Wave Phenomena Physics 15c

Lecture 6 Sound

Transverse Waves

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