the reflected wave, with amplitude B, travels in the −x direction. In the right-hand string segment there is only a transmitted wave going in the +x direction
u 2 (x, t) = Ce i(ω t − k2 x) .
(9)
The waves in the two string segments have different wavenumbers because of the different velocities in the two segments.
The amplitudes of the reflected and transmitted waves are
found using two boundary conditions that the physics of the
string imposes on the solution at the junction x = 0. First, because the two segments at the junction stay joined, the displacement must always be continuous across the junction, so
u 1 (0, t) = u 2 (0, t),
Ae
iωt
+ Be
iωt
= Ce
iωt
.
(10)
For this to occur at all times, the angular frequency of the three
waves must be the same, as we have assumed, and the amplitudes must satisfy
A + B = C.
(11)
Second, the y components of the tension forces acting on
the two sides of the junction must always be equal, or the unequal forces would tear the string apart. Thus, by analogy to
Eqn 2, we have another boundary condition
τ
τ
∂
∂
=
∂
∂
u
t
x
u
t
x
1
2
0
0
( , )
( , ) .
(12)
Taking the derivatives and canceling terms gives
τ k 1 (A − B) = τk 2 C,
(13)
or, because the velocities on the two sides are v i = (τ/ρ i ) 1/2 and
k i = ω /v i ,
ρ 1 v 1 (A − B) = ρ 2 v 2 C.
(14)
We now have two equations (11 and 14) for the three constants A, B, and C, giving the amplitudes of the incident,
reflected, and transmitted waves. We can eliminate C and find
the ratio of the amplitudes of the reflected and incident waves,
known as the reflection coefficient,
R
B
A
v
v
v
v
12
1 1
2 2
1 1
2 2
.
=
=
−
+
ρ
ρ
ρ
ρ
(15)
Similarly, eliminating B yields the transmission coefficient, the
ratio of transmitted and incident wave amplitudes,
T
C
A
v
v
v
12
1 1
1 1
2 2
2
.
=
=
+
ρ
ρ
ρ
(16)
The “12” subscripts indicate that the reflection and transmission coefficients describe a wave incident from segment 1
upon segment 2; the corresponding coefficients for a wave
incident from the right have subscripts “21.” These can be
derived by interchanging the subscripts, showing that
R 12 = −R 21 , T 12 + T 21 = 2.
(17)
The reflection and transmission coefficients depend on the
product of the density and velocity for each string, ρ i v i , a
quantity called the acoustic impedance. Because the amount
reflected depends on the difference in impedances between the
two sides, the strongest reflections occur at boundaries where
properties change significantly. One limiting case is if the
materials on both sides of the junction are identical (ρ 1 = ρ 2 and
v 1 = v 2 ), the reflection coefficient is zero and the transmission
coefficient would be one. Hence, as expected, all the wave is
transmitted, and none reflects. The other limiting case, total
reflection and no transmission, occurs at the end of a string.
The fixed end of a string, where no displacement occurs, can be
treated as a junction with a string of infinite impedance. Hence
the reflection coefficient is
R
v
v
xed
fi
,
=
− ∞
+ ∞
= −
ρ
ρ
1 1
1 1
1
(18)
so the entire incident wave pulse reflects with the opposite
polarity. Similarly, a string whose end is free to move is described by the condition that the derivative ∂u/∂x is zero,
because there is no force applied. This can be treated as a junction with a string of zero impedance, so the reflection coefficient is +1, and the entire incident pulse reflects with the same
polarity. For values between the limiting cases, Eqn 15 shows
that the polarity of the reflection depends upon whether the
wave leaves or enters a string of greater impedance. If the
impedance of segment 2 exceeds that of segment 1, waves
going from segment 1 toward segment 2 reflect with reversed
polarity, whereas waves going the other way reflect without changing polarity. Reflections at free and fixed ends are
extreme cases of this property. Hence the amplitudes of reflections from boundaries can be used to infer changes in physical
properties.
To illustrate these ideas, consider the reflection and transmission of waves on a string divided at x = 10 into two
segments (Fig. 2.2-6). The left segment has ρ 1 = 1, v 1 = 3, and
the right segment has ρ 2 = 4, v 2 = 1.5. At time 0 the string
is plucked for a very short time by a source at the position
marked by the triangle, so waves spread out in either direction.
At time 1, the first time shown, the wave traveling to the right
has just encountered the junction (marked by a vertical dashed
line). The reflection and transmission coefficients depend on the
impedances ρ 1 v 1 = 3 and ρ 2 v 2 = 6. Thus for waves going from
left to right R 12 = −0.33 and T 12 = 0.67. A small reflected pulse
is generated, with a downward polarity opposite that of the
incident pulse, because the reflection coefficient is negative. At
2.2 Waves on a string 33
u 2 (x, t) = Ce i(ω t − k2 x) .
(9)
The waves in the two string segments have different wavenumbers because of the different velocities in the two segments.
The amplitudes of the reflected and transmitted waves are
found using two boundary conditions that the physics of the
string imposes on the solution at the junction x = 0. First, because the two segments at the junction stay joined, the displacement must always be continuous across the junction, so
u 1 (0, t) = u 2 (0, t),
Ae
iωt
+ Be
iωt
= Ce
iωt
.
(10)
For this to occur at all times, the angular frequency of the three
waves must be the same, as we have assumed, and the amplitudes must satisfy
A + B = C.
(11)
Second, the y components of the tension forces acting on
the two sides of the junction must always be equal, or the unequal forces would tear the string apart. Thus, by analogy to
Eqn 2, we have another boundary condition
τ
τ
∂
∂
=
∂
∂
u
t
x
u
t
x
1
2
0
0
( , )
( , ) .
(12)
Taking the derivatives and canceling terms gives
τ k 1 (A − B) = τk 2 C,
(13)
or, because the velocities on the two sides are v i = (τ/ρ i ) 1/2 and
k i = ω /v i ,
ρ 1 v 1 (A − B) = ρ 2 v 2 C.
(14)
We now have two equations (11 and 14) for the three constants A, B, and C, giving the amplitudes of the incident,
reflected, and transmitted waves. We can eliminate C and find
the ratio of the amplitudes of the reflected and incident waves,
known as the reflection coefficient,
R
B
A
v
v
v
v
12
1 1
2 2
1 1
2 2
.
=
=
−
+
ρ
ρ
ρ
ρ
(15)
Similarly, eliminating B yields the transmission coefficient, the
ratio of transmitted and incident wave amplitudes,
T
C
A
v
v
v
12
1 1
1 1
2 2
2
.
=
=
+
ρ
ρ
ρ
(16)
The “12” subscripts indicate that the reflection and transmission coefficients describe a wave incident from segment 1
upon segment 2; the corresponding coefficients for a wave
incident from the right have subscripts “21.” These can be
derived by interchanging the subscripts, showing that
R 12 = −R 21 , T 12 + T 21 = 2.
(17)
The reflection and transmission coefficients depend on the
product of the density and velocity for each string, ρ i v i , a
quantity called the acoustic impedance. Because the amount
reflected depends on the difference in impedances between the
two sides, the strongest reflections occur at boundaries where
properties change significantly. One limiting case is if the
materials on both sides of the junction are identical (ρ 1 = ρ 2 and
v 1 = v 2 ), the reflection coefficient is zero and the transmission
coefficient would be one. Hence, as expected, all the wave is
transmitted, and none reflects. The other limiting case, total
reflection and no transmission, occurs at the end of a string.
The fixed end of a string, where no displacement occurs, can be
treated as a junction with a string of infinite impedance. Hence
the reflection coefficient is
R
v
v
xed
fi
,
=
− ∞
+ ∞
= −
ρ
ρ
1 1
1 1
1
(18)
so the entire incident wave pulse reflects with the opposite
polarity. Similarly, a string whose end is free to move is described by the condition that the derivative ∂u/∂x is zero,
because there is no force applied. This can be treated as a junction with a string of zero impedance, so the reflection coefficient is +1, and the entire incident pulse reflects with the same
polarity. For values between the limiting cases, Eqn 15 shows
that the polarity of the reflection depends upon whether the
wave leaves or enters a string of greater impedance. If the
impedance of segment 2 exceeds that of segment 1, waves
going from segment 1 toward segment 2 reflect with reversed
polarity, whereas waves going the other way reflect without changing polarity. Reflections at free and fixed ends are
extreme cases of this property. Hence the amplitudes of reflections from boundaries can be used to infer changes in physical
properties.
To illustrate these ideas, consider the reflection and transmission of waves on a string divided at x = 10 into two
segments (Fig. 2.2-6). The left segment has ρ 1 = 1, v 1 = 3, and
the right segment has ρ 2 = 4, v 2 = 1.5. At time 0 the string
is plucked for a very short time by a source at the position
marked by the triangle, so waves spread out in either direction.
At time 1, the first time shown, the wave traveling to the right
has just encountered the junction (marked by a vertical dashed
line). The reflection and transmission coefficients depend on the
impedances ρ 1 v 1 = 3 and ρ 2 v 2 = 6. Thus for waves going from
left to right R 12 = −0.33 and T 12 = 0.67. A small reflected pulse
is generated, with a downward polarity opposite that of the
incident pulse, because the reflection coefficient is negative. At
2.2 Waves on a string 33