Wave Mechanics Examples: Mechanical Wave Impedance

This companion article provides exercises for WM22, Mechanical Wave Impedance. The exercises appear first; complete worked solutions follow in Part II.

For an ideal string with Tension T, linear mass density μ, and wave speed

(1)

the characteristic mechanical wave impedance is

(2)

For a pure one-way traveling wave, using the positive-x force convention of WM22,

(3)

for right-moving propagation and

(4)

for left-moving propagation. Correspondingly,

(5)

For a sinusoidal one-way wave,

(6)

At an ideal junction between strings with impedances Z₁ and Z₂, the displacement-amplitude coefficients used in WM22 are

(7)

(8)

with power fractions

(9)

For a lossless ideal junction,

(10)

These relations are standard consequences of continuity of displacement and transverse force at an ideal junction [1, 2, 3, 6].

How to use this problem set

Attempt all problems in Part I before reading Part II. In interface problems, keep four quantities distinct:

• r: signed displacement-amplitude reflection coefficient,
• t: displacement-amplitude transmission coefficient,
• R: reflected power fraction,
• 𝒯 : transmitted power fraction.

In particular, t is not generally the transmitted power fraction.

Part I: Exercises

Exercise 1: Compute wave speed and characteristic impedance

An ideal string has

(11)

Find:

1. the wave speed c,
2. the characteristic impedance Z₀ using μc,
3. the same impedance using ,
4. the SI units of Z₀.

Exercise 2: Parameter scaling

Start with a string having tension T, linear density μ, speed c, and impedance Z₀.

For each change below, determine the multiplicative factor by which c and Z₀ change.

1. T → 4T while μ is unchanged.
2. μ → 4μ while T is unchanged.
3. T → 4T and μ → 4μ simultaneously.

Explain why speed and impedance respond differently to the same parameter changes.

Exercise 3: Force, velocity, and signed power

The following figure summarizes the WM22 sign convention.

A string has

(12)

At one point and instant,

(13)

Find the transverse force F_⊥^(+x) and signed instantaneous power for:

1. a right-moving wave,
2. a left-moving wave.

Explain what the sign of the power means.

Exercise 4: Sinusoidal average power from impedance

A string has

(14)

A right-moving sinusoidal wave has amplitude

(15)

and frequency

(16)

Find:

1. Z₀,
2. ω,
3. v_rms,
4. ⟨P⟩ using Z₀v_rms².

Exercise 5: Required amplitude for a target average power

A sinusoidal right-moving wave travels on a string with

(17)

at frequency

(18)

What displacement amplitude A is required to carry average power

(19)

Exercise 6: Reflection and transmission at an impedance step

A harmonic wave travels from a string with

(20)

into a second string with

(21)

The interface is ideal and lossless.

Find:

1. the displacement reflection coefficient r,
2. the displacement transmission coefficient t,
3. the reflected power fraction R,
4. the transmitted power fraction 𝒯 ,
5. whether the reflected displacement is inverted.

Exercise 7: Read the reflection coefficient from impedance ratio

The reflection coefficient can be written

(22)

For each impedance ratio below, compute r, R, and 𝒯 and state whether the reflected displacement is inverted:

1. q = 0.25,
2. q = 1,
3. q = 2.

Exercise 8: Matched impedance but different speed

Two strings have parameters

(23)

and

(24)

Find:

1. c₁ and c₂,
2. Z₁ and Z₂,
3. the reflection coefficient r,
4. the wavelengths in the two strings if the source frequency is 20 Hz.

Exercise 9: Fixed-like and free-like limits

Using

(25)

answer the following.

1. What limit does r approach when Z₂∕Z₁ →∞?
2. What physical boundary behavior does this resemble?
3. What limit does r approach when Z₂∕Z₁ → 0?
4. What physical boundary behavior does this resemble?

Exercise 10: Derive the amplitude coefficients

At an ideal interface, suppose the displacement amplitudes satisfy

(26)

and the force condition is

(27)

Starting from these two equations, derive

(28)

and

(29)

Then verify directly that

(30)

Exercise 11: Transmission amplitude greater than one

A wave travels from

(31)

into

(32)

Find r, t, R, and 𝒯 .

Then explain why t > 1 does not violate conservation of energy.

Exercise 12: Infer an unknown impedance from a measured reflection

A displacement-amplitude reflection coefficient is measured to be

(33)

for a wave incident from a string with

(34)

Find:

1. the unknown impedance Z₂,
2. the reflected power fraction R,
3. the transmitted power fraction 𝒯 ,
4. the displacement-amplitude transmission coefficient t.

Exercise 13: Same speed, different impedance

String 1 has

(35)

Design string 2 so that it has the same wave speed

(36)

but twice the characteristic impedance of string 1.

Find suitable values of T₂ and μ₂, and then find the displacement reflection coefficient for a wave traveling from string 1 into string 2.

Explain why equal wave speed does not guarantee zero reflection.

Exercise 14: Full impedance-interface synthesis

A sinusoidal wave travels from string 1 into string 2. The string parameters are

(37)

and

(38)

The incident wave has displacement amplitude

(39)

and frequency

(40)

Find:

1. c₁ and c₂,
2. Z₁ and Z₂,
3. λ₁ and λ₂,
4. r and t,
5. A_r and A_t including the sign of A_r,
6. R and 𝒯 ,
7. the incident average power,
8. the magnitudes of reflected and transmitted average power,
9. a numerical check of power conservation.

Part II: Complete Worked Solutions

Solution 1: Compute wave speed and characteristic impedance

The wave speed is

c	=	(41)
	=	(42)
	= 100 m/s .	(43)

Using Z₀ = μc,

Z0	= (0.010)(100)	(44)
	= 1.00 kg/s .	(45)

Using the equivalent formula,

Z0	=	(46)
	=	(47)
	= 1.00 kg/s .	(48)

The units may also be written

(49)

Solution 2: Parameter scaling

The governing relations are

(50)

1. If T → 4T with μ fixed,

   (51)

2. If μ → 4μ with T fixed,

   (52)

3. If both T and μ are multiplied by 4,

   (53)

The difference occurs because speed depends on the ratio T∕μ, whereas impedance depends on the product Tμ.

Solution 3: Force, velocity, and signed power

For the right-moving wave,

(54)

Hence

(55)

The power is

P	= Z0ut2	(56)
	= (1.50)(0.30)2	(57)
	= +0.135 W .	(58)

For the left-moving wave,

(59)

and

(60)

Positive signed power means energy flows toward increasing x; negative signed power means energy flows toward decreasing x.

Solution 4: Sinusoidal average power from impedance

First compute

(61)

The angular frequency is

(62)

The RMS transverse velocity is

vrms	=	(63)
	=	(64)
	≈ 0.1777 m/s .	(65)

Therefore

⟨P⟩	= Z0vrms2	(66)
	= (1.20)(0.1777)2	(67)
	≈ 3.79 × 10−2 W .	(68)

So the average power is approximately

(69)

Solution 5: Required amplitude for a target average power

Start from

(70)

Solve for A:

(71)

Here

(72)

Thus

A	=	(73)
	≈ 1.59 × 10−3 m.	(74)

Therefore

(75)

Solution 6: Reflection and transmission at an impedance step

With Z₁ = 1.0 kg/s and Z₂ = 3.0 kg/s,

r	=	(76)
	=	(77)
	= −0.50 .	(78)

The negative sign means the reflected displacement is inverted.

The displacement transmission coefficient is

t	=	(79)
	=	(80)
	= 0.50 .	(81)

The reflected power fraction is

(82)

The transmitted power fraction is

𝒯	= t2	(83)
	= 3(0.50)2	(84)
	= 0.75 .	(85)

The check is

(86)

Solution 7: Read the reflection coefficient from impedance ratio

Use

(87)

1. For q = 0.25,

   r	= = = 0.60 ,	(88)
   R	= (0.60)2 = 0.36 ,	(89)
   𝒯	= 1 − 0.36 = 0.64 .	(90)

   Since r > 0, there is no displacement inversion.

2. For q = 1,

   (91)

   This is perfect impedance matching.

3. For q = 2,

   r	= = − ,	(92)
   R	= ≈ 0.111,	(93)
   𝒯	= ≈ 0.889.	(94)

   Since r < 0, the reflected displacement is inverted.

Solution 8: Matched impedance but different speed

For string 1,

c1	=	(95)
	= 90 m/s ,	(96)

and

Z1	=	(97)
	= 0.90 kg/s .	(98)

For string 2,

c2	=	(99)
	= 160 m/s ,	(100)

while

Z2	=	(101)
	= 0.90 kg/s .	(102)

Thus

(103)

and therefore

(104)

At f = 20 Hz,

λ1	= = = 4.5 m ,	(105)
λ2	= = = 8.0 m .	(106)

The interface is matched even though the wavelength changes.

Solution 9: Fixed-like and free-like limits

If

(107)

then

(108)

This is the fixed-like limit: the reflected displacement is inverted.

If

(109)

then

(110)

This is the free-like limit: the reflected displacement is not inverted.

Solution 10: Derive the amplitude coefficients

Start with

(111)

Substitute this into the force condition:

(112)

Expand:

(113)

Collect the incident terms on one side and reflected terms on the other:

(114)

Hence

(115)

Since

(116)

divide by A_i:

(117)

Substitute the expression for r:

t	= 1 +	(118)
	=	(119)
	= .	(120)

Thus

(121)

is simply displacement continuity written in coefficient form.

Solution 11: Transmission amplitude greater than one

With

(122)

we obtain

r	=	(123)
	= 0.60 .	(124)

The displacement transmission coefficient is

t	=	(125)
	= 1.60 .	(126)

The power fractions are

(127)

and

𝒯	= (1.60)2	(128)
	= 0.25(2.56)	(129)
	= 0.64 .	(130)

Thus

(131)

The fact that t = 1.60 > 1 does not violate energy conservation because t is a displacement-amplitude ratio. Power contains the impedance factor as well as the square of the amplitude.

Solution 12: Infer an unknown impedance from a measured reflection

Start with

(132)

Solve for Z₂:

(133)

so

(134)

Therefore

(135)

With r = −0.25 and Z₁ = 1.20 kg/s,

Z2	= (1.20)	(136)
	= 2.00 kg/s .	(137)

The reflected power fraction is

(138)

For a lossless junction,

(139)

Finally,

(140)

Solution 13: Same speed, different impedance

For string 1,

(141)

and

(142)

We want

(143)

and

(144)

Using

(145)

we find

(146)

Using

(147)

we obtain

(148)

The reflection coefficient is

r	=	(149)
	= − .	(150)

Thus equal wave speed does not imply zero reflection. The interface reflects because the characteristic impedances are different.

Solution 14: Full impedance-interface synthesis

For string 1,

c1	=	(151)
	= 80 m/s ,	(152)

and

(153)

For string 2,

c2	=	(154)
	= 120 m/s ,	(155)

and

(156)

At f = 30 Hz,

λ1	= = 2.67 m ,	(157)
λ2	= = 4.00 m .	(158)

The amplitude coefficients are

r	=	(159)
	= −0.20 ,	(160)

and

t	=	(161)
	= 0.80 .	(162)

Therefore

Ar	= rAi = (−0.20)(1.5 mm) = −0.30 mm ,	(163)
At	= tAi = (0.80)(1.5 mm) = 1.20 mm .	(164)

The reflected wave is inverted because A_r is negative.

The power fractions are

(165)

and

(166)

The angular frequency is

(167)

The incident average power is

⟨Pi⟩	= Z1Ai2ω2	(168)
	= (0.80)(0.0015)2(60π)2	(169)
	≈ 3.20 × 10−2 W .	(170)

More precisely,

(171)

The reflected power magnitude is

(172)

The transmitted power is

(173)

Finally,

|⟨Pr⟩| + ⟨Pt⟩	≈ 0.00128 + 0.03070	(174)
	≈ 0.03198 W	(175)
	= ⟨Pi⟩.	(176)

Thus the numerical calculation confirms power conservation.

Common mistakes

• Mistake: confusing wave speed c with characteristic impedance Z₀. Speed depends on T∕μ, while impedance depends on Tμ.
• Mistake: using displacement instead of transverse velocity in the force-velocity impedance relation.
• Mistake: dropping the propagation-direction sign in F_⊥^(+x) = ±Z ₀u_t.
• Mistake: treating t as the transmitted power fraction. The power fraction is 𝒯 = (Z₂∕Z₁)t².
• Mistake: assuming t > 1 violates conservation of energy.
• Mistake: assuming equal wave speeds imply matched impedances.
• Mistake: forgetting that a negative displacement reflection coefficient means phase inversion of the reflected displacement.

What WM22E1 reinforces

The exercises connect three parts of the wave-mechanics sequence:

(177)

connects medium parameters to the force-velocity relation,

(178)

connects impedance to directional energy transport, and

(179)

connects impedance mismatch to reflection.

The distinction between amplitude coefficients and power fractions is especially important:

(180)

These ideas transfer directly to later treatments of acoustic impedance, electromagnetic wave impedance, and transmission-line characteristic impedance.

References