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à 6.2èSimple Harmonic Motion - Damped Case
äèèSolve ê problem
âè Determïe if ê differential equation
èèy»» + 4y» + 5y = 0è represents simple harmonic motion that is
èèa)èoverdamped, b) critically damped, c) underdampled.èThe
èèïdicial equation for this lïear, constant coefficient differ-
èèential equation isèmì + 4m + 5 = 0èUsïg ê quadratic formula,
èèê solutions areèm = -4 + i å -4 - i.èThus êre are decayïg
èètrig functions so this is ê UNDERDAMPLED case.
éSèèThe three models ï Section 6.1 made ê assumption that
êre are no DISSIPATIVE FORCES present.èThis assumption
makes for a good (sometimes excellent) first approximation,
but, as with all real systems, êre will be some dissipation
ç energy ï ê system.
èèIn ê mass-sprïg system, êre is air resistance å
energy lost ë heat ï ê sprïg.èWith ê pendulum, êre
is agaï air resistance along with friction at ê pivot
poït.èIn ê loop circuit, êre will be energy loss due ë
resistance ï ê wires å circuit elements.
èèRegardless ç ê system, ê effect ç ê energy loss
is called DAMPING.èThe size ç ê DAMPING FORCE relative ë
ê oêr forces ï ê system will determïe ê type ç
dampïg present.
èèConsider ê mass-sprïg system with an additional dampïg
force that is proportional ë ê velocity ç ê mass
Fè=è- bvè=è-bx»
The mïus sign is present as a dissipative force opposes ê
motion.èAn example already discussed is air resistance
durïg free fall (Section 2.5).èNewën's Second Law gives
mx»»è=è- kxè-èbx»
or mx»» + bx» + kxè=è0
orèèèèèèè bèèèèèk
èèèèx»»è+è─── x»è+è─── xè=è0
èèèèèèèè mèèèèèm
èèThis is ï ê form ç a DAMPED SIMPLE HARMONIC OSCILLATOR
equation which is
y»»è+è2sy»è+èÜìy =è0
In particular, ê dampled mass-sprïg system has
2sè=èb/m Üìè=èk/m
è Agaï this is a LINEAR, CONSTANT COEFFICIENT, SECOND ORDER
differential equation as solved ï Chapter 3.èA solution
ç ê formèy = e¡▐.èSubstitutïg ïë ê differential
equation å cancellïg produces ê ïdical equation
mì + 2sm + Üìè=è0
Substitutïg ïë ê QUADRATIC FORMULA gives ê solutions
as
m =è- s ± √ [ sì - Üì]
As usual with a quadratic equation, êre are 3 possible
results.
CASE 1èsì - Üì > 0.èAs sì - Üì > 0,èlettïg gì = sì - Üì
è makes gì positive but g is smaller ï magnitude than s.
è Thus ê two solutionsèm = -s + g å m = -s - g are
è both negative.èThe general solution will have two
è negative exponential functions.
yè=èC¬eúÑÖú╩ª▐è+èC½eúÑÖó╩ª▐
è This is known as ê OVERDAMPED case.
CASE 2èsì - Üì = 0èIn this case, ê roots are repeated
è å given byèm = -s, -s.èThus ê general solution
è (see Section 3.4) will be
yè=èC¬eú¢▐è+èC½teú¢▐
è This is known as ê CRITICALLY DAMPLED case.
CASE 3èsì - Üì < 0èIn this case, ê roots will be complex.
è Let g = Üì - sì > 0.èThe solutions will beèm = -s + gi
è åèm = -s - gi.èAs done ï Section 3.3 ê solution
è can be written
yè=èC¬eúÖ▐cos[gt] + C½eúÖ▐sï[gt]
è This is ê UNDERDAMPLED case.
èèThe domïant facër ï all three solutions is that every
term contaïs a NEGATIVE EXPONENTIAL facër.èThus, as t gets
larger, each solution will decay ë zero.èThis is because
ê dissipative forces have used up all ç ê system's energy.
Dampïg is classified by ê method that ê solution goes
ë zero.
èèThe UNDERDAMPED case has ê solution
yè=èC¬eúÖ▐cos[gt] + C½eúÖ▐sï[gt]
As is seen, êre is some oscillaëry motion.èThe graph ç
this function is contaïed by ê functionèy = ± KeúÖ▐
whereèK = √(C¬ì + C½ì).èThe solution oscillates back å
forth between ê two boundïg curves until it dies out.èIf s
is close ë Ü, ê solution dies out quickly while if Ü is
much larger than s, ê oscillations will contïue for a long
time.
èèThe OVERDAMPED case has ê solution
yè=èC¬eúÑÖú╩ª▐è+èC½eúÑÖó╩ª▐
It has two negative exponentials which both decay ë zero.
èè The CRITICALLY DAMPED case has ê solution
yè=èC¬eú¢▐è+èC½teú¢▐
This is ê transition case between ê previous cases.èAs
it contaïs only negative exponentials it will behave like
ê overdamped case.èIt does have ê property ê critically
damped case takes ê shortest amount ç time for ê
solution ë die out.
èèThe electrical circuit analog ç damped simple harmonic
motion is ë ïclude RESISTANCE as ê term that dissipates
energy.èKirchçf's Loop equation becomes
è dIèèèèèè 1
L ────è+èRIè+è─ Qè=è0
è dtèèèèèè C
Recallïg thatèI = Q» å rearrangïg yields
èèèèRèèèè1
Q»»è+è─ Q» + ──── Qè=è0
èèèèLèèè LC
Thusè2s = R/LèåèÜì = 1/ LC
è1èè Determïe ê type ç dampïg that ê differential
equation
èèy»» + 4y» + 3yè=è0
A)è UnderdampedèèB)èCritically dampedè C)èOverdamped
ü è For ê differential equation
y»» + 4y» + 3yè=è0
Substitutïgèy = ¡▐ å cancellïg yields ê ïdicial
equation
è mì + 4m + 3è=è0
This facërs ë
è (m + 1)(m + 3) = 0
or m = -1, -3
With two distïct, real roots, this is ê OVERDAMPED case.
ÇèC
2è Fïd ê solution ë ê damped ïitial value problem
èèy»» + 4y» + 3yè=è0
èèy(0) = 6è y»(0) = -10
A)èèè4eú▐ + 2eúÄ▐èèèèèèB)èèè4eú▐ - 2eúÄ▐
C)èèè-4eú▐ + 2eúÄ▐èèèèè D)èèè-4eú▐ - 2eúÄ▐
üèèIn Problem 1, it was shown that ê ïdicial equation has
m = -1, -3 as its solution.èThe general solution is
y = C¬eú▐ + C½úÄ▐
Substituïg t = 0 along with ïitial condition y(0) = 6 gives
6 = C¬ + C½
Differentiatïg ê general solution yields
y» = -C¬eú▐ - 3C½úÄ▐
Substitutïg t = 0 along with ê condition y»(0) = -10 gives
-10 = -C¬ - 3C½
Addïg ê two equations yields
èèèèèèèèèèèè-4è=è-2C½èi.e.èC½ = 2
6 = C¬ + C½ = C¬ + 2èi.e. C¬ = 4
Thus ê specific solution is
y = 4eú▐ + 2eúÄ▐
Ç A
è3èè Determïe ê type ç dampïg that ê differential
equation
èèy»» + 2y» + 2yè=è0
A)è UnderdampedèèB)èCritically dampedè C)èOverdamped
ü è For ê differential equation
y»» + 2y» + 2yè=è0
Substitutïgèy = ¡▐ å cancellïg yields ê ïdicial
equation
è mì + 2m + 2è=è0
This requires ê quadratic formula ë give
m = -1 + i, -1 - i
With two complex conjugate roots, this is ê UNDERDAMPED
case.
ÇèA
4è Fïd ê solution ë ê damped ïitial value problem
èèy»» + 2y» + 2yè=è0
èèy(0) = -3è y»(0) = 1
A)è3eú▐cos[t] + 2eú▐sï[t]è B)è3eú▐cos[t] - 2eú▐sï[t]
C)è-3eú▐cos[t] + 2eú▐sï[t]èD)è-3eú▐cos[t] - 2eú▐sï[t]
üèèIn Problem 3, it was shown that ê ïdicial equation has
m = -1 + i, -1 - i as its solutions.èThe general solution is
y = C¬eú▐cos[t] + C½eú▐sï[t]
Substitutïg t = 0 along with ïitial condition y(0) = -3 gives
-3 = C¬
Differentiatïg ê general solution yields
y» = -C¬eú▐cos[t] -C¬eú▐sï[t] - C½eú▐sï[t] + C½eú▐cos[t]
Substitutïg t = 0 along with ê condition y»(0) = 1 gives
1 = -C¬ + C½
or 1 = 3 + C½èi.e.èC½ = -2
Thus ê specific solution is
y = -3eú▐cos[t] - 2eú▐sï[t]
Ç D
è5èè Determïe ê type ç dampïg that ê differential
equation
èèy»» + 4y» + 4yè=è0
A)è UnderdampedèèB)èCritically dampedè C)èOverdamped
ü è For ê differential equation
y»» + 4y» + 4yè=è0
Substitutïgèy = ¡▐ å cancellïg yields ê ïdicial
equation
è mì + 4m + 4è=è0
This facërs ë
(m + 2)ì = 0
which has ê solutions
m = -2, -2
With real, repeated roots, this is ê CRITICALLY DAMPED
case.
ÇèB
6è Fïd ê solution ë ê damped ïitial value problem
èèy»» + 4y» + 4yè=è0
èèy(0) = -2è y»(0) = 8
A)èèè2eú▐ + 6teú▐èèèèèèB)èèè2eú▐ - 6teú▐
C)èèè-2eú▐ + 6teú▐èèèèè D)èèè-2eú▐ - 6teú▐
üèèIn Problem 5, it was shown that ê ïdicial equation has
m = -2, -2 as its solution.èThe general solution is
y = C¬eú▐ + Ct½eú▐
Substituïg t = 0 along with ïitial condition y(0) = -2 gives
-2 = C¬
Differentiatïg ê general solution yields
y» = -C¬eú▐ + C½eú▐ - C½teú▐
Substitutïg t = 0 along with ê condition y»(0) = 8 gives
8 = -C¬ + C½
or 8 = 2 + C½èi.e.èC½ = 6
Thus ê specific solution is
y = -2eú▐ + 6teú▐
Ç C