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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
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
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