Wave Mechanics Examples: Why Amplitude Is Not Energy

This companion article provides exercises for WM23, Why Amplitude Is Not energy. All exercises appear first; complete worked solutions follow in Part II.

The central mechanical energy-density relation for an ideal stretched string is

(1)

For a right-moving sinusoidal wave

(2)

the average energy density and average power are

(3)

and

(4)

Thus energy and power scale as A² only when the other relevant parameters are fixed. Amplitude itself is a displacement scale, not an energy variable [1, 2, 4, 6].

How to use this problem set

Attempt all exercises in Part I before consulting Part II. In each problem, identify which quantities are being held fixed before using a proportionality such as

(5)

The phrase “proportional to amplitude squared” is meaningful only after those hidden conditions have been specified.

Part I: Exercises

Exercise 1: Amplitude, energy density, and power are different quantities

For a transverse string wave, classify each quantity below by its physical meaning and SI units:

1. displacement amplitude A,
2. local energy density ℰ,
3. total wave energy E,
4. average power ⟨P⟩.

Then explain why the statement “A² is the energy” cannot be literally correct on dimensional grounds.

Exercise 2: The conditional square law

Two sinusoidal traveling waves propagate on the same string at the same frequency. Their amplitudes satisfy

(6)

Find

(7)

and

(8)

State explicitly which quantities had to remain fixed for the A² scaling to apply.

Exercise 3: Equal amplitude, unequal frequency

Two waves on the same ideal string have the same amplitude A but frequencies

(9)

The figure emphasizes that identical displacement amplitude does not imply identical energy transport.

Find

(10)

Would the two waves also have equal average energy density? Explain.

Exercise 4: Equal amplitude and frequency, unequal medium

Two right-moving sinusoidal waves have identical amplitude and frequency. Their characteristic impedances are

(11)

Using

(12)

find

(13)

Why does this problem provide another counterexample to the idea that amplitude alone determines power?

Exercise 5: Displacement phase versus energy phase

For a right-moving sinusoidal wave, define

(14)

Then

(15)

and

(16)

Use the figure below to answer the questions.

For each phase

(17)

state the values of u∕A and ℰ∕ℰ_max. Identify the phases where displacement magnitude is maximum but energy density is zero, and where displacement is zero but energy density is maximum.

Exercise 6: Energy from motion and deformation

At one point on an ideal string,

(18)

and measurements give

(19)

Find:

1. the kinetic energy density 𝒦,
2. the elastic potential energy density 𝒰,
3. the total local energy density ℰ.

Does the displacement u need to be known to answer this problem?

Exercise 7: Sinusoidal average energy and power

A right-moving sinusoidal wave has

(20)

(21)

Find:

1. ω,
2. ⟨ℰ⟩,
3. ⟨P⟩,
4. the maximum instantaneous local energy density ℰ_max.

Compare ℰ_max with ⟨ℰ⟩.

Exercise 8: A standing-wave node can store energy

Consider the Standing Wave

(22)

with

(23)

At a node,

(24)

1. What is the displacement at that point for all time?
2. Show that the maximum elastic energy density at the node is

   (25)

3. Evaluate 𝒰_max numerically.

Exercise 9: Destructive displacement interference does not erase energy

A right-moving component F(x−ct) and a left-moving component G(x + ct) overlap. At one event their displacements cancel exactly,

(26)

but their profile derivatives are

(27)

The string Tension is

(28)

Using

(29)

find the local energy density. Explain why the answer is consistent with zero total displacement at that event.

Exercise 10: Infer amplitude from a target power

A right-moving sinusoidal wave has

(30)

What amplitude is required for

(31)

Use

(32)

Give the answer in meters and millimeters.

Exercise 11: Diagnose conceptual statements

For each statement, decide whether it is correct. If it is incorrect, rewrite it accurately.

1. “If the displacement is zero at a point, the local wave energy must be zero.”
2. “If two waves have equal amplitude, they carry equal power.”
3. “On the same string at the same frequency, doubling amplitude quadruples average power.”
4. “A standing-wave node has zero displacement amplitude, so it can never contain elastic energy.”
5. “Destructive interference of displacement means the component-wave energies have been destroyed.”

Exercise 12: Trade amplitude against frequency at fixed power

Two sinusoidal traveling waves propagate on the same string and must carry the same average power. The second wave has twice the frequency of the first:

(33)

Find the required amplitude ratio

(34)

Explain why this result would be impossible to obtain from amplitude information alone without knowing that the powers are equal and the medium is unchanged.

Exercise 13: Compare two waves when everything changes

Two right-moving sinusoidal waves have the following properties:

μ1	= 0.010 kg/m,	c1	= 80 m/s,	A1	= 2.0 mm,	f1	= 30 Hz,	(35)
μ2	= 0.015 kg/m,	c2	= 100 m/s,	A2	= 1.0 mm,	f2	= 60 Hz.	(36)

Without separately calculating both powers, use a ratio to find

(37)

Which wave carries more average power?

Exercise 14: Synthesis - amplitude is one input, not the whole answer

A sinusoidal traveling wave propagates on a string with

(38)

and has

(39)

Find:

1. the wave speed c,
2. the angular frequency ω,
3. the average energy density ⟨ℰ⟩,
4. the average power ⟨P⟩,
5. the new average power if only the amplitude is doubled,
6. the new average power if only the frequency is doubled,
7. the new average power if both amplitude and frequency are doubled.

Use this example to summarize why amplitude alone cannot determine wave power.

Part II: Complete Worked Solutions

Solution 1: Amplitude, energy density, and power are different quantities

1. The amplitude A is a displacement scale. Its SI unit is

   (40)

2. The local energy density ℰ for a string is energy per unit length:

   (41)

3. Total wave energy has units

   (42)

4. Average power is energy transported per unit time:

   (43)

The quantity A² has units m², which are neither joules, joules per meter, nor watts. Therefore A² cannot literally be energy. It can only appear as one factor inside an energy or power formula that also contains the necessary physical parameters.

Solution 2: The conditional square law

For the same medium and the same frequency,

(44)

and

(45)

Because

(46)

we have

(47)

Therefore

(48)

and

(49)

The frequency ω and the relevant medium quantities, such as μ and c or equivalently Z₀, had to remain fixed.

Solution 3: Equal amplitude, unequal frequency

On the same string at fixed amplitude,

(50)

Thus

	= 2	(51)
	= 2	(52)
	= 9.	(53)

Hence

(54)

The average energy density also scales as f² for fixed A and μ:

(55)

Therefore the second wave also has nine times the average energy density. Equal amplitude is not enough to make the energy measures equal.

Solution 4: Equal amplitude and frequency, unequal medium

With A and ω fixed,

(56)

Therefore

	=	(57)
	=	(58)
	= 2.50.	(59)

Thus

(60)

The displacement amplitude and frequency are identical, but the media present different mechanical impedances. The power therefore differs.

Solution 5: Displacement phase versus energy phase

We use

(61)

At 𝜃 = 0,

(62)

At 𝜃 = π∕2,

(63)

At 𝜃 = π,

(64)

At 𝜃 = 3π∕2,

(65)

Therefore maximum displacement magnitude occurs at 𝜃 = 0 and 𝜃 = π, where the local energy density is zero. Zero displacement occurs at 𝜃 = π∕2 and 3π∕2, where the local energy density is maximum.

Solution 6: Energy from motion and deformation

The kinetic term is

𝒦	= μut2	(66)
	= (0.012)(0.40)2	(67)
	= 9.60 × 10−4 J/m.	(68)

Thus

(69)

The elastic term is

𝒰	= Tux2	(70)
	= (120)(0.025)2	(71)
	= 3.75 × 10−2 J/m.	(72)

Therefore

(73)

The total is

ℰ	= 𝒦 + 𝒰	(74)
	= 0.000960 + 0.0375	(75)
	= 0.03846 J/m.	(76)

Hence

(77)

The displacement u itself was not needed. The local energy depends on the local velocity u_t and slope u_x.

Solution 7: Sinusoidal average energy and power

The angular frequency is

ω	= 2πf	(78)
	= 2π(50)	(79)
	= 100π rad/s	(80)
	≈ 314.16 rad/s.	(81)

Thus

(82)

The average energy density is

⟨ℰ⟩	= μA2ω2	(83)
	= (0.010)(0.0020)2(314.16)2	(84)
	≈ 1.974 × 10−3 J/m.	(85)

Therefore

(86)

The average power is

⟨P⟩	= c⟨ℰ⟩	(87)
	= 100(1.974 × 10−3)	(88)
	≈ 0.197 W.	(89)

Thus

(90)

The instantaneous energy density is proportional to sin ²𝜃, so its maximum is twice its cycle average:

(91)

Hence

(92)

Solution 8: A standing-wave node can store energy

At a node,

(93)

so

(94)

for all time.

The spatial derivative is

(95)

At a node, | sin(kx)| = 1, so

(96)

Therefore

𝒰	= Tux2	(97)
	= 2TA2k2 cos 2(ωt).	(98)

The maximum is

(99)

Numerically,

𝒰max	= 2(80)(0.0015)2(8.0)2	(100)
	= 0.02304 J/m.	(101)

Thus

(102)

The node never moves away from zero displacement, yet the local string can be strongly tilted and therefore store elastic energy.

Solution 9: Destructive displacement interference does not erase energy

Use

(103)

Then

ℰ	= 100(0.040)2 + 100(0.030)2	(104)
	= 100(0.0016 + 0.0009)	(105)
	= 0.25 J/m.	(106)

Therefore

(107)

The condition F + G = 0 concerns the values of the two displacement fields at that event. Energy depends on their derivatives and on the corresponding motion and deformation. Displacement cancellation therefore does not require the energy density to vanish.

Solution 10: Infer amplitude from a target power

Starting from

(108)

solve for A:

(109)

Substituting the data,

A	=	(110)
	≈ 5.93 × 10−3 m.	(111)

Therefore

(112)

The amplitude can be inferred only because the frequency and medium properties are also known.

Solution 11: Diagnose conceptual statements

1. Incorrect. Zero displacement does not imply zero local energy. For a sinusoidal traveling wave, zero displacement occurs when the local energy density is maximum.
2. Incorrect. Equal amplitude does not imply equal power unless the relevant frequency and medium quantities are also equal.
3. Correct. On the same string at the same frequency,

   (113)

   so doubling amplitude multiplies average power by four.

4. Incorrect. A standing-wave node has zero displacement but can have nonzero and even maximum elastic energy density because u_x need not vanish there.
5. Incorrect. Destructive interference cancels the displacement field locally; it does not destroy the energy carried or stored by the component waves in a lossless linear system.

Solution 12: Trade amplitude against frequency at fixed power

On the same string,

(114)

Equal average power therefore requires

(115)

With

(116)

we obtain

(117)

Thus

(118)

and

(119)

The higher-frequency wave needs only half the amplitude to carry the same average power on the same string.

Solution 13: Compare two waves when everything changes

For sinusoidal traveling waves,

(120)

Therefore

	=	(121)
	=	(122)
	= 1.875.	(123)

Hence

(124)

Wave 2 carries more average power even though its amplitude is only half as large. Its higher frequency and different medium more than compensate for the smaller amplitude.

Solution 14: Synthesis - amplitude is one input, not the whole answer

The wave speed is

c	=	(125)
	=	(126)
	=	(127)
	≈ 94.87 m/s.	(128)

Thus

(129)

The angular frequency is

ω	= 2πf	(130)
	= 2π(25)	(131)
	= 50π rad/s	(132)
	≈ 157.08 rad/s.	(133)

So

(134)

The average energy density is

⟨ℰ⟩	= μA2ω2	(135)
	= (0.010)(0.0012)2(157.08)2	(136)
	≈ 1.78 × 10−4 J/m.	(137)

Therefore

(138)

The average power is

⟨P⟩	= c⟨ℰ⟩	(139)
	≈ (94.87)(1.7765 × 10−4)	(140)
	≈ 1.685 × 10−2 W.	(141)

Thus

(142)

If only amplitude doubles, A² increases by four:

(143)

If only frequency doubles, f² also increases by four:

(144)

If both amplitude and frequency double, the two factors multiply:

(145)

Hence

(146)

This synthesis problem shows the central WM23 lesson: amplitude affects power strongly, but it is only one input among several. Frequency and medium properties matter independently.

Common mistakes

• Mistake: saying “energy equals amplitude squared.” The dimensions and the physics are incomplete.
• Mistake: using ⟨P⟩∝ A² while also changing frequency or medium properties.
• Mistake: treating zero displacement as zero local energy.
• Mistake: assuming a standing-wave node contains no energy because its displacement is always zero.
• Mistake: assuming displacement cancellation destroys conserved energy.
• Mistake: forgetting that the energy density of a string depends on u_t and u_x, not directly on u alone.

What WM23E1 reinforces

The central results are

(147)

and, for a sinusoidal traveling wave,

(148)

(149)

These formulas show exactly why the phrase “energy is proportional to amplitude squared” is useful but conditional. The correct broader lesson is

(150)

References