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### Transcript of Where are we? phys16/lectures/sl02.pdf Where are we? Forces of the form F(v) Example: F(v) =...

• 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) = m dv

dt

dt = dv m

F (v)

t− t0 = ∫ t t0

dt′ = ∫ 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) = m dv

dt

dt = dv m

F (v)

t− t0 = ∫ t t0

dt′ = ∫ 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) = m dv

dt

dt = dv m

F (v)

t− t0 = ∫ t t0

dt′ = ∫ 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) = m dv

dt

dt = dv m

F (v)

t− t0 = ∫ t t0

dt′ = ∫ 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) = m dv

dt

dt = dv m

F (v)

t− t0 = ∫ t t0

dt′ = ∫ 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) = m dv

dt

dt = dv m

F (v)

t− t0 = ∫ t t0

dt′ = ∫ 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) = m dv

dt

dt = dv m

F (v)

t− t0 = ∫ t t0

dt′ = ∫ 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 small velocities in viscous fluids.

• ma = m dv

dt = F (v) = −β v = −m Γ v

dt = − dv Γ v

Γ dt = −dv v

∫ t t0

Γ 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 = m dv

dt = F (v) = −β v = −m Γ v

dt = − dv Γ v

Γ dt = −dv v

∫ t t0

Γ 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 = m dv

dt = F (v) = −β v = −m Γ v

dt = − dv Γ v

Γ dt = −dv v

∫ t t0

Γ 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 = m dv

dt = F (v) = −β v = −m Γ v

dt = − dv Γ v

Γ dt = −dv v

∫ t t0

Γ 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 = m dv

dt = F (v) = −β v = −m Γ v

dt = − dv Γ v

Γ dt = −dv v

∫ t t0

Γ 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 = m dv

dt = F (v) = −β v = −m Γ v

dt = − dv Γ v

Γ dt = −dv v

∫ t t0

Γ 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 = m dv

dt = F (v) = −β v = −m Γ v

dt = − dv Γ v

Γ dt = −dv v

∫ t t0

Γ 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 t0

dt′ ẋ(t′)

x(t) = x(t0) + ∫ t t0

dt′ v(t′)

= x(t0) + ∫ t t0

dt′ 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)

1 du

x(t) = x(t0) + v(t0)

Γ

( 1− e−Γ (t−t0))

• x(t) = x(t0) + ∫ t t0

dt′ ẋ(t′)

x(t) = x(t0) + ∫ t t0

dt′ v(t′)

= x(t0) + ∫ t t0

dt′ 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)

1 du

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 Γ

independent 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 through a thin medium like a gas — force arises because the body is bumping into molecules and knocking them out of the way. The force from each collision is ∝ v, and the # per unit time is also ∝ v

• ma = m dv

dt = F (v) = −mβ v2

β dt = −dv v2

∫ t t0

β 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 = m dv

dt = F (v) = −mβ v2

β dt = −dv v2

∫ t t0

β 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 = m dv

dt = F (v) = −mβ v2

β dt = −dv v2

∫ t t0

β 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 = m dv

dt = F (v) = −mβ v2

β dt = −dv v2

∫ t t0

β 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 = m dv

dt = F (v) = −mβ v2

β dt = −dv v2

∫ t t0

β 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 = m dv

dt = F (v) = −mβ v2

β dt = −dv v2

∫ t t0

β 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 = m dv

dt = F (v) = −mβ v2

β dt = −dv v2

∫ t t0

β 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 β

log ( 1 + β v(t0) (t− t0)

)

• Forces of the form F (x)

Lots of examples (if you don’t look too closely) - gravity - the electric force. Many situations in which there is v dependence, but it is small and can be ignored.

• m dv

dt = F (x)

m dv

dt v = F (x)

dx

dt 1

2 m

d

dt v2 = F (x)

dx

dt 1

2 m

∫ t t0

dt′ d

dt′ v(t′)2 =

∫ t t0

dt′ F (x(t′)) dx