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