lecture 5 energy and momentum
TRANSCRIPT
Wave Phenomena Physics 15c
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
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))
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!
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
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
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)
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
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
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
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)
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
Creating Waves
Motor creates waves by moving as ξ0cosωt The power is given by (force) x (velocity) Velocity = −ξ0ω sinωt What is the force?
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)⎡⎣ ⎤⎦
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
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)
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)
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
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
⎛⎝⎜
⎞⎠⎟
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⎛⎝⎜
⎞⎠⎟
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
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
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)
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!
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⎛⎝⎜
⎞⎠⎟
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
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)
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