where the last step used the Taylor series (1 + a 2 ) 1/2 ≈ 1 + a 2 /2
for small a. The potential energy stored in the string is the product of the tension and the strain integrated over the entire
length L,
τ
2
0
2
Ύ
L
u
x
dx
∂
∂
⎛
⎝
⎜
⎞
⎠
⎟
,
(22)
so we can define the average potential energy, PE, in a segment
dx as
PE
u
x
dx .
=
∂
∂
⎛
⎝
⎜
⎞
⎠
⎟
τ
2
2
(23)
We characterize the energy of a traveling wave by the kinetic
and potential energy averaged over a wavelength. If u(x, t)
= A cos (ωt − kx), then the kinetic energy averaged over a
wavelength is
KE
u
t
dx
A
t kx dx
sin (
) .
=
∂
∂
⎛
⎝
⎜
⎞
⎠
⎟
=
−
ρ
λ
ρ ω
λ
ω
λ
λ
2
2
0
2
2 2
0
2
Ύ
Ύ
(24)
The integral of the sinusoid squared over a period is
Ύ
0
2
2
λ
ω
λ
sin (
)
/ ,
t kx dx
−
=
(25)
so the kinetic energy is
KE = A
2
ω
2
ρ/4.
(26)
Similarly, the potential energy averaged over a wavelength is
PE =
τ
λ
τ
λ
λ
λ
2
2
0
2
2 2
0
Ύ
Ύ
∂
∂
u
x
dx
A k
⎛
⎝
⎜
⎞
⎠
⎟
=
sin 2 (ωt − kx)dx,
(27)
which, using Eqn 25, becomes
PE = τA 2 k 2 /4 = A 2 ω 2 ρ/4,
(28)
the same as the kinetic energy.
Hence the total energy transported, averaged over a wavelength, is the sum of the potential and kinetic energies:
E = PE + KE = A 2 ω 2 ρ/2.
(29)
Another way to state this is in terms of the energy flux, the rate
at which the wave transports energy past a point on the string.
The average flux is just the averaged energy times the velocity
Fig. 2.2-7 An incident wave pulse of length λ 1 on a string with velocity v 1
generates a transmitted pulse of length λ 2 in a string with velocity v 2 .
The change in pulse length results from the distance the transmitted
pulse travels while the incident pulse passes through the junction. If the
amplitude of the incident pulse is 1, then the reflected and transmitted
pulses have amplitudes R 12 and T 12 .
1
x = 0
λ 1
λ
λ 1
λ
λ 2
λ
T 12
R 12
2.2 Waves on a string 35
seismic waves in the earth. We will use this approach to show
how we study changes in physical properties at depth in the
earth from the amplitudes of reflected and transmitted waves.
2.2.4 Energy in a harmonic wave
We noted earlier that in some cases the transmission coefficient exceeds 1. To see how this occurs, we consider the energy
transported by the traveling waves. It turns out that although
amplitudes are easier to visualize, energy is often more useful
for understanding wave behavior because energy is conserved,
whereas amplitude is not. Hence, when a result for amplitudes
is hard to understand, considering the energy can provide
insight.
By analogy to the kinetic energy mv 2 /2 of a point mass, the
kinetic energy, KE, of a segment dx of the string is found from
the velocity, the time derivative of the displacement, so
KE
u
t
dx ,
=
∂
∂
⎛
⎝
⎜
⎞
⎠
⎟
ρ
2
2
(20)
because the mass of the string is m = ρdx.
The string also stores potential energy, because it is stretched,
or deformed, from its equilibrium position. We will see shortly
that a measure of the deformation is the strain, e, which for
the string is the ratio of the change in the length to the original
length. Hence for an element of the string (Fig. 2.2-1) with
initial length dx, the strain due to the displacement du is
e
dx
du
dx
dx
du
dx
u
x
(
)
,
/
/
=
+
−
= +
⎛
⎝
⎜
⎞
⎠
⎟
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
− =
⎛
⎝
⎜
⎞
⎠
⎟
2
212
2
1 2
2
1
1
1
2
∂
∂
(21)
for small a. The potential energy stored in the string is the product of the tension and the strain integrated over the entire
length L,
τ
2
0
2
Ύ
L
u
x
dx
∂
∂
⎛
⎝
⎜
⎞
⎠
⎟
,
(22)
so we can define the average potential energy, PE, in a segment
dx as
PE
u
x
dx .
=
∂
∂
⎛
⎝
⎜
⎞
⎠
⎟
τ
2
2
(23)
We characterize the energy of a traveling wave by the kinetic
and potential energy averaged over a wavelength. If u(x, t)
= A cos (ωt − kx), then the kinetic energy averaged over a
wavelength is
KE
u
t
dx
A
t kx dx
sin (
) .
=
∂
∂
⎛
⎝
⎜
⎞
⎠
⎟
=
−
ρ
λ
ρ ω
λ
ω
λ
λ
2
2
0
2
2 2
0
2
Ύ
Ύ
(24)
The integral of the sinusoid squared over a period is
Ύ
0
2
2
λ
ω
λ
sin (
)
/ ,
t kx dx
−
=
(25)
so the kinetic energy is
KE = A
2
ω
2
ρ/4.
(26)
Similarly, the potential energy averaged over a wavelength is
PE =
τ
λ
τ
λ
λ
λ
2
2
0
2
2 2
0
Ύ
Ύ
∂
∂
u
x
dx
A k
⎛
⎝
⎜
⎞
⎠
⎟
=
sin 2 (ωt − kx)dx,
(27)
which, using Eqn 25, becomes
PE = τA 2 k 2 /4 = A 2 ω 2 ρ/4,
(28)
the same as the kinetic energy.
Hence the total energy transported, averaged over a wavelength, is the sum of the potential and kinetic energies:
E = PE + KE = A 2 ω 2 ρ/2.
(29)
Another way to state this is in terms of the energy flux, the rate
at which the wave transports energy past a point on the string.
The average flux is just the averaged energy times the velocity
Fig. 2.2-7 An incident wave pulse of length λ 1 on a string with velocity v 1
generates a transmitted pulse of length λ 2 in a string with velocity v 2 .
The change in pulse length results from the distance the transmitted
pulse travels while the incident pulse passes through the junction. If the
amplitude of the incident pulse is 1, then the reflected and transmitted
pulses have amplitudes R 12 and T 12 .
1
x = 0
λ 1
λ
λ 1
λ
λ 2
λ
T 12
R 12
2.2 Waves on a string 35
seismic waves in the earth. We will use this approach to show
how we study changes in physical properties at depth in the
earth from the amplitudes of reflected and transmitted waves.
2.2.4 Energy in a harmonic wave
We noted earlier that in some cases the transmission coefficient exceeds 1. To see how this occurs, we consider the energy
transported by the traveling waves. It turns out that although
amplitudes are easier to visualize, energy is often more useful
for understanding wave behavior because energy is conserved,
whereas amplitude is not. Hence, when a result for amplitudes
is hard to understand, considering the energy can provide
insight.
By analogy to the kinetic energy mv 2 /2 of a point mass, the
kinetic energy, KE, of a segment dx of the string is found from
the velocity, the time derivative of the displacement, so
KE
u
t
dx ,
=
∂
∂
⎛
⎝
⎜
⎞
⎠
⎟
ρ
2
2
(20)
because the mass of the string is m = ρdx.
The string also stores potential energy, because it is stretched,
or deformed, from its equilibrium position. We will see shortly
that a measure of the deformation is the strain, e, which for
the string is the ratio of the change in the length to the original
length. Hence for an element of the string (Fig. 2.2-1) with
initial length dx, the strain due to the displacement du is
e
dx
du
dx
dx
du
dx
u
x
(
)
,
/
/
=
+
−
= +
⎛
⎝
⎜
⎞
⎠
⎟
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
− =
⎛
⎝
⎜
⎞
⎠
⎟
2
212
2
1 2
2
1
1
1
2
∂
∂
(21)