11/6/2003 PHY 113 -- Lecture 17 1

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3. Today’s lecture –

A few comments about the physics of fluids

The physics of motion (Chap. 17)

11/6/2003 PHY 113 -- Lecture 17 2

Summary of results concerning the physics of fluids --

Bernoulli’s equation: P2 + ½ ρv22 + ρgh2 = P1 + ½ ρv1

2 + 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 3

Streamline flow of air around an airplane wing:

P1

v1

P2v2

P2 + ½ ρv22 + ρgh2 = P1 + ½ ρv1

2 + ρgh1

( )22

212

112

21

21

ρ vvPPvvhh

−+=

>≈

Example:

v1 = 270 m/s, v2 = 260 m/s

ρ = 0.6 kg/m3, A = 40 m2

Flift = 63,600 N

Flift=(P2-P1)A

11/6/2003 PHY 113 -- Lecture 17 4

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.

( )( ) m/s6.1212

21

21

2

20

20

=−

=

=+=++

AaA

Fv

avAVvPVAFP

ρ

ρρ

11/6/2003 PHY 113 -- Lecture 17 5

Another example; v=0

( )dl

lgdglPdlgP wateroilairwateroil +≈++=++ ρρρρρ 00

Potential energy reference

11/6/2003 PHY 113 -- Lecture 17 6Source:http://bb-bird.com/iceburg.html

Bouyant forces:

the tip of the iceburg

%9.93024.1917.0

=≈=

==

water

ice

total

submerged

totalicesubmergedwater

iceB

VV

gVgVgmF

ρρ

ρρ

11/6/2003 PHY 113 -- Lecture 17 7

The wave equation

Wave variable

What does the wave equation mean?

Examples

Mathematical solutions of wave equation and descriptions of waves

2

22

2

2

xyv

ty

∂∂

=∂∂

),( txyposition time

11/6/2003 PHY 113 -- Lecture 17 8

Source: http://www.eng.vt.edu/fluids/msc/gallery/gall.htm

Example: Water waves

Needs more sophistocatedanalysis:

11/6/2003 PHY 113 -- Lecture 17 9

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 10

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 11

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 12

General traveling wave –

t = 0

t > 0

11/6/2003 PHY 113 -- Lecture 17 13

11/6/2003 PHY 113 -- Lecture 17 14

Basic physics behind wave motion --

example: transverse wave on a string with tension T and mass per unit length µ

y⎟⎠

⎞⎜⎝

⎛∆∆

−∆∆

≈∆⇒

∆≈

−=

ABT

dtydx

xm

TTdt

ydm

xy

xyµ

µ

θsinθsin

2

2

AB2

2

θB∆y

∆x

BBB x

y∆∆

=≈ θtanθsin

2

2

0 xy

xy1

xy

xLimABx ∂

∂=⎟

⎠

⎞⎜⎝

⎛∆∆

−∆∆

∆→∆

2

2

2

2

µ xyT

ty

∂∂

=∂∂

11/6/2003 PHY 113 -- Lecture 17 15

The wave equation:

Solutions: y(x,t) = f (x ± vt)

2

22

2

2

xyv

ty

∂∂

=∂∂ string) a(for

µ where Tv ≡

function of any shape

( ) ( )

( ) ( ) ( )

( ) ( ) ( ) 22

22

2

2

2

2

2

22

2

2

2

2

Let :Note

vu

uftu

uuf

tuf

uuf

xu

uuf

xuf

xu

uuf

xuf

vtxu

∂∂

=⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

=∂

∂

∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

∂∂

=∂

∂∂∂

∂∂

=∂

∂±≡

11/6/2003 PHY 113 -- Lecture 17 16

Examples of solutions to the wave equation:

Moving “pulse”:

Periodic wave:

( )2

0),( vtxeytxy −−=

( )( )φsin),( 0 +−= vtxkytxy

“wave vector”not spring constant!!!

ωπ2π2λπ2

===

=

fT

kv

k

vTT

txytxy =⎟⎠⎞

⎜⎝⎛ +⎟

⎠⎞

⎜⎝⎛ −=

λ φλ

π2sin),( 0

phase factor

11/6/2003 PHY 113 -- Lecture 17 17

vTT

txytxy =⎟⎠⎞

⎜⎝⎛ +⎟

⎠⎞

⎜⎝⎛ −=

λ φλ

π2sin),( 0

Periodic traveling waves:

Amplitude

wave length (m)

period (s); T = 1/f

phase (radians)

velocity (m/s)

Combinations of waves (“superposition”)

( )[ ] ( )[ ]BABABA m21

21 cossin2sinsin

: thatNote±=±

11/6/2003 PHY 113 -- Lecture 17 18

( )[ ] ( )[ ]BABABA m21

21 cossin2sinsin ±=±

⎟⎠⎞

⎜⎝⎛ +⎟

⎠⎞

⎜⎝⎛ +=⎟

⎠⎞

⎜⎝⎛ +⎟

⎠⎞

⎜⎝⎛ −= φ

λπ2sin),( φ

λπ2sin),( 00 T

txytxyTtxytxy leftright

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛ +=+

Ttxytxtxy leftrightπ2cosφ

λπ2sin2),(y),( 0

“Standing” wave:

11/6/2003 PHY 113 -- Lecture 17 19

Constraints of standing waves: ( ϕ = 0 )

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=+

Ttxytxtxy leftrightπ2cos

λπ2sin2),(y),( 0

L3,2,1;2:stringfor

== nnLλ

Lnv

Tf

nvLTv

T

21

2 λ

==

=⇒=

11/6/2003 PHY 113 -- Lecture 17 20

The sound of music

String instruments (Guitar, violin, etc.)

nL

n2λ =

Lnvvf

nn 2λ

==

µTv =

(no sound yet.....) ⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=

Ttxytxy π2cos

λπ2sin2),( 0standing

11/6/2003 PHY 113 -- Lecture 17 21

Ln == λ 2

Ln 32 3 == λ

Ln 42 4 == λ

11/6/2003 PHY 113 -- Lecture 17 22

coupling to air

11/6/2003 PHY 113 -- Lecture 17 23

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 24