wave phenomena - harvard university department of...
TRANSCRIPT
What We Did Last Time Studied the energy and momentum carried by waves Energy is distributed non-uniformly over space
Kinetic energy = spring potential energy It travels at the wave velocity Momentum behaves similarly Energy density / velocity = momentum density
Calculated what it takes to create the waves Power needed = energy transfer rate Force needed = momentum transfer rate
We know all about longitudinal waves now
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
F = K
Δll
F
l Δl
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
FA=Y
Δll
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
cw =
Kρl
=YAρvA
=Yρv
Young’s modulus
Volume mass density
cw =
2 ×1011
7800= 5100 m/s
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 Δl
A
l → l + Δl ΔV = AΔl
V →V + ΔV
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
Pressure P =
FA= −MB
ΔVV
K = MBA F = −K
Δll= −K
ΔVV
Sound in Liquid
Elastic modulus K = MB A Linear mass density ρl = ρv A
Wave velocity
Example: water
FA= −MB
ΔVV
= −MB
Δll
cw =
Kρl
=MB
ρv
cw =
MB
ρv
=2.1×109 N/m2
103 kg/m3 = 1.45 ×103 m/s
Bulk modulus
Volume mass density
F
l Δl
A
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, use: ρv =
Mmol
0.0224mass of 1 mole
0°C 1 atm
ρv =
MmolPRT
gas constant 8.31 J/mol⋅K
PV = nRTpressure (N/m2)
temperature (K)
volume (m3) amount (mol)
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.)
P ∝
1V Wrong
P ∝
1V γ 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 …
P ∝
1V γ
dPdV
= −γ PV MB = γP
ΔP = −MB
ΔVV
MB = −VdPdV
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
Mmol = 0.8 ×MN2+ 0.2 ×MO2
= 0.8 × 28 g/mol+ 0.2 × 32 g/mol = 28.8 g/mol
ρv =
0.02880.0224
= 1.29 kg/m3
MB = γP =
75
(1 atm) = 1.4 ×105 N/m2
cw =
MB
ρv
= 330 m/s
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,
ρv =
MmolPRT
MB = γP
cw =
MB
ρv
=γPRTMmolP
=γRTMmol
cw =
1.4 × 8.31×T0.0288
= 20.1 T (K) m/s
cw ∝ T
cw = 20 273 − 60 = 292 m/s = 1050 km/h
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 440 Hz (middle A)
880 Hz is one octave higher 1760 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
That’s 0.01 microns Human ears are very sensitive
12
cwρlω2ξ0
2 =12
(330)(1.29A)(1000 × 2π )2ξ02 = 10−6 A
ρl = 1.29A
ξ0 = 1.1×10−8 m
Where Are We Now? Oscillators (harmonic, damped, forced, coupled)
Waves
Longitudinal Mechanical waves Sound
Electromagnetic waves
Transverse
LC transmission line
Radiation
Velocity String
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 more interesting 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 waves string
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 x ξ(x)
Equation of Motion Consider the piece between x and x + Δx Mass is
The tension at x has a slope
Assuming θ is small, the vertical and horizontal components of the tension are
m = ρlΔx
x x + Dx
ξ(x) ξ(x + Δx) T
T θ
tanθ =
∂ξ∂x
−T sinθ ≈ −T
∂ξ∂x −T cosθ ≈ −T 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
−T
∂ξ(x)∂x
+T∂ξ(x + Δx)
∂x
T∂2ξ(x)∂x2 Δx
ρlΔx
∂2ξ(x,t)∂t 2 =T
∂2ξ(x,t)∂x2 Δx
ρl
∂2ξ(x,t)∂t 2 =T
∂2ξ(x,t)∂x2
We had K here
x x + Dx
ξ(x) ξ(x + Δx) T
T θ
Solution and Wave Velocity The normal mode solutions should look
Throwing it into
we find
Wave velocity is
ξ(x,t) = ξ0 exp(i(kx ±ω t))
ρl
∂2ξ(x,t)∂t 2 =T
∂2ξ(x)∂x2
ρlω2 =Tk 2
cw =
ωk=
Tρl
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
Energy density =
12ρlω
2ξ02
Momentum density =
12ρlω
2ξ02
cw
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
T ξ0cosωt
T cosθ
T sinθ T sinθ ≈ −T
∂ξ∂x
⎛⎝⎜
⎞⎠⎟ x=0
T cosθ ≈T
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
∂(ξ0 cosω t)∂t
= −ξ0ω sinω t
−T
∂(ξ0 cos(kx −ω t))∂x
⎛
⎝⎜⎞
⎠⎟ x=0
= −Tξ0k sinω t
Tξ02kω sin2 ω t
12
Tξ02kω =
12
cwρlω2ξ0
2 cw =
ωk=
Tρl
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
T cosθ ≈T
x x + Δx
ξ(x) ξ(x + Δx) v⊥ =
∂ξ(x,t)∂t
v / / = −v⊥ tanθ = −
∂ξ(x,t)∂t
∂ξ(x,t)∂x
θ
Momentum Density
The momentum density is given by multiplying this with the mass density
Time average gives:
cw =
ωk=
Tρl
Exactly as expected
Energy and momentum carried by transverse waves are identical to those of longitudinal waves.
v / / = −
∂ξ(x,t)∂t
∂ξ(x,t)∂x
= ξ0ω sin(kx −ω t)ξ0k sin(kx −ω t)
dpdx
= ρlωkξ02 sin2(kx −ω t)
dpdx
=12ρlωkξ0
2 =12ρlω
2ξ02
cw
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