Lecture 5 Energy and Momentum

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Wave Phenomena Physics 15c Lecture 5 Energy and Momentum

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Page 1: Lecture 5 Energy and Momentum

Wave Phenomena Physics 15c

Lecture 5 Energy and Momentum

Page 2: Lecture 5 Energy and Momentum

What We Did Last Time Studied (finally!) continuous waves   Started from mass-spring transmission line   Found the wave equation and its normal modes

  Solutions represent waves traveling at constant velocity

  Used Fourier series to show the waveform can be arbitrary

ρl

∂2

∂t 2 ξ(x,t) = K∂2

∂x2 ξ(x,t) ξ(x,t) = Aexp(i(ω t ± kx))

cw =

Kρl

Page 3: Lecture 5 Energy and Momentum

Recap: How We Did It N coupled pendulums turns into continuous transmission line as N infinity   N equations of motion turn into a wave equation

Linear algebra assures that the coupled pendulums have N normal mode solutions   This turns into a factorized solution

Given this tip, we can solve the wave equation to get

Fourier series allows us to break arbitrary function into sin and cos, for which we have wave solutions   Since all solutions travel at the same speed, and because of

linearity, we conclude that arbitrary wave form can be created, and transmitted at the same speed

ξ(x,t) = a(x)eiω t

ξ(x,t) = Aexp(i(ω t ± kx))

Page 4: Lecture 5 Energy and Momentum

Arbitrary Traveling Waves In fact, the wave equation is satisfied by any function of the form ξ(x, t) = f(x ± cwt)

LHS = ρl

∂2

∂t 2 f (x ± cwt)

= ρlcw2 ′′f (x ± cwt)

= K ′′f (x ± cwt)

ρl

∂2

∂t 2 ξ(x,t) = K∂2

∂x2 ξ(x,t)

RHS = K∂2

∂x2 ξ(x ± cwt)

= K ′′f (x ± cwt)

cw =

Kρl

This seems just too simple after all that work!

Page 5: Lecture 5 Energy and Momentum

Subtleties The conclusion appears stupidly simple   We can generate any wave, and it will travel   Examples: sound, radio waves

But there are non-trivial assumptions   Linearity   Constant velocity for all normal modes, i.e. for all ω

  Bad examples: water waves, light passing glass   Those are called dispersive media

  Will come back to this soon

Page 6: Lecture 5 Energy and Momentum

Goals for Today Energy and momentum carried by the waves   Calculate the energy and momentum densities   Find how fast they are transmitted   There is a general relationship between the two

Calculate the power and the force needed to create the waves   Must match the energy and momentum transfer

Page 7: Lecture 5 Energy and Momentum

Energy Density

Consider forward-going normal-mode waves:

Energy is in two forms   Kinetic energy of the masses   Potential energy of the springs

ξ(x − Δx,t) ξ(x,t) ξ(x + Δx,t)

ξ(x,t) = ξ0 cos(kx −ω t)

Page 8: Lecture 5 Energy and Momentum

Kinetic Energy Velocity of the mass at x is

  Kinetic energy is

There is a mass at every Δx   The energy density (in Joules/meter) is

EK =

12

mv 2 =12

m∂ξ(x,t)

∂t⎛⎝⎜

⎞⎠⎟

2

dEK

dx= lim

Δx→0

EK

Δx= lim

Δx→0

12

mΔx

∂ξ(x,t)∂t

⎛⎝⎜

⎞⎠⎟

2

=12ρl

∂ξ(x,t)∂t

⎛⎝⎜

⎞⎠⎟

2

v(x,t) = ∂ξ(x,t)

∂t

Page 9: Lecture 5 Energy and Momentum

Spring Energy The spring between x and x + Δx has an energy of

  Taylor expansion

There is a spring at every Δx   The energy density (in Joules/meter) is

ES =

12

kS ξ(x + Δx,t) − ξ(x,t)( )2

dES

dx= lim

Δx→0

ES

Δx= lim

Δx→0

12

kSΔx∂ξ(x,t)∂x

⎛⎝⎜

⎞⎠⎟

2

=12

K∂ξ(x,t)∂x

⎛⎝⎜

⎞⎠⎟

2

ES =

12

kS

∂ξ(x,t)∂x

Δx⎛⎝⎜

⎞⎠⎟

2

Page 10: Lecture 5 Energy and Momentum

Kinetic + Spring Energy Recall the general solution:

Since for any waveform

  We have not used any specific shape of ξ(x, t)

The total energy density (in J/m) is

dEK

dx=

12ρl

∂ξ(x,t)∂t

⎛⎝⎜

⎞⎠⎟

2

=12ρ

lcw

2 ′f (x ± cwt)( )2

dES

dx=

12

K∂ξ(x,t)∂x

⎛⎝⎜

⎞⎠⎟

2

=12

K ′f (x ± cwt)( )2

ξ(x,t) = f (x ± cwt)

cw =

Kρl

dEK

dx=

dES

dx

Kinetic

Spring

dEdx

= ρl

∂ξ(x,t)∂t

⎛⎝⎜

⎞⎠⎟

2

= K∂ξ(x,t)∂x

⎛⎝⎜

⎞⎠⎟

2

Page 11: Lecture 5 Energy and Momentum

Energy Density in Normal Mode Now consider the normal mode ξ0cos(kx – ωt)

Energy density comes in packets   The packets travel with the

waves, with the velocity cw

x

t

Wave Energy density

dEdx

= ρl

∂ξ0 cos(kx −ω t)∂t

⎝⎜⎞

⎠⎟

2

= ρlω2ξ0

2 sin2(kx −ω t)

Page 12: Lecture 5 Energy and Momentum

Energy Transfer Rate The energy density averages to

This travels at velocity cw   The average rate of energy transfer (in Joules/second = Watts) is

This much of power is needed to create this wave   Let’s check

dEdx

= ρlω2ξ0

2 sin2(kx −ω t)

dEdx

=12ρlω

2ξ02 Average energy density

of a normal-mode wave

dEdt

=12

cwρlω2ξ0

2 =12

Kρlω2ξ0

2

cw =

Kρl

Page 13: Lecture 5 Energy and Momentum

Creating Waves

Motor creates waves by moving as ξ0cosωt The power is given by (force) x (velocity)   Velocity = −ξ0ω sinωt   What is the force?

Page 14: Lecture 5 Energy and Momentum

Creating Waves

Instead of the motor, imagine there were more masses and springs, doing the wave together The spring between x = −Δx and x = 0 is pushing the first “real” mass by

  This is the force you need to keep the wave going

ξ(0,t) ξ(−Δx,t)

F = −kS ξ(0,t) − ξ(−Δx,t)⎡⎣ ⎤⎦

Page 15: Lecture 5 Energy and Momentum

Creating Waves

Multiply this by the velocity

Time average of sin2ωt is 1/2

  This matches the energy transfer rate, as expected

F = −kS ξ(0,t) − ξ(−Δx,t)⎡⎣ ⎤⎦ = −kS

∂ξ(x,t)∂x

⎛⎝⎜

⎞⎠⎟ x=0

Δx = −K∂ξ(x,t)∂x

⎛⎝⎜

⎞⎠⎟ x=0

= −Kξ0k sinω t

ξ(x,t) = ξ0 cos(kx −ω t)

−ξ0ω sinω t

P = Kξ02kω sin2 ω t = Kρlω

2ξ02 sin2 ω t

k =ω

ρl

K

The average power needed to create the waves

P =12

Kρlω2ξ0

2

Page 16: Lecture 5 Energy and Momentum

Example: Audio Speakers Sound is an example of longitudinal waves   Will be discussed in the next lecture

Imagine a round speaker attached to a pipe   Air has volume mass density

1.29 kg/m3 ρl = 1.29π r2 kg/m   Sound velocity cw is 330 m/s

  2-inch speaker moving ±1 mm at 1 kHz 16.5 Watts   12-inch speaker moving ±5 mm at 20 Hz 6.0 Watts

The ω2 factor makes it much more difficult to generate low frequency sound

r

cw =

Kρl

dEdt

=12

Kρlω2ξ0

2 =12

cwρlω2ξ0

2 = 670r 2ω 2ξ02 (Watts)

Page 17: Lecture 5 Energy and Momentum

Momentum Density Momentum is just (mass) x (velocity)   Consider a mass m at position x

  Time average of sinωt is 0 No net momentum

This is wrong The problem: mass density variation

p = m ×

∂ξ(x,t)∂t

= mξ0ω sin(kx −ω t)

Page 18: Lecture 5 Energy and Momentum

Density Wave We are studying longitudinal wave   The direction of movement = the direction of transmission

The movement can be seen as changing density

  To calculate the momentum, we must take this density variation into account

wave

Page 19: Lecture 5 Energy and Momentum

Density Wave Consider the piece between x and x + Δx   Thickness is changed by the wave

  For small Δx,

  This changes the density

x x + Δx

x + ξ(x,t)

x + Δx + ξ(x + Δx,t)

Δx → Δx + ξ(x + Δx,t) − ξ(x,t)

ξ(x + Δx,t) − ξ(x,t) = ∂ξ

∂xΔx

Δx → Δx +

∂ξ∂x

Δx = Δx 1+ ∂ξ∂x

⎛⎝⎜

⎞⎠⎟

Page 20: Lecture 5 Energy and Momentum

Density Wave

Assume small waves

  The density should not change too much…

For

ρl →ρl

1+ ∂ξ∂x

small

ρl → ρl 1− ∂ξ(x,t)

∂x⎛⎝⎜

⎞⎠⎟

ξ(x,t) = ξ0 cos(kx −ω t)

ρl → ρl 1+ ξ0k sin(kx −ω t)( )

Δx → Δx 1+ ∂ξ

∂x⎛⎝⎜

⎞⎠⎟

Page 21: Lecture 5 Energy and Momentum

How It Looks Like

ξ(x,t) = ξ0 cos(kx −ω t)

Δρl = ρlξ0k sin(kx −ω t)

ρl + Δρl

velocity = ξ0ω sin(kx −ω t)

momentum density = mass density × velocity

Page 22: Lecture 5 Energy and Momentum

Momentum Density

Momentum density (in kg/s) is

  Average over time   sin(kx – wt) 0   sin2(kx – wt) ½

The rate of momentum transfer (in newtons) is

Δρl = ρlξ0k sin(kx −ω t) v = ξ0ω sin(kx −ω t)

dpdx

= ρl + Δρl( )v = (ρl + ρlξ0k sin(kx −ω t))ξ0ω sin(kx −ω t)

dpdx

=12ρlωkξ0

2 =12ρl

cw

ω 2ξ02

cw =

ωk

dpdt

= cw

dpdx

=12ρlω

2ξ02

Page 23: Lecture 5 Energy and Momentum

Energy vs. Momentum Energy density and momentum density are related

  This is true for any waveform on this medium   This is true even for different kinds of waves as long as it follows the

non-dispersive wave equation   Electromagnetic waves is one such example   e.g. a 100 MW laser pulse produces a force of

dEdx

=12ρlω

2ξ02

dpdx

=12ρlω

2ξ02

cw

energy densitypropagation velocity

= momentum density

1×108 (W)/3 ×108 (m/s) = 0.3(N)

Page 24: Lecture 5 Energy and Momentum

Force and Wave Creation

Waves carry momentum Motor must give net force to create waves   Let’s see if we can confirm this

Motor movement is ξ0coswt Warning: Tricky math ahead!

Page 25: Lecture 5 Energy and Momentum

Force From the Motor We have already calculated that the force motor must produce was

  Time-averaging this gives 0 (again!)

What’s wrong this time?   Density variation is changing the wavenumber k

  This makes the whole thing non-linear!

F = −K

∂ξ∂x

⎛⎝⎜

⎞⎠⎟ x=0

= −Kξ0k sinω t

cw =

ωk=

Kρl

=kS

mΔx

Δx → Δx 1+ ∂ξ

∂x⎛⎝⎜

⎞⎠⎟

Page 26: Lecture 5 Energy and Momentum

A Bit of Reflection We did not take the effect of density variation into account when we calculated the energy   Why was it allowed?

Whole calculation is based on linear approximation   We always take the first-order in Taylor series   Valid if the wave amplitude x0 is small

If the first-order terms cancel out to 0, we can no longer ignore the higher-order term   This happens when you try to calculate wave momentum

Page 27: Lecture 5 Energy and Momentum

Back to the Force

The force is (to the 2nd-order approximation)

Average over time

F =

12

Kk 2ξ02 =

12ρlω

2ξ02 = momentum transfer rate

k →k

1+ ∂ξ∂x

≈ k 1− ∂ξ∂x

⎛⎝⎜

⎞⎠⎟

cw =

ωk=

Kρl

=kS

mΔx

Δx → Δx 1+ ∂ξ

∂x⎛⎝⎜

⎞⎠⎟

F = −Kξ0k 1−∂(ξ0 cos(kx −ω t))

∂x⎛

⎝⎜⎞

⎠⎟ x=0

⎝⎜

⎠⎟ sinω t

= −Kξ0k sinω t(1− ξ0k sinω t)

Page 28: Lecture 5 Energy and Momentum

Summary 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