📚 Energy of Waves: Propagation and Calculation | 波的能量：传播与计算
Waves are one of the most powerful ways nature transfers energy across space. Whether it is sound carrying a voice from one side of a room to the other, or sunlight travelling 150 million kilometres to warm the Earth, the physics of wave energy is the same: energy is transported without matter being transported. This article explains how wave energy propagates, how to calculate intensity and power, and how to apply these ideas to strings, sound and electromagnetic waves.
波是自然界中最强有力的能量传递方式之一。无论是声音跨过房间传递一个声音，还是阳光穿越1.5亿公里温暖地球，波能量的物理都是相同的：能量被传输，而物质不随之传输。本文将解释波能量如何传播、如何计算强度和功率，以及如何将这些思想应用于弦、声波和电磁波。
1. The Concept of Wave Energy | 波能量的概念
A wave is a disturbance that transfers energy from one point to another. In a mechanical wave, such as a wave on a string or a sound wave in air, particles of the medium oscillate about their equilibrium positions. As they vibrate, they possess kinetic energy because they are moving, and potential energy because the medium is deformed or compressed.
波是一种从一点向另一点传递能量的扰动。在机械波中，例如弦上的波或空气中的声波，介质粒子围绕其平衡位置振动。当它们振动时，由于在运动而具有动能，又因为介质发生形变或压缩而具有势能。
The total energy carried by a wave is the sum of the energies of all the oscillating particles in the region. Even though each particle only moves a small distance, the coordinated motion of many particles can transfer a large amount of energy over long distances.
波所携带的总能量是该区域内所有振动粒子能量的总和。尽管每个粒子只运动很小的距离，但许多粒子的协调运动可以远距离传递大量能量。
For a single particle executing simple harmonic motion, its total energy depends on the square of the amplitude. This fundamental idea leads to the most important relationship in wave energy: energy and intensity are proportional to the square of the amplitude.
对于做简谐运动的单个粒子，其总能量取决于振幅的平方。这个基本思想引出了波能量中最重要的关系：能量和强度与振幅的平方成正比。
2. Energy Transfer Without Transfer of Matter | 能量传递而非物质传递
One essential feature of wave motion is that energy moves through the medium, but the medium itself does not move permanently with the wave. Consider water waves in a pond: a floating cork bobs up and down as the wave passes, but it is not carried along with the wave toward the shore.
波运动的一个基本特征是：能量在介质中传播，但介质本身并不随波永久移动。以池塘中的水波为例：一片浮在水面的软木塞在波经过时上下晃动，但不会随波被带到岸边。
The mechanism is particle-to-particle interaction. In a solid string, each particle pulls on its neighbour through tension. In a gas, particles collide and push each other. These interactions transmit the disturbance while each particle returns to its original equilibrium position after the wave has passed.
其机制是粒子与粒子之间的相互作用。在固体弦上，每个粒子通过张力拉扯其邻居。在气体中，粒子碰撞并互相推动。这些相互作用传递扰动，而每个粒子在波过后返回其原始平衡位置。
For A-Level CIE Physics, you should be able to explain why light and sound can both transfer energy but through different mechanisms: sound requires a medium, while electromagnetic waves can transfer energy through a vacuum.
对于A-Level CIE物理，你应该能够解释光和声音为什么都能传递能量但机制不同：声音需要介质，而电磁波可以在真空中传递能量。
3. Intensity and Amplitude: I ∝ A² | 强度与振幅：I ∝ A²
The intensity of a wave is defined as the power transferred per unit area perpendicular to the wave direction. Its unit is W m⁻². For a point source spreading energy equally in all directions, intensity decreases with distance due to the inverse square law.
波的强度定义为垂直于波传播方向上每单位面积所传递的功率，其单位是W m⁻²。对于向所有方向均匀传播能量的点源，强度随距离减小而遵循平方反比定律。
Why does intensity depend on amplitude squared? The energy of an oscillator is proportional to the square of its amplitude: E ∝ A². Because intensity is energy per unit time per unit area, and the oscillation frequency is fixed by the source, it follows that:
为什么强度与振幅的平方有关？振荡器的能量正比于其振幅的平方：E ∝ A²。因为强度是单位时间单位面积上的能量，而振动频率由波源决定，因此有：
I ∝ A²
Doubling the amplitude quadruples the intensity. This has practical consequences: a sound wave with twice the pressure amplitude carries four times the energy, so its loudness increases by about 12 dB.
振幅加倍，强度变为原来的四倍。这有实际影响：压力振幅加倍的声音波携带四倍能量，因此其响度增加约12 dB。
For sinusoidal waves, the relationship can be made quantitative: average power per unit area is proportional to μ, v, ω², and A² for string waves: I = ½ μ v ω² A² for a wave travelling along a string. This formula is central to many exam calculations.
对于正弦波，关系可以定量化：对于沿弦传播的波，平均功率密度正比于μ、v、ω²和A²：I = ½ μ v ω² A²。这个公式是许多考试计算的核心。
4. Energy Density in a Progressive Wave | 行波中的能量密度
For a wave on a string, imagine a tiny segment of length dx and mass dm = μ dx, where μ is the mass per unit length. The position of this segment is given by a sinusoidal travelling wave: y = A sin(kx − ωt).
对于弦上的波，考虑一个长度为dx、质量为dm = μ dx的微小段，其中μ是单位长度的质量。该段的位置由正弦行波描述：y = A sin(kx − ωt)。
The transverse velocity of the segment is the time derivative: v_y = ∂y/∂t = −ωA cos(kx − ωt). Its instantaneous kinetic energy is ½ dm v_y². When averaged over one full period, cos² averages to ½, giving a mean kinetic energy density of ¼ μ ω² A² per unit length.
该段的横向速度是时间导数：v_y = ∂y/∂t = −ωA cos(kx − ωt)。其瞬时动能为½ dm v_y²。在一个完整周期内取平均，cos²的平均值为½，因此每单位长度的平均动能密度为¼ μ ω² A²。
The potential energy comes from the stretching of the string. The local strain is ∂y/∂x = kA cos(kx − ωt). The average potential energy density has the same value as the average kinetic energy density, because the wave trades energy between kinetic and potential forms exactly in phase.
势能来自弦的拉伸。局部应变为∂y/∂x = kA cos(kx − ωt)。平均势能密度与平均动能密度相同，因为波在动能和势能之间完全同相地互换能量。
Therefore the total average energy per unit length is ¼ μ ω² A² + ¼ μ ω² A² = ½ μ ω² A². This quantity is often called the energy density of the wave.
因此，每单位长度的总平均能量为¼ μ ω² A² + ¼ μ ω² A² = ½ μ ω² A²。这个量通常称为波的能量密度。
5. Calculating Power for a Wave on a String | 弦上波的功率计算
Power is the rate at which energy is transferred. If a wave travels with speed v, then in one second a length v of string passes through a given point. Hence the average power carried by a progressive wave on a string is the average energy per unit length multiplied by the wave speed.
功率是能量传递的速率。如果波以速度v传播，那么在一秒内，长度为v的弦通过一个给定点。因此，弦上行波携带的平均功率等于每单位长度平均能量乘以波速。
P_avg = ½ μ v ω² A²
Here μ is the linear mass density (kg m⁻¹), v is the wave speed (m s⁻¹), ω is the angular frequency (rad s⁻¹), and A is the amplitude (m).
其中μ是线质量密度（kg m⁻¹），v是波速（m s⁻¹），ω是角频率（rad s⁻¹），A是振幅（m）。
Remember that ω = 2πf. If a question gives frequency f, use ω = 2πf rather than converting from wave speed and wavelength, unless the wavelength is more convenient.
请记住ω = 2πf。如果题目给出频率f，应使用ω = 2πf，而不是通过波速和波长换算，除非波长更方便。
It is also useful to write the power in terms of tension and frequency. Because v = √(T/μ), the power can be expressed as P_avg = ½ √(Tμ) ω² A². Both forms appear in exam questions.
用张力和频率表示功率也很有用。因为v = √(T/μ)，功率可以表示为P_avg = ½ √(Tμ) ω² A²。两种形式在考题中都会出现。
6. Sound Waves: Intensity and the Decibel Scale | 声波：强度与分贝标度
Sound is a longitudinal mechanical wave. The energy of a sound wave is carried by compressions and rarefactions of the air. The intensity of sound is the power per unit area, measured in W m⁻².
声音是一种纵机械波。声波的能量由空气的压缩和稀疏携带。声音的强度是单位面积的功率，单位为W m⁻²。
Because the human ear can detect an enormous range of intensities, from about 10⁻¹² W m⁻² at the threshold of hearing to more than 1 W m⁻² at the threshold of pain, a logarithmic scale is used.
由于人耳能探测到极其宽广的强度范围，从听阈约10⁻¹² W m⁻²到痛阈超过1 W m⁻²，因此使用对数标度。
The sound intensity level (SIL) in decibels is defined by:
声强级（dB）定义为：
L = 10 log₁₀ (I / I₀)
where I₀ = 1.0 × 10⁻¹² W m⁻² is the reference intensity, approximately the quietest sound a healthy human ear can hear.
其中I₀ = 1.0 × 10⁻¹² W m⁻²是参考强度，大约是健康人耳能听到的最微弱声音。
A doubling of intensity corresponds to an increase of about 3 dB. A tenfold increase in intensity corresponds to 10 dB, which is perceived as roughly twice as loud.
强度加倍对应约增加3 dB。强度增大到10倍对应10 dB，这被感知为大约响度加倍。
When sound spreads out from a point source in a uniform medium, the intensity obeys the inverse square law: I = P / (4πr²), where P is the total power emitted by the source.
当声音在均匀介质中从点源向外传播时，强度遵循平方反比定律：I = P / (4πr²)，其中P是源发出的总功率。
7. Electromagnetic Waves: Energy in Electric and Magnetic Fields | 电磁波：电场与磁场中的能量
Electromagnetic waves, including light, radio waves and X-rays, store energy in both their electric field and their magnetic field. In vacuum, the energy density of an electric field is ½ ε₀ E², while the energy density of a magnetic field is B² / (2μ₀).
电磁波，包括光、无线电波和X射线，在电场和磁场中都储存能量。在真空中，电场的能量密度为½ ε₀ E²，磁场的能量密度为B² / (2μ₀)。
For a plane electromagnetic wave, the electric and magnetic field amplitudes are related by E = cB. At every instant, the energy density in the electric field equals the energy density in the magnetic field for a wave in vacuum.
对于平面电磁波，电场和磁场振幅满足E = cB。在真空中的任意时刻，电场能量密度等于磁场能量密度。
The average energy density of an EM wave is therefore u_avg = ½ ε₀ E₀², where E₀ is the peak electric field amplitude. Since intensity is the energy passing through unit area per unit time, for a wave travelling at speed c:
因此，电磁波的平均能量密度为u_avg = ½ ε₀ E₀²，其中E₀是电场峰值振幅。因为强度是单位时间通过单位面积的能量，对于以速度c传播的波：
I = c u_avg = ½ c ε₀ E₀²
In terms of the root-mean-square electric field, E_rms = E₀ / √2, the intensity becomes I = c ε₀ E_rms². This is useful when dealing with alternating fields that are detected as oscillating quantities.
用均方根电场表示，E_rms = E₀ / √2，强度变为I = c ε₀ E_rms²。在处理被检测为振荡量的交变电场时很有用。
8. Attenuation and Absorption of Waves | 波的衰减与吸收
When waves travel through a real medium, energy is gradually lost due to friction, heat generation or scattering. This loss is called attenuation. For a plane wave travelling in one direction, the intensity decreases exponentially:
当波在真实介质中传播时，能量因摩擦、发热或散射而逐渐损失。这种损失称为衰减。对于沿一个方向传播的平面波，强度按指数衰减：
I = I₀ e^(−αx)
where I₀ is the initial intensity, x is the distance travelled, and α is the linear absorption coefficient in m⁻¹.
其中I₀是初始强度，x是传播距离，α是线性吸收系数，单位m⁻¹。
In addition to absorption, a wave from a point source loses intensity because it spreads over a larger area. The inverse square law I = P/(4πr²) describes this geometric spreading in three dimensions.
除了吸收，点源发出的波还会因为扩散到更大面积而损失强度。平方反比定律I = P/(4πr²)描述了三维空间的几何扩散。
These two effects are often combined in medical physics and astrophysics problems. For example, gamma radiation from a point source is both geometrically diluted and absorbed by tissue. You must be careful to decide which mechanism is being described in a given question.
这两种效应在医学物理和天体物理问题中常被结合。例如，点源的伽马辐射既发生几何稀释又被组织吸收。你必须在给定题目中仔细判断描述的是哪一种机制。
9. Worked Example: Power of a Wave on a String | 例题：弦上波的功率
A transverse wave travels along a string of linear density μ = 0.020 kg m⁻¹ with a wave speed of v = 12 m s⁻¹. The source oscillates with frequency f = 2.0 Hz and amplitude A = 5.0 cm. Calculate the average power transferred by the wave.
一列横波沿线密度μ = 0.020 kg m⁻¹的弦传播，波速v = 12 m s⁻¹。波源以频率f = 2.0 Hz、振幅A = 5.0 cm振荡。计算波传递的平均功率。
First convert amplitude to metres: A = 0.050 m. The angular frequency is ω = 2πf = 2π × 2.0 = 4π rad s⁻¹. Now substitute into the power formula:
首先将振幅转换为米：A = 0.050 m。角频率为ω = 2πf = 2π × 2.0 = 4π rad s⁻¹。现在代入功率公式：
P_avg = ½ μ v ω² A² = ½ × 0.020 × 12 × (4π)² × (0.050)²
Compute step by step: ½ × 0.020 = 0.010; multiply by 12 gives 0.12; (4π)² = 16π² ≈ 157.9; A² = 0.0025. Therefore P_avg = 0.12 × 157.9 × 0.0025 ≈ 0.047 W. The average power is approximately 0.047 W, or 47 mW.
逐步计算：½ × 0.020 = 0.010；乘以12得到0.12；(4π)² = 16π² ≈ 157.9；A² = 0.0025。因此P_avg = 0.12 × 157.9 × 0.0025 ≈ 0.047 W。平均功率约为0.047 W，即47 mW。
If the frequency were doubled while keeping amplitude the same, the power would increase by a factor of four because P ∝ ω². If the amplitude were doubled, the power would also quadruple. These proportionalities are common exam traps.
如果频率加倍而振幅保持不变，功率将增大到原来的四倍，因为P ∝ ω²。如果振幅加倍，功率也将增大到四倍。这些比例关系是常见的考试陷阱。
10. Worked Example: Intensity and Decibel Level | 例题：强度与分贝
A small speaker emits sound uniformly in all directions with a total power of 0.10 W. Calculate the intensity at a distance of 2.0 m from the speaker, and the corresponding sound intensity level in dB.
一个小型扬声器以总功率0.10 W向所有方向均匀发射声音。计算距离扬声器2.0 m处的强度，以及对应的以dB为单位的声强级。
Using the inverse square law for a point source:
使用点源的平方反比定律：
I = P / (4πr²) = 0.10 / (4π × 2.0²) = 0.10 / (16π) ≈ 1.99 × 10⁻³ W m⁻²
Now convert to decibels. The reference intensity is I₀ = 1.0 × 10⁻¹² W m⁻². Thus:
现在转换为分贝。参考强度为I₀ = 1.0 × 10⁻¹² W m⁻²。因此：
L = 10 log₁₀ (1.99 × 10⁻³ / 1.0 × 10⁻¹²) = 10 log₁₀ (1.99 × 10⁹) ≈ 10 × 9.30 = 93 dB
The intensity is about 2.0 mW m⁻² and the sound level is about 93 dB. This is comparable to a loud motorcycle or a live rock concert at close range.
强度约为2.0 mW m⁻²，声强级约为93 dB。这相当于近距离的摩托车或现场摇滚音乐会。
Note that if you double the distance, the intensity drops to one quarter, and the intensity level decreases by about 6 dB, not 3 dB. Many students incorrectly apply a 3 dB rule to distance changes; 3 dB corresponds to doubling or halving the power or intensity, not the distance.
注意，如果距离加倍，强度降为原来的四分之一，声强级降低约6 dB，而不是3 dB。许多学生错误地将3 dB规则用于距离变化；3 dB对应功率或强度加倍或减半，而不是距离。
11. Common Misconceptions and Exam Tips | 常见误区与考试提示
Misconception 1: “The wave carries particles with it.” In fact, each particle of the medium oscillates about its equilibrium position and does not travel with the wave. The only thing that travels is the disturbance or phase.
误区一：“波携带粒子一起运动。”事实上，介质的每个粒子围绕其平衡位置振动，并不随波传播。唯一传播的是扰动或相位。
Misconception 2: “A larger amplitude means a larger wave speed.” Wave speed in a string is determined by tension and linear density, v = √(T/μ); it does not depend on amplitude. Amplitude controls the energy and intensity.
误区二：“振幅越大，波速越大。”弦上的波速由张力和线密度决定，v = √(T/μ)；与振幅无关。振幅决定能量和强度。
Misconception 3: “Intensity is directly proportional to amplitude.” The correct relationship is I ∝ A². A doubling of amplitude gives four times the intensity, not two.
误区三：“强度与振幅成正比。”正确关系是I ∝ A²。振幅加倍得到四倍强度，而不是两倍。
Exam tip: Always convert amplitude to metres, and ensure frequency is given in Hz before multiplying by 2π. Check whether a question asks for peak amplitude or rms amplitude for electromagnetic waves.
考试提示：始终将振幅转换为米，并确保频率以Hz为单位后再乘以2π。检查电磁波问题问的是峰值振幅还是均方根振幅。
Exam tip: When using the dB formula, your value of I must be in W m⁻², and the ratio I/I₀ must be unitless. Take the logarithm of the ratio, not of I alone.
考试提示：使用分贝公式时，I的值必须是W m⁻²，且比值I/I₀必须无量纲。对比值取对数，而不是单独对I取对数。
Exam tip: For superposition and standing wave questions, remember that energy is not transported through a standing wave; energy is transferred along a progressive wave. This is a frequent distinction in CIE questions.
考试提示：对于叠加和驻波问题，记住驻波并不传输能量；能量沿行波传输。这是CIE考试中常见的区分点。
12. Summary | 总结
Waves transfer energy without transferring matter. The average energy per unit length of a string wave is ½ μ ω² A², and the average power is P_avg = ½ μ v ω² A². Intensity is power per unit area and is proportional to amplitude squared. Sound intensity levels are measured on a logarithmic decibel scale. Electromagnetic waves store energy in electric and magnetic fields, with intensity I = ½ c ε₀ E₀². Wave energy can be lost through geometric spreading and absorption, both of which reduce intensity with distance.
波传递能量而不传递物质。弦波每单位长度的平均能量为½ μ ω² A²，平均功率为P_avg = ½ μ v ω² A²。强度是单位面积的功率，与振幅的平方成正比。声强级以对数分贝标度测量。电磁波在电场和磁场中储存能量，强度为I = ½ c ε₀ E₀²。波能量可以因几何扩散和吸收而损失，两者都随距离减小强度。
When preparing for your CIE A-Level exam, practice deriving the power formula step by step, and always state the units in every calculation. The relationship between intensity, amplitude and distance appears again and again in Section 2 and Paper 5-style questions.
在准备CIE A-Level考试时，请练习逐步推导功率公式，并在每次计算中写出单位。强度、振幅和距离之间的关系在Section 2和类似Paper 5的题目中反复出现。
Published by TutorHao | Physics Revision Series | aleveler.com
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