Wave Mechanics: Power Carried by a 1D Wave

Wave Mechanics: Power Carried by a 1D Wave

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

(1)

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

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

(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

(3)

means energy transport toward increasing x, and

(4)

means energy transport toward decreasing x.

The SI unit of power is

(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

(6)

The transverse velocity of the material at the cut is

(7)

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

P	= Fuvu	(8)
	= (−Tux)ut.	(9)

Therefore

(10)

The geometry and sign convention are summarized below.

3 Dimensional check

The tension has units of newtons,

(11)

while u_x is dimensionless and

(12)

Hence

[Tuxut]	= 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

(16)

Differentiate with respect to time:

(17)

For the ideal string,

(18)

Substitute this into the first term:

ℰt	= Tutuxx + Tuxuxt	(19)
	= T(utux).	(20)

Therefore

(21)

Identifying

(22)

gives the local conservation law

(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:

(24)

Thus

(25)

Equivalently,

(26)

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

6 Power in a pure right-moving wave

Consider

(27)

With

(28)

we have

(29)

and

(30)

Therefore

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

Hence

(34)

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

WM18 showed that the energy density of the same wave is

(35)

Therefore

(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

(37)

we have

(38)

and

(39)

Thus

(40)

Since

(41)

we can write

(42)

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

8 Sinusoidal traveling wave

Consider the right-moving sinusoidal wave

(43)

Define

(44)

Then

(45)

and

(46)

The instantaneous power is therefore

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

Hence

(49)

It is always nonnegative for this right-moving wave.

9 Average power of a sinusoidal wave

Over one phase cycle,

(50)

Therefore

(51)

Use

(52)

and

(53)

Then

(54)

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

Because WM18 found

(55)

we again have

(56)

10 Square-law scaling

For a fixed string and a sinusoidal traveling wave,

(57)

Thus

(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

(59)

let

(60)

Then

(61)

and

(62)

Substitute into the power law:

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

Thus

(65)

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

12 Standing waves and zero average transport

A standing wave can be written

(66)

Then

(67)

and

(68)

Therefore

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

Using double-angle identities,

(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,

(72)

so

(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

(74)

At one event,

(75)

Find the instantaneous power and interpret its sign.

Solution

Use

(76)

Then

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

Thus

(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

(80)

on a string with

(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

(85)

Then

k	=	(86)
	≃ 0.993 rad/m.	(87)

Convert the amplitude:

(88)

Now use

(89)

Thus

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

Therefore

(92)

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

A sinusoidal traveling wave moves on a string with

(93)

At frequency

(94)

what amplitude is required to carry an average power of

(95)

Solution

Start with

(96)

Solve for A:

(97)

The angular frequency is

(98)

Therefore

A	=	(99)
	≃ 1.42 × 10−2 m.	(100)

Thus

(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

(102)

The pulse speed is

(103)

Find the instantaneous power at that event.

Solution

For a pure right-moving wave,

(104)

Therefore

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

Hence

(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

(108)

and the directional profile derivatives have values

(109)

Find the net power.

Solution

Use

(110)

Then

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

Thus

(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

(115)

with

(116)

Find the power at

(117)

when

(118)

Also state the time-averaged power at that position.

Solution

Because

(119)

we have

(120)

The standing-wave power is

(121)

At x = 0.25 m,

(122)

so

(123)

Also

(124)

so

(125)

Therefore

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

Thus

(128)

at this instant.

Over a complete oscillation cycle,

(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

(130)

Together with

(131)

it satisfies the local conservation law

(132)

For pure traveling waves,

(133)

For a right-moving sinusoidal wave,

(134)

and

(135)

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

References

References