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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
6/23/2016
motion (Chap. 17)
PHY 113 -- Lecture 17
1
Summary of results concerning the physics of fluids -Bernoulli’s equation: P2 + ½ rv22 + rgh2 = P1 + ½ rv12 + rgh1
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).
6/23/2016
PHY 113 -- Lecture 17
2
Streamline flow of air around an airplane wing:
P2 + ½ rv22 + rgh2 = P1 + ½ rv12 + rgh1
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
r = 0.6 kg/m3, A = 40 m2
Flift = 63,600 N
6/23/2016
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
 rV 2  P0  rv 2
A 2
2
2F
v
2  12.6m/s
a
Ar 1   A  
P0 
6/23/2016
PHY 113 -- Lecture 17
AV  av
4
Another example; v=0
P0  roil g l  d   P0  r water gl  r air gd
l
roil  r water
ld
Potential energy reference
6/23/2016
PHY 113 -- Lecture 17
5
Bouyant forces:
the tip of the iceburg
FB  mice g
r water gVsubmerged  rice gVtotal
Vsubmerged rice 0.917


 93.9%
Vtotal
r water 1.024
6/23/2016
Source:http://bb-bird.com/iceburg.html
PHY 113 -- Lecture 17
6
The wave equation
y ( x, t )
position
Wave variable
 y
2 y

v
2
2
t
x
2
2
time
What does the wave equation mean?
Examples
Mathematical solutions of wave equation and descriptions
of waves
6/23/2016
PHY 113 -- Lecture 17
7
Example: Water waves
Needs more
sophistocated
analysis:
Source: http://www.eng.vt.edu/fluids/msc/gallery/gall.htm
6/23/2016
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
6/23/2016
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
6/23/2016
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)
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PHY 113 -- Lecture 17
11
General traveling wave –
t=0
t>0
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PHY 113 -- Lecture 17
12
6/23/2016
PHY 113 -- Lecture 17
13
Basic physics behind wave motion -example: transverse wave on a string with tension T and
mass per unit length m
qB
x
y
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
d2y
y 
 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
6/23/2016
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 
u 2
 2 f u   2 f u   u   2 f u  2

v
  
2
2
2
t
u  t 
u
2
6/23/2016
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 

6/23/2016
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 12  A  B 
6/23/2016
PHY 113 -- Lecture 17
17
sin A  sin B  2 sin  12  A  B cos 12  A  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:
2πx
2πt 
yright ( x, t )  yleft ( x, t )  2 y0 sin 
 φ  cos

 λ
  T 
6/23/2016
PHY 113 -- Lecture 17
18
Constraints of standing waves:
(=0)
2πx   2πt 

yright ( x, t )  yleft ( x, t )  2 y0 sin 
 cos

λ
T

 

for string :
2L
  ; n  1,2,3
n
λ
v
T
6/23/2016
PHY 113 -- Lecture 17
2L
T 
nv
1 nv
f  
T 2L
19
The sound of music
String instruments (Guitar, violin, etc.)
2L
λn 
n
v nv
fn 

λ n 2L
T
v
μ
(no sound yet.....)
6/23/2016
2πx   2πt 

ystanding ( x, t )  2 y0 sin 
 cos

λ
T

 

PHY 113 -- Lecture 17
20
n2 L
n  3   23 L
n  4   24 L
6/23/2016
PHY 113 -- Lecture 17
21
coupling to air
6/23/2016
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 =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?
6/23/2016
PHY 113 -- Lecture 17
23
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PHY 113 -- Lecture 17
24