatorrevisited: energyresonance 3 γ ≪ω .3,6 Thus, to the best of our knowledge, an exact ex- ω is(21) condition d⟨E(ω)⟩ ∝ d(ω 02+ω2)A2(ω) =0,andweget ω ⟨E⟩ 2 2(cid:112) 1−(γ/ω 0)2−1 dω dω (cid:114) TheMaclaurinseriesofthisexpressionismissingthefirstfive (cid:113) ω ⟨E⟩= 2ω 0 ω 02−γ2−ω 02 (19) powersinγ/ω 0andisgivenby 1 (cid:18) γ (cid:19)6 19 (cid:18) γ (cid:19)8 fortheresonantfrequencyoftheaverageenergy.Thevalueof R≈ + +..., (22) 16 ω 128 ω theaverageenergyatresonanceisgivenby 0 0 whichshowsthatR≲1%forγ ≲0.65ω .It is 0 erage energy and the amplitudes of potential and kinetic en- easy to show that ω ≈ω ≈ω =ω in the weak damp- ⟨E⟩ A V 0 ergy as functions of ω for γ =0.5ω .to occur between ω and ω , i.e. ω <ω <ω , for anyAswealreadymen- line), and ω ⟨E⟩ (dashed black curve) as functions of the damping tioned,theresonanceofaverageenergywasconsideredinlit- coefficientγ.Inaddition,weshowωAV=(ωA+ωV)/2(dottedblack erature only for γ ≪ω and ω ≈ω was obtained3,6.⟨E(ω ⟨E⟩)⟩−⟨E(ω 0)⟩ ·100% by ω AV =(ω A+ω V)/2.The relative difference of ω AV and 2 ⟨E(ω 0)⟩ Drivendampedharmonicoscillatorrevisited: energyresonance 4 FIG.4.The steady-state instantaneous energy (12) for γ =0.5ω0 and a(ω) (dashed black curve) for γ =0.5ω0.In tally obtained (ω,A(ω)) data, students could calculate the termsofinputenergypercycle,drivingthesystemwithω (ω,ω2A2(ω)),(ω,ω2A2(ω))and(ω,ω2A(ω))dataandplottallyobtainedplotsforseveraldifferentvaluesofγ, students 0 Finally,wenotethataccelerationresonanceisalsonotcon- could gain insight into the behavior of different resonances sidered in standard undergraduate physics textbooks1–6.andtodeterminethevaluesofA(ω), V(ω), and a(ω) that correspond to maximal steady-state av- where a(ω) = ω2A(ω) is the acceleration amplitude.driving frequency that maximizes the acceleration amplitude Weshouldmentionthatlabsetupswhichstudentstypically iseasilydeterminedfromthecondition da(ω) =0whichgives see in their classes when dealing with forced damped oscil- dω lationsofteneasilyachievethedampingcoefficientsγ which ω2 ω are outside the weak damping limit.In Fig.4 we show the scaled displacement amplitude, the scaled velocity amplitude and the scaled acceleration ampli- tudeasfunctionsofω forγ =0.5ω .".Your output should use the following template:\n\n## {Subtitle01}\n- {Bulletpoint01}\n- {Bulletpoint02}\n## {Subtitle02}\n- {Bulletpoint03}\n- {Bulletpoint04}\n\nSummarize the giving topic to generate a mind map (as many subtitles as possible, with a minimum of three subtitles) structure markdown. Do not include anything in the response, that is not the part of mindmap.\n  Most Importantly your output must use language "English" and each point or pointer should include no more than 9 words.
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