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Waves on Strings: Reflection, Transmission, and Mechanical Impedance

(Topic)

Waves on Strings: Reflection, Transmission, and Mechanical Impedance

M06-08 developed traveling and Standing Waves on a uniform stretched string.

A new phenomenon appears when a traveling wave encounters a change in the string.

At an interface, the incoming wave generally separates into:

  • a reflected wave traveling back into the first string,
  • a transmitted wave traveling into the second string.

The amount reflected and transmitted is controlled by the mechanical impedance of the two strings.

For an ideal transverse string,

-----------------------
|     T         ∘  --- |
|Z =  -- = μc =    Tμ, |
------c----------------|
(1)

where

For a lossless junction between strings with impedances

Z1
(2)

and

Z2,
(3)

the displacement-amplitude coefficients are

|-------------------|
r = Ar- =  Z1 −-Z2, |
|    Ai    Z1 + Z2  |
---------------------
(4)

and

|------------------|
|   At      2Z1    |
t = --- = --------.|
----Ai----Z1-+--Z2--
(5)

The corresponding Power fractions are

|---------|
ℛ  = |r|2, |
-----------
(6)

and

|------------|
|     Z2-  2 |
|𝒯 =  Z  |t| .|
--------1----
(7)

For lossless strings,

|------------|
|ℛ +  𝒯 =  1.|
-------------
(8)

These relations unify ordinary string reflection, fixed and free boundaries, impedance matching, and the formation of standing waves.

PIC

Figure 1. A right-moving incident wave reaches a junction between two strings. In general, part of the wave is reflected into string 1 and part is transmitted into string 2.

1 Two joined strings

Let the interface be at

x = 0.
(9)

String 1 occupies

x < 0
(10)

and has parameters

T1,    μ1,     c1.
(11)

String 2 occupies

x > 0
(12)

and has

T2,    μ2,     c2.
(13)

Their wave speeds are

|----∘-----|
|          |
cj =    Tj.|
|       μj |
------------
(14)

Their characteristic mechanical impedances are

|----------∘-------|
Zj =  Tj-=    Tjμj.|
------cj------------
(15)

For the common textbook case of two segments of one stretched string, the static tension is the same on both sides:

T1 =  T2 = T.
(16)

2 Incident, reflected, and transmitted waves

Use complex notation and take the physical displacement to be the real part.

The incident wave is

|----------------|
|        i(k1x− ωt) |
-yi =-Aie-------.-
(17)

The reflected wave is

|------------------|
|yr = Arei(− k1x− ωt).|
--------------------
(18)

The transmitted wave is

|----------------|
|yt = Atei(k2x− ωt).|
------------------
(19)

All three waves have the same angular frequency

ω.
(20)

The wave numbers are

|---------------------|
|     ω            ω  |
k1 = c- ,    k2 = c-. |
------1------------2---
(21)

The wavelength can change across the junction even though the frequency does not.

3 First boundary condition: displacement continuity

If the two strings remain attached at the junction, the junction has one transverse displacement.

Therefore

|------------------|
|y1(0,t) = y2(0,t). |
-------------------
(22)

At

x = 0,
(23)

Ai + Ar = At.
(24)

Thus

|----------|
|1 + r = t,|
-----------
(25)

where

    Ar-         At-
r = Ai ,    t = Ai .
(26)

4 Second boundary condition: transverse force continuity

For small slope, the transverse component of tension is proportional to

  ∂y-
T ∂x .
(27)

At a massless junction with no external transverse force, the transverse force must be continuous:

|------|-----------|----|
|  ∂y1 |        ∂y2|    |
T1 ----||  =  T2 ---||  . |
----∂x--0−-------∂x--0+---
(28)

On string 1,

y1 = yi + yr.
(29)

Therefore

    |
∂y1-|| = ik (A  − A  )e− iωt.
∂x  |0     1   i    r
(30)

On string 2,

∂y2||
---|| =  ik2Ate−iωt.
∂x  0
(31)

Hence

T1k1(Ai − Ar) = T2k2At.
(32)

Since

T k  = ω Tj-=  ωZ ,
  j j    cj       j
(33)

|--------------------|
Z  (A  − A ) = Z  A .|
--1--i-----r-----2-t--
(34)

Equivalently,

|----------------|
|Z1(1 − r) = Z2t.|
------------------
(35)

PIC

Figure 2. An ideal massless junction imposes two conditions: the string displacement is continuous and the transverse tension force is continuous. These two conditions determine the reflected and transmitted amplitudes.

5 Displacement reflection coefficient

Use

t = 1 + r.
(36)

Then

Z1(1 − r) = Z2(1 + r).
(37)

Therefore

Z1 − Z2 = (Z1 +  Z2)r.
(38)

Thus

|-------------|
r =  Z1 −-Z2. |
-----Z1 +-Z2---
(39)

This coefficient includes the phase of the reflected displacement.

For positive real impedances:

  • Z2 > Z1 gives r < 0,
  • Z2 < Z1 gives r > 0,
  • Z2 = Z1 gives r = 0.

6 Displacement transmission coefficient

Since

t = 1 + r,
(40)

t = 1 + Z  − Z
-1-----2
Z1 + Z2 (41)
= --2Z1---
Z1 + Z2. (42)

Therefore

|------------|
|   --2Z1--- |
t = Z  +  Z .|
------1----2--
(43)

The displacement transmission amplitude can exceed unity.

That does not imply that transmitted power exceeds incident power, because the power also depends on the impedance.

7 Mechanical impedance of a traveling string wave

Consider a right-moving sinusoidal wave

y = A cos(kx − ωt ).
(44)

The transverse velocity is

     ∂y
vy = --- = ωA  sin(kx − ωt ).
     ∂t
(45)

The transverse force magnitude associated with the local slope is

         ||   ||
|Fy | = T |∂y-|.
         |∂x |
(46)

For the sinusoid,

|   |
||∂y-||=  kA |sin(kx −  ωt)|.
|∂x |
(47)

Thus the force-to-velocity amplitude ratio is

T-kA-   T-
 ωA   =  c .
(48)

Therefore

|--------|
|     T- |
|Z =  c .|
---------
(49)

Using

    ∘ -----
c =   T ∕μ,
(50)

|----∘-----------|
|Z =    Tμ =  μc.|
------------------
(51)

Its SI units are

|--------------------|
|[Z ] = kg∕s = N s∕m. |
----------------------
(52)

8 Instantaneous power carried by a string wave

The local mechanical energy density is

    1   2   1   2
u = -μy t + -T yx.
    2       2
(53)

Using the wave equation, one obtains the energy continuity relation

|--------------|
|∂u-+  ∂P--= 0,|
-∂t----∂x-------
(54)

where the instantaneous power flow in the positive

x
(55)

direction is

|-------------|
P--=-−-Tyxyt.-|
(56)

This sign convention makes

P > 0
(57)

for a right-moving traveling wave.

9 Average power of a sinusoidal wave

For

y = A cos(kx − ωt ),
(58)

yx = − kA sin(kx − ωt ),
(59)

and

yt = ωA sin(kx − ωt ).
(60)

Therefore

P =  TkωA2  sin2(kx −  ωt).
(61)

Since

    2    1-
⟨sin ⟩ = 2,
(62)

-----------------
|       1        |
|⟨P⟩ =  -T kωA2. |
--------2--------|
(63)

Using

Tk =  ωZ,
(64)

|----------------|
|       1        |
|⟨P ⟩ = -Z ω2A2. |
--------2--------
(65)

For a left-moving wave, the power flux has the opposite sign.

10 Power reflection coefficient

The incident average power magnitude is

⟨P ⟩ =  1Z ω2 |A  |2.
   i    2 1     i
(66)

The reflected wave carries power in the negative

x
(67)

direction with magnitude

        1-   2    2
|⟨Pr⟩| = 2 Z1ω |Ar |.
(68)

Therefore the reflected power fraction is

|------------------|
|     |⟨P  ⟩|       |
|ℛ =  ---r--=  |r|2.|
-------⟨Pi⟩--------
(69)

For real positive string impedances,

|------------------|
|    ( Z  −  Z )2  |
ℛ  =   --1----2   .|
-------Z1-+--Z2-----
(70)

11 Power transmission coefficient

The transmitted average power is

        1    2   2
⟨Pt⟩ =  -Z2ω  |At |.
        2
(71)

Therefore

𝒯 = ⟨Pt⟩
----
⟨Pi⟩ (72)
= Z2-
Z1|t|2. (73)

Hence

|----------------|
|       4Z1Z2    |
|𝒯 =  ---------2.|
------(Z1-+--Z2)---
(74)

12 Energy conservation at the junction

For a lossless ideal junction,

ℛ + 𝒯 = (Z  − Z )2
--1----2-2-
(Z1 + Z2) +   4Z  Z
-----1-2--2
(Z1 + Z2 ) (75)
= (Z1 + Z2)2
(Z--+-Z-)2-
  1    2 (76)
= 1. (77)

Therefore

|------------|
-ℛ-+--𝒯-=--1.|
(78)

Amplitude conservation is not the correct energy statement.

Power conservation is.

13 Same tension but different linear densities

Suppose the two strings have the same tension:

T  =  T =  T.
  1    2
(79)

Then

Z  = ∘T--μ-.
 j        j
(80)

The reflection coefficient becomes

|----√------√----|
|    --μ1 −---μ2-|
r =  √ μ1 + √ μ2.|
------------------
(81)

The transmission coefficient is

|--------√-------|
|t = √--2--μ1√---.|
|      μ1 +  μ2  |
------------------
(82)

Since

Z =  T∕c,
(83)

the reflection coefficient can also be written

|------------|
|    c2-−-c1 |
|r = c1 + c2 |
-------------
(84)

for equal tension.

14 Phase reversal

If the wave travels from lower impedance to higher impedance,

Z2 >  Z1,
(85)

then

r < 0.
(86)

The reflected displacement is inverted.

This is a phase change of

π.
(87)

If the wave travels from higher impedance to lower impedance,

Z2 <  Z1,
(88)

then

r > 0.
(89)

The reflected displacement is not inverted.

The sign of

r
(90)

therefore contains direct physical information.

15 Fixed-end reflection

A perfectly fixed end requires

|----------|
y(0,t) = 0.|
------------
(91)

Thus at the end,

Ai +  Ar = 0.
(92)

Therefore

-------------
|r    =  − 1.|
--fixed-------|
(93)

The reflected displacement has the same magnitude as the incident wave but is inverted.

No average power passes through the fixed boundary:

|------------------|
-ℛ-=-1,-----𝒯--=-0.-
(94)

The fixed end is a displacement node.

16 Free-end reflection

At a free end, the transverse force vanishes:

|--------|
T y  = 0.|
---x------
(95)

For nonzero

T,
(96)

|-------|
yx-=-0.--
(97)

This requires

Ai −  Ar = 0.
(98)

Therefore

|----------|
-rfree-=-+1.--
(99)

The reflected displacement is not inverted.

The free end is a displacement antinode.

PIC

Figure 3. A fixed end reflects displacement with a sign reversal, while a free end reflects without inversion. Both ideal boundaries reflect all incident power.

17 Fixed and free ends as impedance limits

The junction formula gives an intuitive impedance interpretation.

For

Z2 →  ∞,
(100)

r = Z1-−--Z2 →  − 1.
    Z1 + Z2
(101)

This is the fixed-end limit.

For

Z  →  0,
 2
(102)

r → +1.
(103)

This is the free-end limit.

Thus fixed and free boundaries are the extreme high- and low-impedance limits of the same reflection physics.

18 Impedance ratio

Define

|--------|
|    Z2  |
|η = ---.|
-----Z1---
(104)

Then

|----------|
|    1 − η |
r =  -----,|
-----1 +-η--
(105)

and

|----------|
|    --2---|
|t = 1 + η.|
------------
(106)

These forms make the dependence on impedance mismatch especially clear.

PIC

Figure 4. Displacement reflection and transmission amplitudes versus impedance ratio η = Z2∕Z1. The reflected displacement changes sign when the second string has the larger impedance.

19 Power coefficients versus impedance ratio

In terms of

η,
(107)

|----------------|
|     ( 1 − η )2 |
|ℛ =    ------  ,|
--------1 +-η-----
(108)

and

|--------------|
|     ---4η--- |
|𝒯 =  (1 + η)2.|
---------------
(109)

Both are nonnegative.

At

η = 1,
(110)

ℛ = 0,     𝒯  = 1.
(111)

For very large or very small impedance ratio,

ℛ  → 1.
(112)

PIC

Figure 5. Reflected and transmitted power fractions for a lossless junction. Transmission is complete when the impedances match, while extreme mismatch approaches total reflection.

20 Impedance matching

If

|--------|
Z1 =  Z2,|
----------
(113)

then

|------|
-r =-0,-
(114)

and

|------|
-t =-1.|
(115)

Therefore

|------------------|
|ℛ = 0,     𝒯  = 1.|
--------------------
(116)

The interface becomes reflectionless in this ideal model.

Matching the wave speed is not the fundamental criterion.

Matching the mechanical impedance is.

For two strings with different tensions and densities, it is possible to have

c1 ⁄= c2
(117)

while still having

Z1 =  Z2.
(118)

PIC

Figure 7. When the string impedances are equal, the interface produces no reflected wave. The transmitted wavelength may still differ if the two wave speeds differ.

21 Example 1: wave from a light string into a heavy string

Let the common tension be

T  = 100 N.
(119)

Take

μ1 = 0.0100 kg∕m,
(120)

and

μ2 = 0.0400 kg∕m.
(121)

Then

     ∘  -100---  |---------|
c1 =    -------= |100 m ∕s,|
        0.0100   ----------
(122)

while

     ∘ -------
        -100---  |---------|
c2 =    0.0400  = -50.0-m-∕s.-
(123)

The impedances are

      ∘ ------------  |----------|
Z1 =    100(0.0100) = -1.00kg-∕s,|
(124)

and

      ∘ ------------  |----------|
Z2 =    100(0.0400) = -2.00kg-∕s.|
(125)

Therefore

            |----|
    1-−-2-  |  1-|
r = 1 + 2 = |− 3.|
            ------
(126)

The reflected displacement is inverted.

The transmitted amplitude coefficient is

            |--|
t = 2(1)- = |2.|
    1 + 2   -3--
(127)

The power coefficients are

|-----1--|
|ℛ  = --,|
------9--|
(128)

and

|--------|
|     8- |
|𝒯  = 9 .|
---------
(129)

Most of the power is transmitted even though the transmitted displacement amplitude is smaller.

22 Example 2: wave from heavy string into light string

Reverse the direction of incidence.

Now

Z1 =  2.00kg ∕s,    Z2 = 1.00 kg∕s.
(130)

Then

|--------|
|      1-|
-r =-+-3.-
(131)

The reflected displacement is not inverted.

The transmitted displacement coefficient is

|------|
|    4-|
|t = 3.|
--------
(132)

The transmitted displacement amplitude exceeds the incident amplitude.

Nevertheless,

|--------------------|
|ℛ  = 1-,    𝒯  = 8-.|
------9-----------9--|
(133)

The power fractions are unchanged when the direction is reversed for this lossless two-string junction.

23 Why t > 1 does not violate energy conservation

In Example 2,

    4-
t = 3.
(134)

If amplitude alone measured power, that result would look impossible.

But

          2
⟨P ⟩ ∝ ZA  .
(135)

The transmitted string has the smaller impedance.

Therefore a larger displacement amplitude can carry less power per unit amplitude squared.

The correct energy test is always

ℛ +  𝒯 =  1.
(136)

24 Pulse interpretation

The sinusoidal coefficients also describe sufficiently narrowband pulses locally.

A pulse moving from low impedance toward high impedance produces:

  • an inverted reflected pulse,
  • a transmitted pulse with reduced displacement amplitude.

A pulse moving from high impedance toward low impedance produces:

  • a noninverted reflected pulse,
  • a transmitted pulse that can have a larger displacement amplitude.

The pulse shapes remain idealized only when the media are nondispersive and the interface itself has no frequency-dependent dynamics.

25 Standing waves from reflection

Suppose a right-moving incident wave and its reflected wave occupy the same uniform string:

        i(kx−ωt)      i(−kx−ωt)
y = Aie        + Are        .
(137)

If

|r| = 1,
(138)

the two waves have equal amplitude.

Their interference produces a pure standing wave.

For a fixed end,

r = − 1.
(139)

The end is a node.

For a free end,

r = +1.
(140)

The end is an antinode.

Thus boundary reflection phase directly determines the standing-wave geometry.

PIC

Figure 6. Reflection phase determines whether the boundary is a displacement node or antinode. A fixed end has r = −1, while a free end has r = +1.

26 Normal-mode conditions from round-trip phase

Reflection provides another derivation of the normal-mode spectra from M06-08.

A wave traveling across a string of length

L
(141)

and back accumulates propagation phase

2kL.
(142)

It also acquires reflection phases at the two ends.

A self-consistent normal mode requires the total round-trip phase to be an integer multiple of

2π.
(143)

27 Fixed-fixed resonance condition

Each fixed reflection contributes phase

π.
(144)

Two fixed reflections contribute

2π,
(145)

which is equivalent to zero phase modulo

2π.
(146)

Therefore

2kL =  2πn.
(147)

Hence

|----------|
|      nπ  |
|kn =  --. |
-------L---
(148)

This reproduces the fixed-fixed normal-mode result.

28 Fixed-free resonance condition

A fixed reflection contributes

π,
(149)

while a free reflection contributes zero displacement phase shift.

Therefore

2kL +  π = 2πn.
(150)

Hence

      (      )
            1-
kL =   n −  2  π.
(151)

Equivalently,

|----------------|
|     (2n − 1)π  |
|kn = ----------.|
----------2L------
(152)

This reproduces the fixed-free spectrum.

29 Partial reflection and imperfect standing waves

If

|r| < 1,
(153)

the counter-propagating waves do not have equal amplitudes.

The resulting pattern contains both:

  • spatial interference,
  • net energy transport.

It is not a pure standing wave.

Nodes no longer remain exact zero-displacement points unless special additional conditions apply.

This distinction becomes important in resonators with lossy or partially transmitting boundaries.

30 Multiple reflections

If a wave enters a finite segment bounded by two impedance changes, it can reflect repeatedly.

Each round trip produces:

  • propagation phase,
  • reflection amplitudes,
  • reflection phase shifts.

The outgoing wave is then a superposition of many delayed contributions.

This mechanism underlies:

  • echoes,
  • resonant buildup,
  • interference fringes,
  • frequency-selective transmission.

A later wave-analysis article can treat these effects using complex amplitudes and geometric series.

PIC

Figure 8. A finite string segment between two mismatched interfaces produces repeated internal reflections. Successive transmitted contributions can interfere constructively or destructively.

31 Connection to other wave systems

The impedance principle is much broader than strings.

Analogous ideas occur in:

The detailed definition of impedance changes from system to system, but the general principle remains:

Reflection is controlled by mismatch between the wave impedances presented on the two sides of a boundary.

32 Common mistakes

  1. Confusing wave speed with mechanical impedance.
  2. Writing Z = Tc instead of Z = T∕c.
  3. Forgetting that Z = √ ---
  T μ for the ideal string.
  4. Applying displacement continuity but forgetting transverse-force continuity.
  5. Using Ai − Ar = At for displacement continuity instead of Ai + Ar = At.
  6. Losing the minus sign in the reflected-wave spatial derivative.
  7. Confusing displacement amplitude coefficients with power coefficients.
  8. Assuming 𝒯 = |t|2 when the two impedances differ.
  9. Interpreting t > 1 as a violation of energy conservation.
  10. Forgetting that low-to-high impedance reflection is inverted.
  11. Forgetting that high-to-low impedance reflection is noninverted.
  12. Reversing the fixed-end and free-end reflection signs.
  13. Calling a partially reflected pattern a pure standing wave.
  14. Assuming equal wave speeds are required for impedance matching.
  15. Forgetting that the frequency remains the same across a stationary linear interface while wavelength can change.

33 Practice exercises

  1. Derive the mechanical impedance Z = T∕c = √T-μ- of an ideal string.
  2. Derive the displacement and transverse-force boundary conditions at a massless junction.
  3. Starting from those two conditions, derive r and t.
  4. Derive the average power ⟨P⟩ = 1
2Zω2A2 of a sinusoidal traveling wave.
  5. Derive the power coefficients ℛ and 𝒯 .
  6. Verify algebraically that ℛ + 𝒯 = 1.
  7. For equal tension, express r in terms of μ1 and μ2.
  8. For equal tension, express r in terms of c1 and c2.
  9. Derive the fixed-end and free-end reflection coefficients directly from their boundary conditions.
  10. Recover the fixed and free limits from Z2 →∞ and Z2 → 0.
  11. Reproduce both numerical examples in this article.
  12. Show explicitly why reversing the direction of incidence changes the sign of r but leaves ℛ unchanged.
  13. Derive the fixed-fixed normal-mode condition using round-trip propagation and reflection phases.
  14. Derive the fixed-free normal-mode condition using the same reflection-phase method.
  15. Explain why impedance matching, rather than wave-speed matching, is the correct reflectionless condition.

34 Summary

The mechanical impedance of an ideal transverse string is

|---------------∘------|
|Z =  T- = μc =    Tμ. |
------c----------------|
(154)

At an ideal junction,

|--------------|
|Ai + Ar = At, |
---------------
(155)

and

|--------------------|
Z1 (Ai − Ar) = Z2At. |
----------------------
(156)

Therefore

|-------------|
|    Z1 −-Z2- |
r =  Z +  Z , |
------1----2---
(157)

and

|------------|
|   --2Z1--- |
t = Z  +  Z .|
------1----2--
(158)

The average power carried by a sinusoidal wave is

|----------------|
|⟨P ⟩ = 1Z ω2A2. |
--------2--------|
(159)

Hence

|----(---------)2--|
ℛ  =   Z1-−--Z2   ,|
|      Z1 +  Z2    |
--------------------
(160)

and

|----------------|
|       4Z1Z2    |
|𝒯 =  ---------2.|
------(Z1-+--Z2)---
(161)

For a lossless junction,

|------------|
|ℛ +  𝒯 =  1.|
-------------
(162)

A fixed end has

|--------|
-r-=-−-1,|
(163)

while a free end has

|--------|
-r-=-+1.-|
(164)

An impedance-matched junction has

|------|
-r =-0.-
(165)

Reflection phase then connects directly to the standing-wave boundary conditions and normal-mode spectra developed in M06-08.

References

References

[1]   PhysicsLibrary, M06-08, Continuous Strings, Standing Waves, and Normal Modes.

[2]   PhysicsLibrary, M06-06, Discrete Oscillator Chains and Dispersion.

[3]   A. P. French, Vibrations and Waves, W. W. Norton, 1971.

[4]   J. R. Taylor, Classical Mechanics, University Science Books, 2005.


"Waves on Strings: Reflection, Transmission, and Mechanical Impedance" is owned by bloftin.
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Also defines:  string mechanical impedance, displacement reflection coefficient, displacement transmission coefficient, power reflection coefficient, power transmission coefficient, impedance matching
Keywords:  waves on strings, reflection, transmission, mechanical impedance, wave impedance, power flow, impedance matching, fixed-end reflection, free-end reflection, standing waves, boundary conditions

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