Wave Mechanics: Power Carried by a 1D Wave

WM18 developed the mechanical energy density of an ideal stretched string,

1 1 ℰ(x, t) = --μu2t + --Tu2x. 2 2

(1)

Energy density answers the question, “How much mechanical energy is stored near this position?” The next question is different:

How rapidly does wave energy cross a fixed position?

That rate of energy transfer is the wave power. With positive power defined as energy flow toward increasing x, the ideal-string result is

|------------------| |P(x, t) = − T uxut.| --------------------

(2)

The sign matters. A positive value means net mechanical energy crosses the chosen position toward +x; a negative value means the net flow is toward −x. This local expression is the one-dimensional energy-flux law associated with the string wave equation and is consistent with standard wave-energy treatments [1, 2, 3, 4].

WM19 derives the result mechanically and from local energy conservation, applies it to arbitrary traveling profiles and sinusoidal waves, derives the time-averaged sinusoidal power, and explains what happens when right- and left-moving waves are both present.

1 Power is energy crossing a position per unit time

Choose a fixed position x. During a short time interval dt, mechanical energy can cross that position even though the material of the string merely oscillates transversely.

We define P(x,t) so that

P > 0

(3)

means energy transport toward increasing x, and

P < 0

(4)

means energy transport toward decreasing x.

The SI unit of power is

J- W = s.

(5)

This is not yet intensity. Intensity is power per unit area and will be introduced later when waves spread through higher-dimensional space [3].

2 Mechanical derivation at a cut in the string

Imagine cutting the mathematical description of the string at one fixed position. The material just to the left exerts a Tension force on the material just to the right.

For a small-slope string, the transverse component of the force exerted by the left side on the right side is approximately

Fu = − T ux.

(6)

The transverse velocity of the material at the cut is

vu = ut.

(7)

Mechanical power delivered from the left portion into the right portion is force times velocity:

P	= F_uv_u	(8)
	= (−Tu_x)u_t.	(9)

Therefore

|--------------| |P = − Tuxut. | ---------------

(10)

The geometry and sign convention are summarized below.

Figure. At a fixed cut, the local slope determines the transverse component of tension and u_t gives the local material velocity. Their product determines the mechanical work rate across the cut. Positive P is defined as energy transport toward +x.

3 Dimensional check

The tension has units of newtons,

[T ] = N,

(11)

while u_x is dimensionless and

[ut] = m/s.

(12)

Hence

[Tu_xu_t]	= N	(13)
	=	(14)
	= W.	(15)

So the expression has the correct dimensions for power.

4 The same result from local energy conservation

The energy-density formula from WM18 is

1 1 ℰ = -μu2t + -T u2x. 2 2

(16)

Differentiate with respect to time:

ℰt = μututt + Tuxuxt.

(17)

For the ideal string,

μutt = T uxx.

(18)

Substitute this into the first term:

ℰ_t	= Tu_tu_xx + Tu_xu_xt	(19)
	= T(u_tu_x).	(20)

Therefore

∂ ℰt + --(− T utux) = 0. ∂x

(21)

Identifying

|------------| P = − T uxut| --------------

(22)

gives the local conservation law

|------------| |ℰt + Px = 0.| --------------

(23)

This compact equation says that a local change in stored energy is caused by an imbalance of power flow into and out of the region.

5 Energy balance on a finite interval

Integrate the local conservation law from x = a to x = b:

∫ ∫ b b ℰtdx + Px dx = 0. a a

(24)

Thus

∫ b d- ℰ dx + P (b,t) − P (a,t) = 0. dt a

(25)

Equivalently,

|------------------------------| |d ∫ b | |-- ℰ dx = P (a,t) − P(b,t).| -dt--a-------------------------

(26)

Power entering at the left boundary increases the energy stored in the interval; power leaving at the right boundary decreases it.

Figure. Energy conservation on a fixed interval: rate of change of stored energy equals incoming power minus outgoing power.

6 Power in a pure right-moving wave

Consider

u(x,t) = F (x − ct).

(27)

With

ξ = x − ct,

(28)

we have

u = F′(ξ) x

(29)

and

′ ut = − cF (ξ).

(30)

Therefore

P	= −Tu_xu_t	(31)
	= −TF′(ξ)[−cF′(ξ)]	(32)
	= Tc[F′(ξ)]².	(33)

Hence

|-----------′--------2-----| -P→--=-T-c[F-(x-−-ct)]-≥--0.|

(34)

The power is nonnegative because this wave carries energy toward +x.

WM18 showed that the energy density of the same wave is

′ 2 ℰ = T[F (x − ct)].

(35)

Therefore

|----------| -P→--=-cℰ-.|

(36)

This is physically intuitive: a packet of energy density ℰ translating at speed c carries energy past a fixed point at rate cℰ.

7 Power in a pure left-moving wave

For

u(x, t) = G (x + ct),

(37)

we have

ux = G ′

(38)

and

u = cG ′. t

(39)

Thus

|----------------------------| | ′ 2 | -P←--=-−-T-c[G-(x-+-ct)]-≤--0.

(40)

Since

′ 2 ℰ = T[G (x + ct)],

(41)

we can write

|------------| -P-←-=-−-cℰ.-|

(42)

The sign is a direction marker: the energy moves toward decreasing x.

8 Sinusoidal traveling wave

Consider the right-moving sinusoidal wave

u(x,t) = A cos(kx − ωt + ϕ).

(43)

Define

𝜃 = kx − ωt + ϕ.

(44)

Then

ux = − Ak sin 𝜃

(45)

and

ut = A ωsin 𝜃.

(46)

The instantaneous power is therefore

P	= −T(−Ak sin 𝜃)(Aω sin 𝜃)	(47)
	= TA²kω sin ²𝜃.	(48)

Hence

|----------------------------------| P (x,t) = T A2kω sin2(kx − ωt + ϕ).| ------------------------------------

(49)

It is always nonnegative for this right-moving wave.

Figure. For a right-moving sinusoidal wave, normalized instantaneous power varies as sin ²𝜃. Power is largest where the displacement passes through zero and vanishes at displacement extrema.

9 Average power of a sinusoidal wave

Over one phase cycle,

⟨sin2𝜃 ⟩ = 1. 2

(50)

Therefore

|----------------| | 1 | |⟨P⟩ = -T A2kω. | --------2--------

(51)

Use

ω = ck

(52)

and

T = μc2.

(53)

Then

|----------------| | 1 2 2 | |⟨P ⟩ = --μA ω c.| -------2----------

(54)

This is the standard time-averaged power of a sinusoidal transverse wave on an ideal string [3, 1].

Because WM18 found

1- 2 2 ⟨ℰ ⟩ = 2μA ω ,

(55)

we again have

|------------| |⟨P ⟩ = c⟨ℰ ⟩. -------------

(56)

10 Square-law scaling

For a fixed string and a sinusoidal traveling wave,

⟨P ⟩ = 1-μA2 ω2c. 2

(57)

Thus

------------- | 2 2 | -⟨P⟩-∝-A--ω--|

(58)

when the string properties are fixed.

Consequently:

doubling A multiplies average power by 4;
doubling f or ω multiplies average power by 4;
doubling both amplitude and frequency multiplies average power by 16.

OpenStax emphasizes this same amplitude-squared and frequency-squared scaling for sinusoidal mechanical waves [3].

11 Both propagation directions at once

For the general two-direction solution

u(x,t) = F (x − ct) + G (x + ct),

(59)

let

ξ = x − ct, η = x + ct.

(60)

Then

′ ′ ux = F (ξ) + G (η)

(61)

and

′ ′ ut = − cF (ξ) + cG (η).

(62)

Substitute into the power law:

P	= −T[F′ + G′][−cF′ + cG′]	(63)
	= Tc.	(64)

Thus

|--------------------------------------| |P = Tc ([F ′(x − ct)]2 − [G ′(x + ct)]2) .| ---------------------------------------

(65)

The net power is the right-moving contribution minus the left-moving contribution.

Figure. Directional power adds with a sign. Right-moving energy contributes positive power and left-moving energy contributes negative power under the chosen convention.

12 Standing waves and zero average transport

A Standing Wave can be written

u (x,t) = B sin(kx )cos(ωt).

(66)

Then

u = Bk cos(kx)cos(ωt) x

(67)

and

ut = − B ωsin(kx )sin (ωt ).

(68)

Therefore

P	= −Tu_xu_t	(69)
	= TB²kω cos(kx) sin(kx) cos(ωt) sin(ωt).	(70)

Using double-angle identities,

|------------------------------------| | 1- 2 | |P (x, t) = 4T B k ωsin(2kx )sin (2ωt).| -------------------------------------

(71)

The instantaneous local power is generally not zero. Its sign reverses as energy moves back and forth within the standing-wave pattern.

However, over a complete time cycle,

⟨sin(2ωt)⟩ = 0,

(72)

|--------------| ⟨P ⟩standing = 0.| ----------------

(73)

A perfect standing wave therefore has no net time-averaged transport through a fixed position, even though energy can flow locally and instantaneously.

13 Worked Example 1: Power from local slope and velocity

A string has tension

T = 100 N.

(74)

At one event,

ux = 0.020, ut = − 0.30 m/s.

(75)

Find the instantaneous power and interpret its sign.

Solution

Use

P = − Tuxut.

(76)

Then

P	= −(100)(0.020)(−0.30)	(77)
	= 0.60 W.	(78)

Thus

|--------------| -P-=-+0.60--W.--

(79)

The positive sign means net energy is crossing the selected position toward increasing x at that instant.

14 Worked Example 2: Average power of a sinusoidal wave

A right-moving sinusoidal wave has

A = 5.0 mm, f = 10 Hz,

(80)

on a string with

T = 80 N, μ = 0.020 kg/m.

(81)

Find c, ω, k, and the time-averaged power.

Solution

The wave speed is

c	=	(82)
	=	(83)
	≃ 63.25 m/s.	(84)

The angular frequency is

ω = 2πf = 62.83 rad/s.

(85)

Then

k	=	(86)
	≃ 0.993 rad/m.	(87)

Convert the amplitude:

A = 0.0050 m.

(88)

Now use

1- 2 2 ⟨P ⟩ = 2 μA ω c.

(89)

Thus

⟨P⟩	= (0.020)(0.0050)²(62.83)²(63.25)	(90)
	≃ 6.24 × 10⁻² W.	(91)

Therefore

|----------------| -⟨P-⟩-≃-0.0624-W.--

(92)

15 Worked Example 3: Required amplitude for a desired average power

A sinusoidal traveling wave moves on a string with

μ = 0.010 kg/m, c = 100 m/s.

(93)

At frequency

f = 50Hz,

(94)

what amplitude is required to carry an average power of

⟨P ⟩ = 10 W?

(95)

Solution

Start with

1 ⟨P ⟩ = --μA2 ω2c. 2

(96)

Solve for A:

∘ ------ A = 2-⟨P-⟩. μ ω2c

(97)

The angular frequency is

ω = 2π(50) = 314.16 rad/s.

(98)

Therefore

A	=	(99)
	≃ 1.42 × 10⁻² m.	(100)

Thus

|--------------| -A-≃--14.2-mm.--|

(101)

16 Worked Example 4: Power carried by a translating pulse from its energy density

At one location in a pure right-moving pulse, the instantaneous energy density is

ℰ = 0.12 J/m.

(102)

The pulse speed is

c = 50 m/s.

(103)

Find the instantaneous power at that event.

Solution

For a pure right-moving wave,

P = cℰ .

(104)

Therefore

P	= (50)(0.12)	(105)
	= 6.0 W.	(106)

Hence

|-------------| P = +6.0 W. | --------------

(107)

If the same energy-density profile were traveling toward −x, the power would instead be −6.0 W.

17 Worked Example 5: Net power with two propagation directions

At one event in a two-direction wave field, suppose

T = 50 N, c = 40 m/s,

(108)

and the directional profile derivatives have values

F ′ = 0.060, G ′ = 0.040.

(109)

Find the net power.

Solution

Use

( ′2 ′2) P = T c [F ] − [G ] .

(110)

Then

P	= (50)(40)	(111)
	= 2000(0.0036 − 0.0016)	(112)
	= 4.0 W.	(113)

Thus

|-------------| P = +4.0 W. | --------------

(114)

The right-moving contribution is larger than the left-moving contribution, so the net energy transport is toward +x.

18 Worked Example 6: Instantaneous power in a standing wave

Consider

u(x,t) = B sin(kx)cos(ωt )

(115)

with

T = 60N, B = 0.010 m, k = πrad/m, c = 30 m/s.

(116)

Find the power at

x = 0.25 m

(117)

when

π- ωt = 8 .

(118)

Also state the time-averaged power at that position.

Solution

Because

ω = ck,

(119)

we have

ω = 30π rad/s.

(120)

The standing-wave power is

1- 2 P = 4 TB kω sin (2kx)sin(2ωt ).

(121)

At x = 0.25 m,

π 2kx = 2π (0.25) = 2,

(122)

sin(2kx) = 1.

(123)

Also

π 2ωt = 4,

(124)

√ -- --2- sin(2ωt) = 2 .

(125)

Therefore

P	= (60)(0.010)²(π)(30π)	(126)
	≃ 0.314 W.	(127)

Thus

|--------------| P--≃-+0.314-W---

(128)

at this instant.

Over a complete oscillation cycle,

|--------| ⟨P ⟩ = 0.| ----------

(129)

The local energy flow reverses later in the cycle, so there is no net time-averaged transport through the point.

19 Common mistakes

Mistake: dropping the minus sign in P = −Tu_xu_t. The sign carries propagation-direction information.
Mistake: assuming positive displacement means positive power. Power depends on slope and velocity, not displacement alone.
Mistake: treating a standing wave as having zero instantaneous power everywhere. Its time-averaged net transport is zero, but instantaneous local power can be nonzero.
Mistake: confusing power with energy density. Their units differ: watts versus joules per meter.
Mistake: confusing one-dimensional power with intensity. Intensity requires division by an area.
Mistake: applying P = cℰ to an arbitrary superposition. That relation holds directly for a single pure right-moving wave; a left-moving wave gives P = −cℰ, and a two-direction field requires the signed difference of the directional contributions.

20 Summary

For an ideal stretched string, the instantaneous mechanical power crossing a fixed position is

|--------------| |P = − Tuxut. | ---------------

(130)

Together with

1- 2 1- 2 ℰ = 2μu t + 2T ux,

(131)

it satisfies the local conservation law

|------------| |ℰ + P = 0.| --t----x------

(132)

For pure traveling waves,

|---------------------------| P-→-=-+c-ℰ-,----P←--=-−-cℰ.-|

(133)

For a right-moving sinusoidal wave,

------------------------------- | 2 2 | -P-=--TA--kω-sin-(kx-−-ωt-+-ϕ)-|

(134)

and

|-------1----------1---------| |⟨P⟩ = -T A2kω = -μA2 ω2c. | --------2----------2---------|

(135)

The next step is to turn these results into a systematic treatment of average power, intensity, and wave impedance.

References

[1] A. P. French, Vibrations and Waves, M.I.T. Introductory Physics Series, W. W. Norton & Company, 1971.

[2] Frank S. Crawford, Jr., Waves, Berkeley Physics Course, Volume 3, McGraw-Hill, 1968.

[3] William Moebs, Samuel J. Ling, and Jeff Sanny, University Physics, Volume 1, OpenStax, 2016, Section 16.4, “Energy and Power of a Wave.”

[4] Howard Georgi, The Physics of Waves, Benjamin/Cummings, 1992, continuum and traveling-wave chapters; also distributed through MIT OpenCourseWare 8.03SC.

[5] Massachusetts Institute of Technology, 8.03SC Physics III: Vibrations and Waves, Fall 2016, mechanical-wave lectures, notes, and Problem Set 5, MIT OpenCourseWare.

[6] Richard P. Feynman, Robert B. Leighton, and Matthew Sands, The Feynman Lectures on Physics, Volume I, Chapter 47, “Sound. The wave equation.”

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