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2. Physics colloquium today at 3 PM in Olin101
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3. Today’s lecture –
A few comments about the physics of fluids
The physics of
11/6/2003
motion (Chap. 17)
PHY 113 -- Lecture 17
1
Summary of results concerning the physics of fluids -Bernoulli’s equation: P2 + ½ ρv22 + ρgh2 = P1 + ½ ρv12 + ρgh1
Applies to incompressible fluids or
fluids in steamline flow.
Assumes no friction or turbulent
flow.
Can also be used to analyze static
fluids (vi = 0).
11/6/2003
PHY 113 -- Lecture 17
2
Streamline flow of air around an airplane wing:
P2 + ½ ρv22 + ρgh2 = P1 + ½ ρv12 + ρgh1
h1 ≈ h2
Flift=(P2-P1)A
v1 > v2
v1
P1
P2 = P1 + 12 ρ(v12 − v22 )
Example:
P2
v2
v1 = 270 m/s, v2 = 260 m/s
ρ = 0.6 kg/m3, A = 40 m2
ÎFlift = 63,600 N
11/6/2003
PHY 113 -- Lecture 17
3
A hypodermic syringe contains a medicine with the density of
water. The barrel of the syringe has a cross-sectional area
A=2.5x10-5m2, and the needle has a cross-sectional area a=
1.0x10-8m2. In the absence of a force on the plunger, the
pressure everywhere is 1 atm. A force F of magnitude 2 N acts
on the plunger, making the medicine squirt horizontally from
the needle. Determine the speed of the medicine as leave the
needle’s tip.
F 1
1
+ ρV 2 = P0 + ρv 2
2
A 2
2F
v=
2 = 12.6 m/s
a
Aρ (1 − ( A ) )
P0 +
11/6/2003
PHY 113 -- Lecture 17
AV = av
4
Another example; v=0
P0 + ρoilg(l + d) = P0 + ρwatergl+ ρairgd
l
ρoil ≈ ρwater
l +d
Potential energy reference
11/6/2003
PHY 113 -- Lecture 17
5
Bouyant forces:
the tip of the iceburg
FB = miceg
ρwatergVsubmerged= ρicegVtotal
Vsubmerged ρice 0.917
=
≈
= 93.9%
Vtotal ρwater 1.024
11/6/2003
Source:http://bb-bird.com/iceburg.html
PHY 113 -- Lecture 17
6
2
The wave equation
y ( x, t )
position
Wave variable
time
2
∂ y
2 ∂ y
=
v
2
2
∂t
∂x
¾What does the wave equation mean?
¾Examples
¾Mathematical solutions of wave equation and descriptions
of waves
11/6/2003
PHY 113 -- Lecture 17
7
Example: Water waves
Needs more
sophistocated
analysis:
Source: http://www.eng.vt.edu/fluids/msc/gallery/gall.htm
11/6/2003
PHY 113 -- Lecture 17
8
Waves on a string:
Typical values for v:
3x108 m/s light waves
~1000 m/s wave on a string
331 m/s sound in air
11/6/2003
PHY 113 -- Lecture 17
9
Peer instruction question
Which of the following properties of a wave are characteristic
of the medium in which the wave is traveling?
(A) Its frequency
(B) Its wavelength
(C) Its velocity
(D) All of the above
11/6/2003
PHY 113 -- Lecture 17
10
Mechanical waves occur in continuous media. They are characterized
by a value (y) which changes in both time (t) and position (x).
Example -- periodic wave
y(t0,x)
y(t,x0)
11/6/2003
PHY 113 -- Lecture 17
11
General traveling wave –
t=0
t>0
11/6/2003
PHY 113 -- Lecture 17
12
11/6/2003
PHY 113 -- Lecture 17
13
Basic physics behind wave motion -example: transverse wave on a string with tension T and
mass per unit length µ
y
θB
∆x
∆y
∆y
sin θ B ≈ tan θ B =
∆x B
d2y
m 2 = T sin θ B − T sin θ A
dt
m ≈ µ∆x
∆y ⎞
d2y
⎛ ∆y
−
⇒ µ∆x 2 ≈ T ⎜
⎟
dt
⎝ ∆x B ∆x A ⎠
1 ⎛ ∆y
∆y ⎞ ∂ 2 y
−
⎟= 2
Lim ⎜
∆x A ⎠ ∂x
∆x →0 ∆x ⎝ ∆x B
∂2 y T ∂2 y
=
2
µ ∂x 2
∂t
11/6/2003
PHY 113 -- Lecture 17
14
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
Note: Let u ≡ x ± vt
∂f (u) ∂f (u) ∂u
=
∂x
∂u ∂x
2
∂2 f (u) ∂2 f (u) ⎛ ∂u ⎞ ∂2 f (u)
=
⎟ =
2
2 ⎜
∂x
∂u ⎝ ∂x ⎠
∂u2
∂2 f (u) ∂2 f (u) ⎛ ∂u ⎞ ∂2 f (u) 2
=
v
⎟ =
2
2 ⎜
2
∂u
∂t
∂u ⎝ ∂t ⎠
2
11/6/2003
PHY 113 -- Lecture 17
15
Examples of solutions to the wave equation:
Moving “pulse”:
y ( x , t ) = y0 e
−( x −vt )2
phase factor
Periodic wave: y ( x, t ) = y0 sin (k ( x − vt ) + φ )
“wave vector”
2π
k=
not spring constant!!!
λ
2π
kv =
= 2 πf = ω
T
λ
⎛ ⎛x t ⎞
⎞
y ( x, t ) = y0 sin ⎜ 2 π⎜ − ⎟ + φ ⎟
=v
T
⎠
⎝ ⎝λ T ⎠
11/6/2003
PHY 113 -- Lecture 17
16
phase (radians)
Periodic traveling waves:
⎛ ⎛x t ⎞
⎞
y ( x, t ) = y0 sin ⎜ 2 π⎜ − ⎟ + φ ⎟
⎠
⎝ ⎝λ T ⎠
λ
=v
T
velocity (m/s)
period (s); T = 1/f
wave length (m)
Amplitude
Combinations of waves (“superposition”)
Note that :
sin A ± sin B = 2 sin[ 1 2 ( A ± B )]cos[ 1 2 ( A m B )]
11/6/2003
PHY 113 -- Lecture 17
17
sin A ± sin B = 2 sin[ 1 2 ( A ± B )]cos[ 1 2 ( A m B )]
⎛ ⎛x t ⎞
⎞
yright ( x, t ) = y0 sin ⎜ 2 π⎜ − ⎟ + φ ⎟
⎠
⎝ ⎝λ T ⎠
⎛ ⎛x t ⎞
⎞
yleft ( x, t ) = y0 sin ⎜ 2 π⎜ + ⎟ + φ ⎟
⎠
⎝ ⎝λ T ⎠
“Standing” wave:
2 πx
2 πt ⎞
yright ( x, t ) + y left ( x, t ) = 2 y0 sin ⎛⎜
+ φ ⎟⎞ cos⎛⎜
⎟
⎝ λ
⎠ ⎝ T ⎠
11/6/2003
PHY 113 -- Lecture 17
18
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
11/6/2003
PHY 113 -- Lecture 17
2L
⇒T =
nv
1 nv
f = =
T 2L
19
The sound of music
String instruments (Guitar, violin, etc.)
λn =
2L
n
v nv
fn =
=
λ n 2L
v=
(no sound yet.....)
11/6/2003
T
µ
2 πx ⎞ ⎛ 2 πt ⎞
⎛
ystanding ( x, t ) = 2 y0 sin ⎜
⎟
⎟ cos⎜
λ
T
⎝
⎠ ⎝
⎠
PHY 113 -- Lecture 17
20
n=2 λ=L
n = 3 λ = 23 L
n = 4 λ = 24 L
11/6/2003
PHY 113 -- Lecture 17
21
coupling to air
11/6/2003
PHY 113 -- Lecture 17
22
Peer instruction question
Suppose you pluck the “A” guitar string whose fundamental
frequency is f=440 cycles/s. The string is 0.5 m long so the
wavelength of the standing wave on the string is λ=1m.
What is the velocity of the wave on string?
(A) 1/220 m/s
(B) 1/440 m/s (C) 220 m/s
(D) 440 m/s
If you increased the tension of the string, what would happen?
11/6/2003
PHY 113 -- Lecture 17
23
11/6/2003
PHY 113 -- Lecture 17
24