Where are we?
Forces of the form F (v)
Example: F (v) = −m Γ v
Another example: F (v) = −mβ v2
Forces of the form F (x)
Review of the harmonic oscillator
Linearity and Time Translation Invariance
Back to F (v) = −m Γ v
Forces of the form F (v)
Formal solution of F = ma
F (v) = ma
F (v) = mdv
dt
dt = dvm
F (v)
t− t0 =∫ t
t0dt′ =
∫ v(t)
v(t0)dv′
m
F (v′)solve this for v(t) as a function of t
Forces of the form F (v)
Formal solution of F = ma
F (v) = ma
F (v) = mdv
dt
dt = dvm
F (v)
t− t0 =∫ t
t0dt′ =
∫ v(t)
v(t0)dv′
m
F (v′)solve this for v(t) as a function of t
Forces of the form F (v)
Formal solution of F = ma
F (v) = ma
F (v) = mdv
dt
dt = dvm
F (v)
t− t0 =∫ t
t0dt′ =
∫ v(t)
v(t0)dv′
m
F (v′)solve this for v(t) as a function of t
Forces of the form F (v)
Formal solution of F = ma
F (v) = ma
F (v) = mdv
dt
dt = dvm
F (v)
t− t0 =∫ t
t0dt′ =
∫ v(t)
v(t0)dv′
m
F (v′)solve this for v(t) as a function of t
Forces of the form F (v)
Formal solution of F = ma
F (v) = ma
F (v) = mdv
dt
dt = dvm
F (v)
t− t0 =∫ t
t0dt′ =
∫ v(t)
v(t0)dv′
m
F (v′)solve this for v(t) as a function of t
Forces of the form F (v)
Formal solution of F = ma
F (v) = ma
F (v) = mdv
dt
dt = dvm
F (v)
t− t0 =∫ t
t0dt′ =
∫ v(t)
v(t0)dv′
m
F (v′)solve this for v(t) as a function of t
Forces of the form F (v)
Formal solution of F = ma
F (v) = ma
F (v) = mdv
dt
dt = dvm
F (v)
t− t0 =∫ t
t0dt′ =
∫ v(t)
v(t0)dv′
m
F (v′)solve this for v(t) as a function of t
Example: F (v) = −β v = −m Γ v
Frictional force appropriate for smallvelocities in viscous fluids.
ma = mdv
dt= F (v) = −β v = −m Γ v
dt = − dv
Γ vΓ dt = −dv
v∫ t
t0Γ dt′ = −
∫ v(t)
v(t0)dv′
1
v′
Γ (t− t0) = − ln v(t) + ln v(t0)
eΓ (t−t0) = v(t0)/v(t)
v(t) = v(t0) e−Γ (t−t0)
ma = mdv
dt= F (v) = −β v = −m Γ v
dt = − dv
Γ vΓ dt = −dv
v∫ t
t0Γ dt′ = −
∫ v(t)
v(t0)dv′
1
v′
Γ (t− t0) = − ln v(t) + ln v(t0)
eΓ (t−t0) = v(t0)/v(t)
v(t) = v(t0) e−Γ (t−t0)
ma = mdv
dt= F (v) = −β v = −m Γ v
dt = − dv
Γ vΓ dt = −dv
v∫ t
t0Γ dt′ = −
∫ v(t)
v(t0)dv′
1
v′
Γ (t− t0) = − ln v(t) + ln v(t0)
eΓ (t−t0) = v(t0)/v(t)
v(t) = v(t0) e−Γ (t−t0)
ma = mdv
dt= F (v) = −β v = −m Γ v
dt = − dv
Γ vΓ dt = −dv
v∫ t
t0Γ dt′ = −
∫ v(t)
v(t0)dv′
1
v′
Γ (t− t0) = − ln v(t) + ln v(t0)
eΓ (t−t0) = v(t0)/v(t)
v(t) = v(t0) e−Γ (t−t0)
ma = mdv
dt= F (v) = −β v = −m Γ v
dt = − dv
Γ vΓ dt = −dv
v∫ t
t0Γ dt′ = −
∫ v(t)
v(t0)dv′
1
v′
Γ (t− t0) = − ln v(t) + ln v(t0)
eΓ (t−t0) = v(t0)/v(t)
v(t) = v(t0) e−Γ (t−t0)
ma = mdv
dt= F (v) = −β v = −m Γ v
dt = − dv
Γ vΓ dt = −dv
v∫ t
t0Γ dt′ = −
∫ v(t)
v(t0)dv′
1
v′
Γ (t− t0) = − ln v(t) + ln v(t0)
eΓ (t−t0) = v(t0)/v(t)
v(t) = v(t0) e−Γ (t−t0)
ma = mdv
dt= F (v) = −β v = −m Γ v
dt = − dv
Γ vΓ dt = −dv
v∫ t
t0Γ dt′ = −
∫ v(t)
v(t0)dv′
1
v′
Γ (t− t0) = − ln v(t) + ln v(t0)
eΓ (t−t0) = v(t0)/v(t)
v(t) = v(t0) e−Γ (t−t0)
1 2 3 4 5
0.2
0.4
0.6
0.8
1
↑v(t)v(t0)
Γ (t−t0) →
x(t) = x(t0) +∫ t
t0dt′ x(t′)
x(t) = x(t0) +∫ t
t0dt′ v(t′)
= x(t0) +∫ t
t0dt′ v(t0) e−Γ (t′−t0)
u = e−Γ (t′−t0) du = −dt′ Γ e−Γ (t′−t0)
x(t) = x(t0)− v(t0)
Γ
∫ e−Γ (t−t0)
1du
x(t) = x(t0) +v(t0)
Γ
(1− e−Γ (t−t0)
)
x(t) = x(t0) +∫ t
t0dt′ x(t′)
x(t) = x(t0) +∫ t
t0dt′ v(t′)
= x(t0) +∫ t
t0dt′ v(t0) e−Γ (t′−t0)
u = e−Γ (t′−t0) du = −dt′ Γ e−Γ (t′−t0)
x(t) = x(t0)− v(t0)
Γ
∫ e−Γ (t−t0)
1du
x(t) = x(t0) +v(t0)
Γ
(1− e−Γ (t−t0)
)
1 2 3 4 5
0.2
0.4
0.6
0.8
1
↑x(t)−x(t0)
v(t0)/Γ
Γ (t−t0) →
More general but still very simple case
F (v) = F0 −m Γ v
More general but still very simple case
F (v) = F0 −m Γ v
v(t) = v(t0) e−Γ (t−t0) +F0
m Γ
(1− e−Γ (t−t0)
)
More general but still very simple case
F (v) = F0 −m Γ v
v(t) = v(t0) e−Γ (t−t0) +F0
m Γ
(1− e−Γ (t−t0)
)
v(∞) =F0
m Γindependent of v0
Arisotelian physics if Γ is large - v ∝ F0
1 2 3 4 5
0.5
1
1.5
2
2.5
3
↑v(t)
F0/mΓ
Γ (t−t0) →
v0 > F0
mΓ
v0 < F0
mΓ
Another example: F (v) = −mβ v2
Frictional force for rapid movement througha thin medium like a gas — force arisesbecause the body is bumping into moleculesand knocking them out of the way. Theforce from each collision is ∝ v, and the #per unit time is also ∝ v
ma = mdv
dt= F (v) = −mβ v2
β dt = −dv
v2
∫ t
t0β dt′ = −
∫ v(t)
v(t0)
dv′
v′ 2
1
v(t0)+ β (t− t0) =
1
v(t)− 1
v(t0)− 1
v(t0)
v(t) =v(t0)
1 + β v(t0) (t− t0)
ma = mdv
dt= F (v) = −mβ v2
β dt = −dv
v2
∫ t
t0β dt′ = −
∫ v(t)
v(t0)
dv′
v′ 2
1
v(t0)+ β (t− t0) =
1
v(t)− 1
v(t0)− 1
v(t0)
v(t) =v(t0)
1 + β v(t0) (t− t0)
ma = mdv
dt= F (v) = −mβ v2
β dt = −dv
v2
∫ t
t0β dt′ = −
∫ v(t)
v(t0)
dv′
v′ 2
1
v(t0)+ β (t− t0) =
1
v(t)− 1
v(t0)− 1
v(t0)
v(t) =v(t0)
1 + β v(t0) (t− t0)
ma = mdv
dt= F (v) = −mβ v2
β dt = −dv
v2
∫ t
t0β dt′ = −
∫ v(t)
v(t0)
dv′
v′ 2
β (t− t0) =1
v(t)− 1
v(t0)
v(t) =v(t0)
1 + β v(t0) (t− t0)
ma = mdv
dt= F (v) = −mβ v2
β dt = −dv
v2
∫ t
t0β dt′ = −
∫ v(t)
v(t0)
dv′
v′ 2
β (t− t0) =1
v(t)− 1
v(t0)
1
v(t0)+ β (t− t0) =
1
v(t)
ma = mdv
dt= F (v) = −mβ v2
β dt = −dv
v2
∫ t
t0β dt′ = −
∫ v(t)
v(t0)
dv′
v′ 2
1 + v(t0)β (t− t0)
v(t0)=
1
v(t)
1
v(t0)+ β (t− t0) =
1
v(t)
ma = mdv
dt= F (v) = −mβ v2
β dt = −dv
v2
∫ t
t0β dt′ = −
∫ v(t)
v(t0)
dv′
v′ 2
1 + v(t0)β (t− t0)
v(t0)=
1
v(t)
v(t) =v(t0)
1 + β v(t0) (t− t0)
F (v) = −β v = −m Γ v
⇒ v(t) = v(t0) e−Γ (t−t0)
F (v) = −mβ v2
⇒ v(t) =v(t0)
1 + β v(t0) (t− t0)
F x(t)
F0 ⇒ x(t0) + v(t0) (t− t0) +F0
2m(t− t0)
2
−m Γ v ⇒ x(t0) +v(t0)
Γ
(1− e−Γ (t−t0)
)
−mβ v2 ⇒ x(t0) +1
βlog
(1 + β v(t0) (t− t0)
)
Forces of the form F (x)
Lots of examples (if you don’t look tooclosely) - gravity - the electric force. Manysituations in which there is v dependence,but it is small and can be ignored.
mdv
dt= F (x)
mdv
dtv = F (x)
dx
dt1
2m
d
dtv2 = F (x)
dx
dt1
2m
∫ t
t0dt′
d
dt′v(t′)2 =
∫ t
t0dt′ F (x(t′))
dx
dt′1
2m
(v(x)2 − v2
0
)=
∫ x
x0
dx′ F (x′)
We use a dirty trick - that we will eventuallysee is related to conservation of energy
mdv
dt= F (x)
mdv
dtv = F (x)
dx
dt1
2m
d
dtv2 = F (x)
dx
dt1
2m
∫ t
t0dt′
d
dt′v(t′)2 =
∫ t
t0dt′ F (x(t′))
dx
dt′1
2m
(v(x)2 − v2
0
)=
∫ x
x0
dx′ F (x′)
mdv
dt= F (x)
mdv
dtv = F (x)
dx
dt1
2m
d
dtv2 = F (x)
dx
dt1
2m
∫ t
t0dt′
d
dt′v(t′)2 =
∫ t
t0dt′ F (x(t′))
dx
dt′1
2m
(v(x)2 − v2
0
)=
∫ x
x0
dx′ F (x′)
mdv
dt= F (x)
mdv
dtv = F (x)
dx
dt1
2m
d
dtv2 = F (x)
dx
dt1
2m
∫ t
t0dt′
d
dt′v(t′)2 =
∫ t
t0dt′ F (x(t′))
dx
dt′1
2m
(v(x)2 − v2
0
)=
∫ x
x0
dx′ F (x′)
mdv
dt= F (x)
mdv
dtv = F (x)
dx
dt1
2m
d
dtv2 = F (x)
dx
dt1
2m
∫ t
t0dt′
d
dt′v(t′)2 =
∫ t
t0dt′ F (x(t′))
dx
dt′1
2m
(v(x)2 − v2
0
)=
∫ x
x0
dx′ F (x′)
mdv
dt= F (x)
mdv
dtv = F (x)
dx
dt1
2m
d
dtv2 = F (x)
dx
dt1
2m
∫ t
t0dt′
d
dt′v(t′)2 =
∫ t
t0dt′ F (x(t′))
dx
dt′1
2m
(v(x)2 − v2
0
)=
∫ x
x0
dx′ F (x′)
Once you know v(x), you can find x(t) byintegration
dx
dt= v(x)
Once you know v(x), you can find x(t) byintegration
dx
dt= v(x)
dt =dx
v(x)
Once you know v(x), you can find x(t) byintegration
dx
dt= v(x)
dt =dx
v(x)∫ t
t0dt′ = t− t0 =
∫ x(t)
x0
dx′
v(x′)
Once you know v(x), you can find x(t) byintegration
dx
dt= v(x)
dt =dx
v(x)∫ t
t0dt′ = t− t0 =
∫ x(t)
x0
dx′
v(x′)This can now be solved IN PRINCIPLE forx(t)
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-¾
Review of the harmonic oscillator
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x →|
equilibrium at x = 0
F = −K x
K is the spring constant the trajectories arethe solutions of this
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x →|
equilibrium at x = 0
F = −K x
K is the spring constant the trajectories arethe solutions of this
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x →|
equiilbrium at x = 0
F = −K x
F = ma = md2
dt2x = m x = −K x
the trajectories are the solutions of this
m x = −K x
x(t) = a cos ωt + b sin ωt for ω =√
K/m
m x = −K x
x(t) = a cos ωt + b sin ωt for ω =√
K/m
check that this is right
d
dzcos z = − sin z
d
dzsin z = cos z
chain rule
d
dtcos ωt = −ω sin ωt
d
dtsin ωt = ω cos ωt
m x = −K x
x(t) = a cos ωt + b sin ωt for ω =√
K/m
check that this is right
d
dzcos z = − sin z
d
dzsin z = cos z
chain rule
d
dtcos ωt = −ω sin ωt
d
dtsin ωt = ω cos ωt
m x = −K x
x(t) = a cos ωt + b sin ωt for ω =√
K/m
check that this is right
d
dzcos z = − sin z
d
dzsin z = cos z
chain rule
d
dtcos ωt = −ω sin ωt
d
dtsin ωt = ω cos ωt
m x = −K x
x(t) = a cos ωt + b sin ωt for ω =√
K/m
x(t) = −aω sin ωt + bω cos ωt
m x = −K x
x(t) = a cos ωt + b sin ωt for ω =√
K/m
x(t) = −aω sin ωt + bω cos ωt
x(t) = −aω2 cos ωt− bω2 sin ωt
m x = −K x
x(t) = a cos ωt + b sin ωt for ω =√
K/m
x(t) = −aω sin ωt + bω cos ωt
x(t) = −aω2 cos ωt− bω2 sin ωt
= −ω2(a cos ωt + b sin ωt) = −ω2 x(t)
m x(t) = −ω2 mx(t) = −K x(t)
m x = −K x
x(t) = a cos ωt + b sin ωt for ω =√
K/m
x(t) = −aω sin ωt + bω cos ωt
x(t) = −aω2 cos ωt− bω2 sin ωt
= −ω2(a cos ωt + b sin ωt) = −ω2 x(t)
m x(t) = −ω2 mx(t) = −K x(t)
m x = −K x
x(t) = a cos ωt + b sin ωt for ω =√
K/m
constant ω is called the Angular Frequency
m x = −K x
x(t) = a cos ωt + b sin ωt for ω =√
K/m
constant ω is called the Angular Frequency
Two constants (a and b) label thetrajectories - as expected - two initialconditions for one degree of freedom -related to initial x and v
m x = −K x
x(t) = a cos ωt + b sin ωt
setting t = 0 gives
x(0) = a cos 0 + b sin 0 = a
⇒ a is the position of the mass at t = 0
m x = −K x
x(t) = a cos ωt + b sin ωt
v(t) = x(t) = −aω sin ωt + bω cos ωt
setting t = 0 gives
v(0) = x(0) = −aω sin 0 + bω cos 0 = bω
⇒ bω is the velocity of the mass at t = 0
x(t) = x(0) cos ωt +v(0)
ωsin ωt
for arbitrary initial time t = t0
= x(t0) cos[ω(t− t0)] +v(t0)
ωsin[ω(t− t0)]
F x(t)
F0 ⇒ x(t0) + v(t0) (t− t0) +F0
2m(t− t0)
2
−m Γ v ⇒ x(t0) +v(t0)
Γ
(1− e−Γ (t−t0)
)
−mβ v2 ⇒ x(t0) +1
βlog
(1 + β v(t0) (t− t0)
)
−mω2 x ⇒ x(t0) cos[ω(t− t0)
]+
v(t0)
ωsin
[ω(t− t0)
]
x(t) = a cos ωt + b sin ωt = c cos(ωt− φ)
x(t) = a cos ωt + b sin ωt = c cos(ωt− φ)
= c (cos ωt cos φ + sin ωt sin φ)
= c cos φ cos ωt + c sin φ sin ωt
x(t) = a cos ωt + b sin ωt = c cos(ωt− φ)
= c (cos ωt cos φ + sin ωt sin φ)
= c cos φ cos ωt + c sin φ sin ωt
a = c cos φ b = c sin φ
x(t) = a cos ωt + b sin ωt = c cos(ωt− φ)
= c (cos ωt cos φ + sin ωt sin φ)
= c cos φ cos ωt + c sin φ sin ωt
a = c cos φ b = c sin φ
x(t) = a cos ωt + b sin ωt = c cos(ωt− φ)
= c (cos ωt cos φ + sin ωt sin φ)
= c cos φ cos ωt + c sin φ sin ωt
a = c cos φ b = c sin φ
c =√
a2 + b2 φ = arctanb
a
x(t) = a cos ωt + b sin ωt = c cos(ωt− φ)
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c
..
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..
..
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..
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..
..
..
..
..
..
..
..
..
..
..
..
..
..
.. slope is b ω
0 2πω
5ω
10ω
φω
0
a
t →
↑x
sin and cos periodic with period 2π
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c
..
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..
..
..
..
..
..
..
..
..
.. slope is b ω
0 2πω
5ω
10ω
φω
0
a
t →
↑x
sin and cos periodic with period 2π
x(t + 2π/ω) = x(t)
motion repeats after a time τ = 2π/ω called“the period of the oscillation” — frequency(rather than angular frequency) is
ν =ω
2π=
1
τ
Remembering where the 2πs go —
ωt — argument of sin and cos is an angle
ω has unitsradians
sec
Remembering where the 2πs go —
ωt — argument of sin and cos is an angle
ω has unitsradians
secνt is the number of repeats or “cycles”
ν has unitscycles
sec
Remembering where the 2πs go —
ωt — argument of sin and cos is an angle
ω has unitsradians
secνt is the number of repeats or “cycles”
ν has unitscycles
sec
ωradians
sec= 2π ν
cycles
sec
Linearity and Time TranslationInvariance
equations of motion of the formm x + K x = 0 are ubiquitous
appears in many different mechanicalsystems, and even in electrical systems
What is so special about this???
Linearity and Time TranslationInvariance
equations of motion of the formm x + K x = 0 are ubiquitous
appears in many different mechanicalsystems, and even in electrical systems
What is so special about this???
Linearity and Time TranslationInvariance
equations of motion of the formm x + K x = 0 are ubiquitous
appears in many different mechanicalsystems, and even in electrical systems
What is so special about this???
m d2
dt2x + K x = 0
1: Time translation invariance — noexplicit dependence on t — d/dt but no t
m d2
dt2x + K x = 0
1: Time translation invariance — noexplicit dependence on t — d/dt but no t
if x(t) is a solution then so is x(t + a)
m d2
dt2x + K x = 0
1: Time translation invariance — noexplicit dependence on t — d/dt but no t
if x(t) is a solution then so is x(t + a)
2: Linearity — all terms ∝ one power of x
or its derivatives
m d2
dt2x + K x = 0
1: Time translation invariance — noexplicit dependence on t — d/dt but no t
if x(t) is a solution then so is x(t + a)
2: Linearity — all terms ∝ one power of x
or its derivatives
new solutions as linear combinations of oldones — if x1(t) and x2(t) are solutionsthen so is Ax1(t) + B x2(t)
Time translation invariance — laws ofphysics don’t change with time or do so veryslowly
Linearity — what does it have to do withphysics? Why should many systems beapproximately linear?
Oscillations about equilibrium
d2
dt2x = F(x)
Oscillations about equilibrium
d2
dt2x = F(x)
equilibrium ⇒ F(0) = 0
Oscillations about equilibrium
d2
dt2x = F(x)
equilibrium ⇒ F(0) = 0
F(x) = F(0) + xF ′(0) +1
2x2F ′′(0) + · · ·
Oscillations about equilibrium
d2
dt2x = F(x)
equilibrium ⇒ F(0) = 0
F(x) = F(0) + xF ′(0) +1
2x2F ′′(0) + · · ·
= xF ′(0) +1
2x2F ′′(0) + · · ·
Oscillations about equilibrium
d2
dt2x = F(x)
equilibrium ⇒ F(0) = 0
F(x) = F(0) + xF ′(0) +1
2x2F ′′(0) + · · ·
= xF ′(0) +1
2x2F ′′(0) + · · ·
small for sufficiently small x – unless F ′(0)
Oscillations about equilibrium
d2
dt2x = F(x)
equilibrium ⇒ F(0) = 0
F(x) = F(0) + xF ′(0) +1
2x2F ′′(0) + · · ·
= xF ′(0) +1
2x2F ′′(0) + · · ·
Only the linear term relevant for sufficientlysmall oscillations
Back to F (v) = −m Γ v
It’s Linear!
Smoothness - reversibility - unlike −β v2
friction
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