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

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

.. 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

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

.. 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