Wave Mechanics Examples: Power Carried by a 1D Wave

This companion article provides exercises for WM19, wave mechanics: power Carried by a 1D Wave. All exercises are stated first so they can be attempted without seeing the answers. Complete worked solutions follow in Part II.

The central WM19 result is the instantaneous signed power crossing a fixed position on an ideal stretched string,

(1)

Positive power denotes net mechanical-energy transport toward increasing x, while negative power denotes net transport toward decreasing x.

For a pure right-moving wave,

(2)

and for a pure left-moving wave,

(3)

For a right-moving sinusoidal wave,

(4)

the instantaneous and average powers are

(5)

and

(6)

For a general two-direction solution,

(7)

WM19 found

(8)

These are the standard ideal-string power relations developed in WM19 and in standard wave-mechanics treatments [1, 2, 3, 5].

How to use this problem set

Attempt all exercises in Part I before consulting Part II. Throughout the set, keep three distinctions explicit:

• power is an energy-transfer rate, not an energy density;
• the sign of power indicates direction, not whether the string has “positive” or “negative” energy;
• a Standing Wave can have nonzero instantaneous local power even though its time-averaged power is zero.

Part I: Exercises

Exercise 1: Instantaneous power from local measurements

A string has Tension

(9)

At one event,

(10)

Find:

1. the instantaneous power;
2. the direction of energy flow;
3. the new power if u_t changes sign while u_x does not.

Exercise 2: Sign diagnostics

Without doing any arithmetic beyond signs, determine the sign of P and the corresponding direction of energy flow for each case:

1. u_x > 0, u_t > 0;
2. u_x > 0, u_t < 0;
3. u_x < 0, u_t > 0;
4. u_x < 0, u_t < 0;
5. either u_x = 0 or u_t = 0.

Use the sign diagram above to check your reasoning.

Exercise 3: Dimensional consistency

Show that

(11)

has units of watts. Explicitly identify the units of T, u_x, and u_t.

Exercise 4: Pure traveling wave and the relation P = ±cℰ

At one event, a pure traveling wave has energy density

(12)

and wave speed

(13)

Find the signed power if the wave is:

1. purely right-moving;
2. purely left-moving.

Explain why the magnitudes are equal.

Exercise 5: Instantaneous sinusoidal power

A right-moving sinusoidal wave is

(14)

1. Derive the instantaneous power formula from P = −Tu_xu_t.
2. At what phase values in one cycle is the instantaneous power zero?
3. At what phase values is the instantaneous power maximum?
4. Explain why maximum power occurs when the displacement itself passes through zero.

Exercise 6: Average power of a sinusoidal wave

A string has

(15)

A right-moving sinusoidal wave on the string has amplitude

(16)

and frequency

(17)

Find:

1. the wave speed;
2. the angular frequency;
3. the average power.

Exercise 7: Required amplitude for a target power

The same string as in Exercise 6 must carry an average power of

(18)

at frequency

(19)

What sinusoidal amplitude is required?

Exercise 8: Square-law scaling

For a fixed string, a sinusoidal traveling wave initially has average power P₀.

Find the new average power in terms of P₀ if:

1. amplitude is doubled while frequency is unchanged;
2. frequency is tripled while amplitude is unchanged;
3. amplitude is halved and frequency is doubled;
4. both amplitude and frequency are doubled.

Exercise 9: Energy balance on a finite interval

The power entering a fixed interval [a,b] at one instant is

(20)

while the power leaving at the right boundary is

(21)

1. Find the instantaneous rate of change of the total wave energy stored in [a,b].
2. Is the stored energy increasing or decreasing?
3. How would the answer change if the two powers were interchanged?

Exercise 10: Competing right- and left-moving waves

At one event in a two-direction field,

(22)

and the local profile derivatives are

(23)

Use

(24)

to find:

1. the right-moving power contribution;
2. the signed left-moving contribution;
3. the net power;
4. the net direction of energy flow.

Exercise 11: Equal counter-propagating components

Suppose at one event the two directional derivatives satisfy

(25)

1. What is the net instantaneous power according to the two-direction formula?
2. Does zero net power imply zero local energy density?
3. Explain how this relates to a perfect standing wave built from equal counter-propagating components.

Exercise 12: Instantaneous power in a standing wave

A standing wave is

(26)

1. Derive

   (27)

2. Show that the time average at any fixed x is zero.
3. Name two different reasons the instantaneous power can vanish at a particular event.
4. Explain why “zero average transport” does not mean “no local energy motion.”

Exercise 13: Power carried by a localized pulse

At one point in a purely right-moving pulse, the local energy density is

(28)

and the propagation speed is

(29)

1. Find the instantaneous power crossing that point.
2. If the same local energy density belonged to a purely left-moving pulse, what would the signed power be?
3. Why can this problem be solved without knowing the detailed pulse shape?

Exercise 14: Synthesis from string properties to transported power

A stretched string has

(30)

It carries a right-moving sinusoidal wave with

(31)

Find:

1. the wave speed c;
2. the angular frequency ω;
3. the wavelength λ;
4. the Wavenumber k;
5. the average energy density ⟨ℰ⟩;
6. the average power using ⟨P⟩ = c⟨ℰ⟩;
7. the average power again using ⟨P⟩ = TA²kω, and verify agreement.

Part II: Complete Worked Solutions

Solution 1: Instantaneous power from local measurements

Use

(32)

With

(33)

we obtain

P	= −(80)(−0.030)(0.50)	(34)
	= 1.20 W .	(35)

Because P > 0, energy flows toward increasing x.

If u_t changes sign while the slope does not, then

(36)

and

P	= −(80)(−0.030)(−0.50)	(37)
	= −1.20 W .	(38)

The magnitude is unchanged, but the energy-flow direction reverses.

Solution 2: Sign diagnostics

Since T > 0, the sign of P is the opposite of the sign of u_xu_t.

1. u_x > 0 and u_t > 0 give u_xu_t > 0, so

   (39)

   meaning energy flows toward −x.

2. u_x > 0 and u_t < 0 give u_xu_t < 0, so

   (40)

   meaning energy flows toward +x.

3. u_x < 0 and u_t > 0 also give u_xu_t < 0, so

   (41)

4. u_x < 0 and u_t < 0 give u_xu_t > 0, so

   (42)

5. If either factor vanishes, then

   (43)

Solution 3: Dimensional consistency

The units are

(44)

The slope u_x is dimensionless because it is displacement per horizontal distance, meter per meter.

Therefore

[P]	= [T][ux][ut]	(45)
	= N	(46)
	=	(47)
	=	(48)
	= W .	(49)

Solution 4: Pure traveling wave and the relation P = ±cℰ

For a pure right-moving wave,

(50)

Hence

P→	= (35)(0.24)	(51)
	= 8.4 W .	(52)

For a pure left-moving wave,

(53)

so

(54)

The magnitudes are equal because the energy density and propagation-speed magnitude are the same; only the direction of transport differs.

Solution 5: Instantaneous sinusoidal power

Let

(55)

Then

(56)

(57)

and

(58)

Therefore

P	= −Tuxut	(59)
	= −T(−Ak sin 𝜃)(Aω sin 𝜃)	(60)
	= TA2kω sin 2𝜃 .	(61)

The power is zero when

(62)

so in one cycle

(63)

It is maximum when

(64)

which occurs at

(65)

At these phases the displacement is zero because cos 𝜃 = 0, but both the local slope magnitude and transverse-speed magnitude are maximal. Since power depends on their product rather than on displacement itself, the power is largest there.

Solution 6: Average power of a sinusoidal wave

First find the wave speed:

c	=	(66)
	=	(67)
	=	(68)
	≈ 94.9 m/s .	(69)

The angular frequency is

ω	= 2πf	(70)
	= 2π(25)	(71)
	= 50π rad/s	(72)
	≈ 157.1 rad/s.	(73)

Convert the amplitude:

(74)

Now use

(75)

Thus

⟨P⟩	= (0.010)(0.0040)2(157.1)2(94.9)	(76)
	≈ 0.187 W .	(77)

Solution 7: Required amplitude for a target power

Solve

(78)

for A:

(79)

Using

(80)

we obtain

A	≈	(81)
	≈ 0.0131 m.	(82)

Therefore

(83)

Solution 8: Square-law scaling

For a fixed string,

(84)

Therefore:

1. Doubling A multiplies power by 2² = 4:

   (85)

2. Tripling f multiplies power by 3² = 9:

   (86)

3. Halving A gives a factor 1∕4, while doubling f gives a factor 4. The factors cancel:

   (87)

4. Doubling both gives

   (88)

   so

   (89)

Solution 9: Energy balance on a finite interval

Use

(90)

Therefore

	= 3.2 − 1.1	(91)
	= 2.1 W .	(92)

The stored energy is increasing because more power enters than leaves.

If the powers are interchanged,

(93)

so the same magnitude of energy is being lost from the interval.

Solution 10: Competing right- and left-moving waves

The right-moving contribution is

P→	= Tc(F′)2	(94)
	= (50)(20)(0.040)2	(95)
	= 1.60 W .	(96)

The signed left-moving contribution is

P←	= −Tc(G′)2	(97)
	= −(50)(20)(0.030)2	(98)
	= −0.90 W .	(99)

Hence the net power is

P	= 1.60 − 0.90	(100)
	= 0.70 W .	(101)

Because the net power is positive, the net energy transport is toward increasing x.

Solution 11: Equal counter-propagating components

If

(102)

then

(103)

Therefore

(104)

This does not imply that the local energy density is zero. Each component can carry nonzero energy, but their signed directional power contributions cancel.

A perfect standing wave is produced by equal-amplitude counter-propagating components, but the two local derivative magnitudes need not be equal at every event because the two components are sampled at different phases. Thus a standing wave can have nonzero instantaneous local power. The equality considered here is a special event at which the opposing directional contributions cancel exactly; over a full cycle, a perfect standing wave has zero time-averaged transport.

Solution 12: Instantaneous power in a standing wave

For

(105)

we have

(106)

and

(107)

Thus

P	= −Tuxut	(108)
	= TB2kω cos(kx) sin(kx) cos(ωt) sin(ωt).	(109)

Use

(110)

twice to obtain

(111)

At fixed x, the time average of sin(2ωt) over a full cycle is zero, so

(112)

Instantaneous power can vanish because the spatial factor is zero,

(113)

or because the temporal factor is zero,

(114)

Zero average transport means that there is no net energy carried permanently in one direction over a cycle. It does not mean that energy is locally stationary; energy moves back and forth within the standing-wave pattern.

Solution 13: Power carried by a localized pulse

For a pure right-moving wave,

(115)

Hence

P	= (48)(0.075)	(116)
	= 3.60 W .	(117)

For a pure left-moving wave with the same local energy density,

(118)

The detailed pulse shape is unnecessary because the pure-traveling-wave identity P = ±cℰ already contains the local energy-flow rate.

Solution 14: Synthesis from string properties to transported power

The string properties are

(119)

The wave amplitude and frequency are

(120)

(a) Wave speed

c	=	(121)
	=	(122)
	=	(123)
	≈ 89.4 m/s .	(124)

(b) Angular frequency

ω	= 2πf	(125)
	= 80π rad/s	(126)
	≈ 251.3 rad/s .	(127)

(c) Wavelength

λ	=	(128)
	=	(129)
	≈ 2.24 m .	(130)

(d) Wavenumber

k	=	(131)
	≈	(132)
	≈ 2.81 rad/m .	(133)

(e) Average energy density

⟨ℰ⟩	= μA2ω2	(134)
	= (0.015)(0.0030)2(251.3)2	(135)
	≈ 4.26 × 10−3 J/m .	(136)

(f) Average power from c⟨ℰ⟩

⟨P⟩	= c⟨ℰ⟩	(137)
	= (89.4)(4.26 × 10−3)	(138)
	≈ 0.381 W .	(139)

(g) Check using TA²kω

⟨P⟩	= TA2kω	(140)
	= (120)(0.0030)2(2.81)(251.3)	(141)
	≈ 0.381 W .	(142)

The two independent forms agree, as they must.

Common mistakes

• Mistake: dropping the minus sign in P = −Tu_xu_t. The sign is what records the direction of transport.
• Mistake: treating power as energy density. Energy density has units of joules per meter; power has units of watts.
• Mistake: assuming P = 0 whenever u = 0. For a sinusoidal traveling wave, power is actually maximum when the displacement crosses zero.
• Mistake: forgetting to convert millimeters to meters before using the average-power formula.
• Mistake: adding right- and left-moving power magnitudes without signs. They contribute with opposite signs to the net power.
• Mistake: saying that a standing wave has zero instantaneous power everywhere. Its time-averaged transport is zero, but instantaneous local power can be nonzero.

What WM19E1 reinforces

The main result is

(143)

For a pure traveling wave,

(144)

where the sign gives the propagation direction. For a sinusoidal traveling wave,

(145)

For a fixed interval,

(146)

and for equal counter-propagating components the signed directional power can cancel even though substantial wave energy remains in the system.

These ideas prepare the next stage of the course, where average power, intensity, flux, and impedance are developed more systematically.

References

References