Announcements
1. Physics lecture today 4 PM – Professor Brian Matthews
from U. Oregon – “Tolerance and Intolerance in Protein
Structure and Function”
2. Third exam on Chaps. 15-20 – Tuesday, Nov. 25th
Bring:
Equation sheet (8.5″ x 11″)
Calculator
Clear head
3. Extra problem solving session ?
Sunday night 10-11 PM ?
Monday evening 6-7 PM ?
Other ?
11/20/2003
PHY 113 -- Lecture Review 15-20
1
Advice for preparing for exam:
Review lecture notes and text chapters
Prepare equation sheet
Work problems using equation sheet and calculator
From homework
From previous exams
From on line quizes
Topics:
Simple harmonic motion & resonance
Wave motion, standing waves, sound waves
The physics of fluids
Thermodynamics & the ideal gas law
11/20/2003
PHY 113 -- Lecture Review 15-20
2
Simple harmonic motion & resonance
d 2x
Hooke’s law: F = − kx = ma = m 2
dt
d 2x
k
⇒ 2 =− x
dt
m
Simple
pendulum:
Solution :
k
x(t ) = A cos(ωt + φ); ω =
m
d 2Θ
τ = mgL sin Θ = − Iα = -I 2
dt
d 2Θ
mgL
mgL
sin Θ ≈ −
⇒ 2 =−
Θ
I
I
dt
Solution (for small Θ) :
mgL
Θ(t ) = A cos(ωt + φ); ω =
I
11/20/2003
PHY 113 -- Lecture Review 15-20
3
The notion of resonance:
Suppose F=−kx+F0 sin(Ωt)
According to Newton’s second law:
d 2x
− kx + F0 sin(Ωt ) = m 2
dt
Differential equation (" inhomogeneous" ) :
F0
d 2x
k
= − x + sin(Ωt )
2
dt
m
m
A solution to this equation takes the form :
F0 / m
F0 / m
x(t ) =
sin(Ωt ) ≡ 2
sin(Ωt )
2
2
k /m−Ω
ω −Ω
11/20/2003
PHY 113 -- Lecture Review 15-20
4
Physics of a “driven” harmonic oscillator:
d 2x
− kx + F0 sin(Ωt ) = m 2
dt
“driving” frequency
F0 / m
F0 / m
x(t ) =
sin(Ωt ) ≡ 2
sin(Ωt )
2
2
k /m−Ω
ω −Ω
“natural” frequency
F0 / m
ω2 − Ω 2
(Ω=2rad/s)
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(mag)
ω
PHY 113 -- Lecture Review 15-20
5
2
The wave equation
y ( x, t )
position
11/20/2003
time
2
∂ y
2 ∂ y
=
v
2
2
∂t
∂x
PHY 113 -- Lecture Review 15-20
6
The wave equation:
2
∂2 y
∂
y
2
=v
2
∂t
∂x 2
T
where v ≡
(for a string)
µ
Solutions: y(x,t) = f (x ± vt)
function of any shape
Moving “pulse”: y ( x, t ) = y e −( x −vt )2
0
“wave vector”
not spring constant!!!
Periodic wave: y ( x, t ) = y0 sin (k ( x − vt ) + φ )
⎛ ⎛x t ⎞
⎞
y ( x, t ) = y0 sin ⎜ 2 π⎜ − ⎟ + φ ⎟
⎠
⎝ ⎝λ T ⎠
11/20/2003
PHY 113 -- Lecture Review 15-20
λ
=v
T
7
Superposition of waves:
⎛ ⎛x t ⎞
⎛ ⎛x t ⎞
⎞
⎞
yright ( x, t ) = y0 sin ⎜ 2 π⎜ − ⎟ + φ ⎟ yleft ( x, t ) = y0 sin ⎜ 2 π⎜ + ⎟ + φ ⎟
⎝ ⎝λ T ⎠
⎝ ⎝λ T ⎠
⎠
⎠
sin A ± sin B = 2 sin[ 12 ( A ± B )]cos[ 12 ( A m B )]
“Standing” wave:
2 πx
2 πt ⎞
⎛
⎞
⎛
yright ( x, t ) + y left ( x, t ) = 2 y0 sin ⎜
+ φ ⎟ cos⎜
⎟
λ
T
⎝
⎠ ⎝
⎠
11/20/2003
PHY 113 -- Lecture Review 15-20
8
Constraints of standing waves:
(ϕ=0)
2 πx ⎞ ⎛ 2 πt ⎞
⎛
yright ( x, t ) + y left ( x, t ) = 2 y0 sin ⎜
⎟ cos⎜
⎟
λ
T
⎝
⎠ ⎝
⎠
for string :
2L
λ = ; n = 1,2,3L
n
λ
=v
T
11/20/2003
PHY 113 -- Lecture Review 15-20
2L
⇒T =
nv
1 nv
f = =
T 2L
9
The sound of music
String instruments (Guitar, violin, etc.)
2L
λn =
n
v nv
=
fn =
λ n 2L
v=
T
µ
2 πx ⎞ ⎛ 2 πt ⎞
⎛
ystanding ( x, t ) = 2 y0 sin ⎜
⎟ Î Couples to sound in
⎟ cos⎜
⎝ λ ⎠ ⎝ T ⎠ the air
11/20/2003
PHY 113 -- Lecture Review 15-20
10
Sound waves
Longitudinal waves propagating in a fluid or solid
2
∂2 y
2∂ y
y ( x, t ) describes density or pressure variations
=v
2
2
∂x
∂t
γP
T
B ⎛ compressibility ⎞
⎟
⎜
=
≈ 343 m/s
v=
⇒
⎟
⎜
µ
ρ ⎝
density
ρ
⎠
11/20/2003
PHY 113 -- Lecture Review 15-20
11
Periodic sound wave
In terms of pressure:
2 πx 2 πt ⎞
±
P ( x, t ) = P0 + ∆Pmax sin ⎛⎜
⎟
T ⎠
⎝ λ
Sound intensity: (energy/(unit time ⋅ unit area))
I≡
(∆Pmax )2
2ρv
Decibel scale:
⎛I ⎞
β ≡ 10 log⎜⎜ ⎟⎟
⎝ I0 ⎠
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I 0 = 10 −12 W/m 2
PHY 113 -- Lecture Review 15-20
12
“Wind” instruments (standing waves in air)
2L
λn =
; n = 1,2,3, K
n
v nv
=
=
fn
λ n 2L
γP
v=
≈ 343 m/s
ρ
4L
λn =
; n = 1,3,5, K
n
v nv
=
fn =
λ n 4L
v=
11/20/2003
PHY 113 -- Lecture Review 15-20
γP
≈ 343 m/s
ρ
13
The “Doppler” effect
v=sound velocity
observer moving, source stationary
vt1 = d − vO t1
v(t 2 − T ) = d − vO t 2
observer
source
1
v
=T
t 2 − t1 =
v + vO
fO
v + vO
fO = f S
v
vS=0
d
Summary of sound Doppler effect :
11/20/2003
v ± vO
fO = f S
v m vS
PHY 113 -- Lecture Review 15-20
toward
away
14
The physics of fluids.
•Fluids include liquids (usually “incompressible) and gases (highly
“compressible”).
•Fluids obey Newton’s equations of motion, but because they
move within their containers, the application of Newton’s laws to
fluids introduces some new forms.
¾Pressure: P=force/area
1 (N/m2) = 1 Pascal
¾Density: ρ =mass/volume 1 kg/m3 = 0.001 gm/ml
Note: In this chapter P≡pressure (NOT MOMENTUM)
11/20/2003
PHY 113 -- Lecture Review 15-20
15
Density: ρ =mass/volume
Effects of the weight of a fluid:
P(y+∆y)
ρg∆y
y
P(y)
F ( y ) = F ( y + ∆y ) + mg
F ( y ) F ( y + ∆y ) mg
=
+
A
A
A
P( y ) = P ( y + ∆y ) + ρg∆y
P ( y + ∆y ) − P( y ) dP
lim
=
∆y → 0
∆y
dy
dP
⇒
= −ρg
dy
Note: In this formulation +y is
defined to be in the up direction.
11/20/2003
PHY 113 -- Lecture Review 15-20
16
For an “incompressible” fluid (such as mercury):
ρ = 13.585 x 103 kg/m3 (constant)
dP
= −ρg ⇒ P = P0 − ρg ( y − y0 )
dy
Example:
P0
h = y − y0 =
ρg
1.013 ×105 Pa
=
13.595 × 103 kg/m 3 ⋅ 9.8 m/s 2
= 0.76 m
ρ = 13.595 x 103 kg/m3
11/20/2003
PHY 113 -- Lecture Review 15-20
17
Buoyant force for fluid acting on a solid:
FB=ρfluidVdisplacedg
P( y ) = P ( y + ∆y ) + ρ fluid g∆y
Buoyant force :
FB = Fbottom − Ftop
FB = {P ( y ) − P ( y + ∆y )}A = ρ fluid g∆yA = ρ fluid gVsubmerged
−
FB
A
∆y
ρfluidVsubmergedg − ρsolidVsolidg = 0
Vsubmerged
mg
11/20/2003
mg = 0
Vsolid
=
ρ solid
ρ fluid
PHY 113 -- Lecture Review 15-20
18
Bouyant forces:
the tip of the iceburg
FB = miceg
ρwatergVsubmerged= ρicegVtotal
Vsubmerged ρice 0.917
=
≈
= 93.9%
Vtotal ρwater 1.024
11/20/2003
Source:http://bb-bird.com/iceburg.html
PHY 113 -- Lecture Review 15-20
19
Effects of the weight of a “compressible” fluid on pressure.
ρ0
dP
= −ρg
For a gas, ρ = P
dy
P0
Solution :
P( y ) = P0 e
ρ g
− 0 ( y − y0 )
P0
≈ P0 e
−
y − y0
8000 m
≈ P0 e
−
y-y0(mi)
⎛ρ g ⎞
dP
= − P⎜⎜ 0 ⎟⎟
dy
⎝ P0 ⎠
P (atm)
11/20/2003
PHY 113 -- Lecture Review 15-20
20
y − y0
5 mi
Energetics of fluids:
∆x1 =v1∆ t
∆x2 =v2∆ t
A1 ∆x1= A2 ∆x2
m = ρ A1 ∆x1
K2 + U2 = K1 + U1 + W12
½ mv22 + mgh2 = ½ mv12 + mgh1 + (P1 A1 ∆x1– P2 A2 ∆x2)
ÎP2 + ½ ρv22 + ρgh2 = P1 + ½ ρv12 + ρgh1
11/20/2003
PHY 113 -- Lecture Review 15-20
21
Bernoulli’s equation:
P2 + ½ ρv22 + ρgh2 = P1 + ½ ρv12 + ρgh1
Example:
Suppose we know A1, A2, ρ, P, y1, y2
P + ½ ρv22 + ρgy2 =
P0 + ½ ρv12 + ρgy1
A2 v2 = A1v1
v1 =
11/20/2003
2( gh − ( P0 − P ) / ρ )
1− (
PHY 113 -- Lecture Review 15-20
A1
)
2
A2
22
Thermodynamic statement of conservation of energy –
First Law of Thermodynamics
∆Eint = Q - W
Work done by system
Heat added to system
“Internal” energy of system
11/20/2003
PHY 113 -- Lecture Review 15-20
23
How is temperature related to Eint?
Consider an ideal gas
Î Analytic expressions for physical variables
Î Approximates several real situations
Ideal Gas Law:
PV=nRT
volume (m3)
pressure (Pa)
temperature (K)
gas constant (8.31 J/(mole ⋅K))
number of moles
11/20/2003
PHY 113 -- Lecture Review 15-20
24
Review of results from ideal gas analysis in terms of the
specific heat ratio γ ≡ CP/CV:
n
∆Eint =
R∆T = nCV ∆T
γ-1
γR
CP =
γ-1
R
CV =
γ-1
CP
Note : γ =
CV
11/20/2003
PHY 113 -- Lecture Review 15-20
25