Pro One: Differential Equations

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SECTON 6.1èSimple Harmonic Oscillaër - Undamped äèSolve ê problem â èèFïd ê period ç a pendulum that is 50 cm = 0.50 m long.è For a pendulum, ê period is given by èèT = 2π √(l/g)è=è2π √[0.50 m/9.8 m súì] = 1.42 sec éSèè Consider a mass m that has been suspended from a vertical sprïg å allowed ë come ë rest.èThis is known as ê equilibrium position å is ê place where ê upward sprïg force balances ê downward force ç gravity.èIf ê mass is moved eiêr upward or downward, ê sprïg will exert a force so as ë move ê mass back ë its equilbrium poït.èAlso, ê farêr ê mass is displaced from ê equilibrium position, ê stronger ê resërïg force ç ê sprïg.èThe simplest model for a sprïg force is Fè=è- kx whereèx = displacement from equilbrium èèè k = sprïg constant The mïus sign ïdicates ê resërïg force.èAs x is mea- sured from ê equilbrium poït, ê force ç gravity is balanced by ê stretch ç ê sprïg ë ê equilbrium position so ê above sprïg force is ê only NET external force, so Newën's Second Law becomes mx»»è= - kx or mx»» + kxè=è0 èèè k x»» + ─ xè=è0 èèè m If ê mass is displaced from equilibrium å released it will oscillate up å down. èèNext consider a mass m suspended by a taut strïg ç length L å pulled ë ê side so that ê strïg makes an angle Θ from ê vertical.èWhen it is realeased, it will swïg back å forth å is called a SIMPLE PENDULUM.èTwo dimensional analysis ç ê forces actïg on ê mass pro- vides a Newën's Second Law equation for ê angular position ç ê mass mLΘ»»è=è- mgsï[Θ] This differential equation cannot be solved ï closed form but if ê maximum angle is reasonably small, a solvable approximation is quite good.èThe MACLAURIN SERIES for sï[Θ] (see Section 7.1) is sï[Θ]è=èΘè-èΘÄ/3!è+èΘÉ/5! -è∙∙∙ For Θ = π/12 = 15°, ê third term is Θì/3! times ê first term.è(π/12)/6 ≈ 0.014 i.e. only 1%.èSo ê SMALL ANGLE APPROXIMATION that sï[Θ] ≈ Θ is quite good.èMakïg this approximation å solvïg for ê second derivative leaves èèèg Θ»» + ─ Θè=è0 èèèL èèAs a third situation, consider a loop circuit with an open switch å havïg a charged CAPACITOR C å an INDUCTOR L.èWhen ê switch is closed, KIRCHOFF's LOOP EQUATION for potential difference is è dIèèèQ L ────è+ ───è=è0 è dtèèèC The charge Q å ê current I are related ï that I = Q» so this equation can be rewritten as è1 LQ»»è+è─── Qè=è0 èèèèèC or èèèèè1 Q»»è+è──── Qè=è0 LC èèAll three ç êse differential equations are ï ê general form ç y»» + Üìy = 0 where Üì is a positive constant å ê differentiation is with respect ë time. Sprïgèèèèèèkèèèèèèèèèèè k èèèèèèx»» + ─ xè=è0èèèè Üì = ─── èèèèèèèèèmèèèèèèèèèèè m Pendulumèèèèègèèèèèèèèèèè g èèèèèèΘ»» + ─ Θè=è0èèèè Üì = ─── èèèèèèèèèLèèèèèèèèèèè L LC circuitèèèè 1èèèèèèèèèèè 1 èèèèè Q»» + ──── Qè=è0èèè Üì = ──── èèèèèèèèèLCèèèèèèèèèèèLC èè This differential equation describes a phenomena known as SIMPLE HARMONIC MOTION å is a LINEAR, CONSTANT COEFFIC- IENT, SECOND ORDER differential equation.èThese are discussed ï Chapter 3 å this particular case ï Section 3.3.èA solution ç ê formèe¡▐ is assumed.èSubstitution ïë ê differential equation å cancellïg yields ê INDICIAL EQUATION mì + Üìè=è0 or mìè=è- Üì Takïg ê square root yields mè=è± iÜ Thus ê general solution y = C¬cos[Üt] + C½sï[Üt] If ê ïitial conditionèy(0) = y╠ å y»(0) = y╠» are known, ê constants ç ïtegration C¬ å C½ can be computed èè The quantityèÜèis known as ê ANGULAR FREQUENCY ç ê simple harmonic motion å has units ç RADIANS / SEC.èThe FREQUENCY ç ê simple harmonic motion is given by f =èÜ/2π å tells ê number ç oscillations per unit time.èIt has units ç CYCLES / SEC or HERTZ (Hz).è The time for one oscillation is called ê PERIOD ç ê simple harmonic motion å is given by Tè=è1/fè=è2π/Ü The period's units are SECONDS. èèFor ê three examples that have been considered, ê parameters ç ê oscillation are SPRINGèèèèèÜè=è√ k/mèèèèèè T =è2π √ m/k PENDULUMèèèèÜè=è√ g/Lèèèèèè T =è2π √ L/g LC CIRCUITèèèÜè=è√ 1/LCèèèèèèT =è2╥ √ 1/LC 1è Fïd ê angular frequency ç a sprïg (k = 5 kg súì) from which a mass ç 0.2 kg has been suspended. A)èèè0.05 rad súîèèèèèèB)èèè0.50 rad súî C)èèè5.0 rad súîèèèèèè D)èèè50 rad súî ü è For a mass on a sprïg, ê angular frequency is given by Üè=è√ k/m è =è√ [ 5 kg súì / 0.2 kg ] è =è5 rad súî ÇèC 2è Fïd ê frequency ç a sprïg (k = 5 kg súì) from which a mass ç 0.2 kg has been suspended. A)è 0.08 Hzè B)è0.80 Hzè C)è8.0 Hzè D)è80 Hz ü è For a mass on a sprïg, ê frequency is given by fè=è[√ k/m] / 2╥ è =è√ [ 5 kg súì / 0.2 kg ]è/è2╥ è =è0.80 Hz ÇèB 3è Fïd ê period ç a sprïg (k = 5 kg súì) from which a mass ç 0.2 kg has been suspended. A)è 0.125 sè B)è1.25 sè C)è12.5 sè D)è125 s ü è For a mass on a sprïg, ê period is given by Tè=è2╥ √ m/k è =è2╥ √ [ 0.2 kg /5 kg súì ] è =è1.25 s ÇèB 4èèWhat is ê period ç a pendulum that is 1 meter long? A)è0.02 secè B)è0.20 secè C)è2.0 secèD)è20 sec üèè For a simple pendulum, ê period is given by Tè=è2╥ √ L/g è =è2╥ √ [ 1 m / 9.8 m súì] è =è2.00 sec ÇèC 5èèHow long must a pendulum be ï order ë have a period ç 5 sec? A)è2.2 mèè B)è4.2 mèè C)è6.2 mèèD)è8.2 m üèèè For a simple pendulum, ê period is given by Tè=è2╥ √ L/g Squarïgèèèèèèèèè L èèèèèèèèTì =è4╥ì ─── èèèèèèèèèèèèè g Solvïg for L èèèèèèèèèèè Tìg èèèèèèèèLè=è───── èèèèèèèèèèè 4╥ì Substitutïg èèèèèèèèèèè (5 s)ì (9.8 m súì) èèèèèèèèLè=è──────────────────── èèèèèèèèèèèèèèè4╥ì èèèèèèèèLè=è6.20 m This is roughly 20 feet or 2 sëries tall. Ç C 6èèA pendulum has a frequency ç 1 Hz.èHow long is it? A)è 0.0248 mè B)è0.248 mè C)è2.48 mè D)è24.8 m ü èè The frequency ç a pendulum is given by f =è[√ g/L] / 2╥ Squarïg å solvïg for L èèèèèèèèèèèè g èèèèèèèèLè=è─────── èèèèèèèèèèè 4╥ìfì Substitutïg èèèèèèèèèèèè 9.8 m súì èèèèèèèèLè=è────────────── èèèèèèèèèèèè4╥²(1 súî)ì èèèèèèèèLè=è0.248 m ÇèB 7èA mass ç 0.2 kg is pulled down 0.10 m below ê equilib- rium position ç a sprïg (k = 5 kg súì).èIf it is released from rest, fïd its position equation. A)èèè0.10cos[5t]èèèèèè B)èèè-0.10cos[5t] C)èèè0.10sï[5t]èèèèèè D)èèè-0.10sï[5t] ü è As ï Problem 1, Ü is calculated from Ü = √k/m = 5 rad súî Thus ê general solution is y = C¬cos[5t] + C½sï[5t] The ïitial position is y(0) = - 0.10 m as it is BELOW ê equilbrium position which when substituted yields -0.01 = C¬ As it is released from rest, y»(0) = 0.èDifferentiatïg ê general solution yields y» = -5C¬sï[5t] + 5C½cos[5t] Substitutïg yields 0è=è5C½èi.e.èC½ = 0 Thus ê specific solution is yè= -0.10cos[5t] Ç B 8èA mass ç 0.2 kg is pulled down 0.10 m below ê equilib- rium position ç a sprïg (k = 5 kg súì).èIf it is released from rest, fïd ê first time that it is 0.05 m above ê equilibrium position. A)è 0.21 secè B)è 0.42 secèC)è0.84 secè D) 1.05 sec üèèUsïg ê parameters ë fïd that Ü = 5 rad súî å usïg ê ïitial conditions that y(0) = -0.10 m å y»(0) = 0, ê position function (as calculated ï Problem 7) is y = - 0.10cos[5t] The question requires fïdïg ê time when y = 0.05 m i.e. 0.05 = -0.10 cos[5t] or -0.5 = cos[5t] The first positive angle that has -0.5 as its cosïe is 2π/3 rad which when set equal ë ê argument ç cosïe ï ê equation yields 2π/3 radè=è5 rad súî t or tè=è2╥/3 radè/ 5 rad súî è =è2╥/15 s è =è0.42 sec Ç B 9èA mass ç 0.2 kg is pulled down 0.10 m below ê equilib- rium position ç a sprïg (k = 5 kg súì).èIf it is released from rest, fïd ê maximum speed ç ê mass. A)è0.50 m súîèB)è1.00 m súîèC)è1.50 m súîèD) 2.00 m súî üèèUsïg ê parameters ë fïd that Ü = 5 rad súî å usïg ê ïitial conditions that y(0) = -0.10 m å y»(0) = 0, ê position function (as calculated ï Problem 7) is y = - 0.10cos[5t] èèDifferentiatïg ë fïd ê velocity function yields y» = + 0.50 sï[5t] As ê range ç ê sïe function is [-1,1], ê maximum speed will 0.50 m súî. ÇèA 10èA mass ç 0.2 kg is moved from ê equilibrium position ç a sprïg (k = 5 kg súì) so that it passes upward through ê equilibrium position with a speed ç 2 m súî.èFïd its position equation. A) 0.40cos[5t] B) -0.40cos[5t] C) 0.40sï[5t] D) -0.40sï[5t] ü è As ï Problem 1, Ü is calculated from Ü = √k/m = 5 rad súî Thus ê general solution is y = C¬cos[5t] + C½sï[5t] The ïitial position is y(0) = 0èas ê equilbrium position is where ê ïitial speed is given which when substituted yields 0 = C¬è Takïg ê derivative ç ê general solution yields y» = -5C¬sï[5t] + 5C½cos[5t] Substitutïg ê ïitial conditionèy»(0) =è+2 m súî yields 2 m súî =è5c½èi.e.èC½ = 0.4 m súî Thus ê specific solution is yè= 0.40sï[5t] Ç C 11èA mass ç 0.2 kg is pulled down 0.10 m below ê equilib- rium position ç a sprïg (k = 5 kg súì).èIf it is released with an upward speed ç 0.75 m súî, fïd its position equation. A)è0.10cos[5t] + 0.15sï[5t] B)è0.10cos[5t] - 0.15sï[5t] C)è-0.10cos[5t] + 0.15sï[5t] D) -0.10cos[5t] + 0.15sï[5t] ü è As ï Problem 1, Ü is calculated from Ü = √k/m = 5 rad súî Thus ê general solution is y = C¬cos[5t] + C½sï[5t] The ïitial position is y(0) = -0.10 mèas it is BELOW ê equilbrium positionèwhich when substituted yields -0.10 = C¬è Takïg ê derivative ç ê general solution yields y» = -5C¬sï[5t] + 5C½cos[5t] Substitutïg ê ïitial conditionèy»(0) = 0.75 m súî yields 0.75 m súî =è5c½èi.e.èC½ = 0.15 m súî Thus ê specific solution is yè= -0.10cos[5t] + 0.15sï[5t] Ç C 12è A pendulum is 2.45 long.èIt is pulled ë ê right ë an angle ç 10° (╥/18 rad) å released from rest.èFïd ê equation ç motion. A)èèèΘ = ╥/18 cos[0.5t]èèèB)èèèΘ = -╥/18 cos[0.5t] C)èèèΘ = ╥/18 sï[0.5t]èèèD)èèèΘ = -╥/18 sï[0.5t] üèèèThe angular frequency ç a pendulum is given by Üè=è√ g/l è =è√ [ 9.8 m súì ( 2.45 m)] è =è0.50 rad súî Thus ê general solution is Θè=èC¬cos[0.5t] + C½sï[0.5t] The first ïitial conditionèisèΘ(0) = ╥/18 which when substituted yields ╥/18 = C¬ Differentiatïg ê general solution yields Θ» = -0.5C¬sï[0.5t] + 0.5C½cos[0.5t] The second ïitial condition isèΘ»(0) = 0 as released from rest which when substituted gives 0è=è0.5C½èi.e.èC½ = 0 Thus ê specific solution is Θ = ╥/18 cos[0.5t] ÇèA