Wave PhenomenaPhysics 15c

Lecture 7Sound

(H&L Chapter 5)

Transverse waves(H&L Section 4.6)

What We Did Last Time

! Studied the energy and momentum carried by waves! Energy and momentum are distributed over space

! They travel at the wave velocity! Kinetic energy = spring potential energy

! Energy density / velocity = momentum density

! Calculated what it takes to create the waves! Power needed = energy transfer rate! Force needed = momentum transfer rate

Goals for Today

! Sound in solid, liquid and gas! Apply what we did in the last two lectures

! Transverse waves on string! Different kind of mechanical waves! Define the wave equation and solve it! Calculate velocity, energy density and momentum density

Sound

! Sound is longitudinal waves in various material! Medium can be solid, liquid or gas

! Sound travels in 3 dimensions! But much of the physics can be analyzed in 1 dimension

! We already know everything about 1-dimensional longitudinal waves! All we need to find out is the linear mass density ρl and the

elastic modulus K

! This is going to be an easy lecture…

Sound in Solid

! A solid rod can be considered as a (very stiff) spring! It can be compressed or stretched by force! What is the elastic modulus K?

! The elastic modulus is defined by

! K depends on the material and how thick the rod is! It is proportional to the cross section of the rod

llKF ∆

=

F

l ∆l

Young’s Modulus

! We can write K = YA! A is the cross section area (in m2)

! Y is Young’s modulus of this material! Unit is Newtons/m2

! Hooke’s law can be rewritten as! LHS: force per unit cross section! RHS: Young’s modulus × relative deformation! This is the microscopic law of elasticity, which is

independent of the specific shape of the rod

llY

AF ∆=

Mass Density

! Linear mass density ρl is easy to calculate! Assume we know the volume mass density ρv

! The mass of the rod is ρv × A × l! Divide by the length " ρl = ρv × A

! Now we can calculate everything! Just look up the formulas from the last lecture

Sound Velocity in Solid

! Sound velocity in solid is

! This is a material constant! Example: steel has Y = 2×1011 N/m2, ρv = 7800 kg/m3

! See H&L p. 80 for more examples

vvlw

YA

YAKcρρρ

=== Young’s modulusVolume mass density

m/s 51007800

102 11

=×

=wc

Sound in Liquid

! Liquid transmits sound the same way as solid! Consider a pipe filled with liquid

! Force F changes the volume of a small section of the liquid

F

l ∆l

A

lll ∆+→lAV ∆=∆

VVV ∆+→

Bulk Modulus

! Elasticity of liquid is best described by the change of volume in response to pressure! Liquid has no fixed shape " Can’t use “length” or “area”

! MB is “bulk modulus” of the liquid! We can calculate K from MB

VVM

AFP B

∆−== Pressure

BK M A=l VF K K

l V∆ ∆

= − = −

Sound in Liquid

! Elastic modulus K = MB A! Linear mass density ρl = ρv A

! Wave velocity

! Example: water

llM

VVM

AF

BB∆

−=∆

−= F

l ∆l

A

v

B

lw

MKcρρ

==

m/s 1045.1kg/m 10

N/m 101.2 333

29

×=×

==v

Bw

Mcρ

Bulk modulusVolume mass density

Sound in Gas

! Sound in gas can be analyzed in the same way! Gas is much more compressible, and much lighter

! For gasses, we know a bit more about the constants! Most gasses at normal T & P are very close to ideal gas

! Ideal gas is made of infinitely small molecules that are not interacting with each other

! Properties of ideal gas is determined the molecular mass! MB and ρv can be calculated

! No such luck with solid and liquid " Y, MB, ρv must be measured (or looked up in the tables)

Volume Density of Ideal Gas

! 1 mole of ideal gas occupies 22.4 liter under STP.

! If not STP, use0224.0

molMv =ρ mass of 1 mole 0°C

1 atm

RTPM

vmol=ρ

gas constant 8.31 J/K

nRTPV =

pressure (N/m2)temperature (K)

volume (m2) amount (moles)

Compressibility of Ideal Gas

! How the pressure of gas reacts to compression?

! Ideal gas law PV = nRT suggests

! Temperature goes up when gas is compressed! This increases the pressure even more

! γ = 5/3 for monoatomic gasses (He, Ne, etc.)! γ = 7/5 for diatomic gasses (H2, N2, O2, etc.)

VP 1∝ Wrong

γVP 1∝ γ = ratio of specific heats > 1

Bulk Modulus of Ideal Gas

! The bulk modulus MB is defined by

! Using

! We can now calculate sound velocity, etc., for any gas as long as it’s close to ideal gas! That is, it’s not too compressed or too cold

! We just need to know the mass of 1 mole! And γ …

γVP 1∝

VP

dVdP γ−= PM B γ=

VVMP B

∆−=∆

dVdPVM B −=

Let’s try with air

Sound Velocity in Air

! Air is about 80% N2 + 20% O2

! 1 mole weighs

! ρv at STP is

! Both N2 and O2 are diatomic " So is air

g/mol 28.8g/mol 322.0g/mol 288.0

2.08.022 ONmol

=×+×=

×+×= MMM

3kg/m 29.10224.00288.0

==vρ

25 N/m 101.4atm) 1(57

×=== PM B γ m/s 330==v

Bw

Mcρ

Sound Velocity in Air

! If not at STP?

! Mmol = 0.0288 kg/m3, γ = 7/5 gives

! Sound velocity of any (ideal) gas! It does not depend on the pressure P! Next time you fly, check the outside temperature

! If T = −60°C,

RTPM

vmol=ρ

PM B γ=

molmol MRT

PMPRTMc

v

Bw

γγρ

===

m/s (K) 1.200288.0

31.84.1 TTcw =××

=

wc T∝

km/h 1050m/s 2926027320 ==−=wc

Sound Intensity

! Intensity of sound (how “loud”) is given by the energy carried by the sound! Since sound can spread out, we need the density of energy

per unit area

How much power(in Watts) goes

through this area?

Human Sensitivity to Sound

! Human can hear sound in 20 Hz – 20 kHz! The intensity range is 10-12 W/m2 – 1 W/m2

! Normal conversation ~10-6 W/m2

! We feel both frequency and intensity in log scale! Relative to 400 Hz

! 800 Hz is one octave higher! 1600 Hz is one octave higher again

! Relative to 10-6 W/m2

! 10-5 W/m2 is loud! 10-4 W/m2 is twice as loud

Sound Wave Amplitude

! What is the amplitude of 1 kHz sound wave with intensity 10-6 W/m2?! Imagine the sound is transmitted inside a pipe of cross-

section A m2

! Density of air at STP is 1.29 kg/m3 "

! Energy transfer rate is

AAc lw62

022

02 10)21000)(29.1)(330(

21

21 −=×= ξπξωρ

Al 29.1=ρ

m 101.1 80

−×=ξ0.01 microns!

Where Are We Now?

Oscillators (harmonic, damped, forced, coupled)

Waves

LongitudinalMechanical waves Sound

Electromagnetic waves

Transverse

LC transmission line

Radiation

VelocityString

Energy

Momentum

Where Do We Go Next?

! Wave velocity, energy and momentum are the most important characteristics for all waves! We have studied them only for longitudinal waves! We must explore other types of waves

! We will then go on and study moreinteresting features of the wave phenomena! e.g. reflection and standing waves! We will use the 3 types of waves to illustrate them

Study two other types of wavesstring

LC trans-mission line

Transverse Waves on String

! A string is stretched by tension T! The string’s linear mass density is ρl

! We vibrate the string vertically! Transverse to the direction of the string! Let’s call the displacement at x as ξ(x)

T T xξ(x)

Equation of Motion

! Consider the piece between x and x + ∆x! Mass is

! The tension at x has a slope

! Assuming θ is small, thevertical and horizontal components of the tension are

xm l∆= ρ

x x + ∆x

ξ(x) ξ(x + ∆x)T

Tθ

x∂∂

=ξθtan

xTT∂∂

−≈−ξθsin TT −≈− θcos Cancels between

the two ends

Equation of Motion

! Total vertical force is

! Equation of motion:

! Looks identical to the longitudinal waves! Solution must be the same

x x + ∆x

ξ(x) ξ(x + ∆x)T

Tθ

xxxT

xxT

∂∆+∂

+∂

∂−

)()( ξξ

xx

xT ∆∂

∂2

2 )(ξ

xx

txTt

txxl ∆∂

∂=

∂∂

∆ 2

2

2

2 ),(),( ξξρ 2

2

2

2 ),(),(x

txTt

txl ∂

∂=

∂∂ ξξρ

We had K here

Solution and Wave Velocity

! The normal mode solutions should look

! Throwing it into

we find

! Wave velocity is

))(exp(),( 0 tkxitx ωξξ ±=

2

2

2

2 )(),(x

xTt

txl ∂

∂=

∂∂ ξξρ

22 Tkl =ωρ

lw

Tk

cρ

ω==

Tension (N)Mass density (kg/m)

Energy and Momentum

! We can imagine that the energy and momentum densities are same as longitudinal waves with K " T

! We must check these

! First calculate how much power is needed to create transverse waves on the string! Conservation of energy assures that this is identical to the

energy transfer rate! You will calculate the latter in the problem set

20

2

21densityEnergy ξωρl=

w

l

c

20

2

21density Momentum ξωρ

=

Creating Transverse Waves

! To create ξ(x, t) = ξ0cos(kx – ωt), we drive the left end of the string by ξ0cosωt! What is the force against which we must work?

ξ(x,t)ξ0cosωt

Tξ0cosωt

θcosT

θsinT0

sin=

∂∂

−≈xx

TT ξθ

TT ≈θcos

Power (Energy Transfer Rate)

! The power is given by (force) x (velocity)! Velocity is vertical

! Vertical component of the force is

! Multiply them and take time average

tt

t ωωξωξ sin)cos(0

0 −=∂

∂

tkTx

tkxTx

ωξωξ sin))cos((0

0

0 −=

∂−∂

−=

tkT ωωξ 220 sin 2

022

0 21

21 ξωρωξ lwckT = l

wT

kc

ρω==

Exactly as expected

Momentum Density

! Can transverse waves carry longitudinal momentum?! We need horizontal component of the movement! It appears zero because we approximated

! The string do move horizontally! Direction of motion is orthogonal

to the string! Vertical velocity

! Horizontal velocity

TT ≈θcos

x x + ∆x

ξ(x) ξ(x + ∆x)t

txv∂

∂=⊥

),(ξ

//( , ) ( , )tan x t x tv vt x

ξ ξθ⊥

∂ ∂= − = −

∂ ∂

θ

Momentum Density

! The momentum density is given by multiplying this with the mass density

! Time average gives:

lw

Tk

cρ

ω==

Exactly as expected

Energy and momentum carried by transverse waves areidentical to those of longitudinal waves.

)sin()sin(),(),(00// tkxktkx

xtx

ttxv ωξωωξξξ

−−=∂

∂∂

∂−=

2 20 sin ( )l

dp k kx tdx

ρ ω ξ ω= −

2 22 00

1 12 2

ll

w

dp kdx c

ρ ω ξρ ω ξ= =

Summary

! Studied sound in solid, liquid and gas! Young’s modulus, bulk modulus and volume density

determine everything! Ideal gas is particularly simple: only need molecular mass

! Analyzed transverse waves on string! Wave equation looks the same as longitudinal waves

! Just replace elastic modulus K with tension T! Solution has the same characteristics as well! Energy and momentum densities are also identical

! Next lecture: LC transmission line