📚 Wave Energy | 波的能量
Waves are a primary mechanism for energy transfer in physics. Whether it is a vibrating string, a sound wave in air, or an electromagnetic wave from the Sun, energy propagates through the medium (or vacuum) without the net movement of matter. Understanding the energy carried by waves allows us to analyse everything from the loudness of sound to the brightness of light. This article explores the fundamental concepts of wave energy, including power, intensity, and the crucial relationship between amplitude and energy, following the CIE A-Level Physics syllabus.
波是物理学中能量传递的主要方式。无论是振动的琴弦、空气中的声波，还是来自太阳的电磁波，能量都在介质（或真空）中传播，而没有物质的净位移。理解波携带的能量使我们能够分析从声音的响度到光的亮度等各种现象。本文基于 CIE A-Level 物理大纲，探讨波的能量基本概念，包括功率、强度以及振幅与能量的关键关系。
1. The Nature of Wave Energy | 波能量的本质
In a mechanical wave, the particles of the medium oscillate about their equilibrium positions. Each oscillating particle possesses both kinetic energy and potential energy. As the wave travels, energy is handed from one particle to the next through the coupling forces between them. Importantly, the particles themselves do not travel with the wave; they merely vibrate. The net transfer of energy occurs without net transport of matter.
在机械波中，介质粒子围绕其平衡位置振动。每个振动的粒子同时具有动能和势能。当波传播时，能量通过粒子间的相互作用力从一个粒子传递到下一个。重要的是，粒子本身并不随波移动，它们只是振动。能量的净传递是在没有物质净输运的情况下发生的。
2. Kinetic and Potential Energy in a Travelling Wave | 行波中的动能与势能
re μ is the mass per unit length. At an instant when the displacement is y, the transverse velocity of the element is vy = ∂y/∂t. Its kinetic energy is therefore ½ (Δm) vy². The potential energy arises from the stretching of the string as it is displaced from equilibrium; for small amplitudes, it can be shown that the instantaneous potential energy equals the instantaneous kinetic energy. Thus, the total mechanical energy of the element is twice its kinetic energy.
考虑一根张紧弦上的横波。弦上长度为 Δx 的小质元的质量为 Δm = μ Δx，其中 μ 是单位长度的质量。在位移为 y 的瞬间，该质元的横向速度 vy = ∂y/∂t。因此其动能为 ½ (Δm) vy²。势能来源于弦偏离平衡位置时的拉伸；对于小振幅，可以证明瞬时势能等于瞬时动能。因此，该质元的总机械能是其动能的两倍。
Using the wave function y(x,t) = A sin(kx – ωt), we find the transverse velocity vy = –Aω cos(kx – ωt). Then the instantaneous kinetic energy of the element is (½ μ Δx) A²ω² cos²(kx – ωt), and the total energy is μ Δx A²ω² cos²(kx – ωt).
利用波函数 y(x,t) = A sin(kx – ωt)，我们得到横向速度 vy = –Aω cos(kx – ωt)。那么该质元的瞬时动能为 (½ μ Δx) A²ω² cos²(kx – ωt)，总能量为 μ Δx A²ω² cos²(kx – ωt)。
3. Power Transmitted by a Wave on a String | 弦上波传递的功率
egment Δx passes a point in time Δt = Δx / v, where v is the wave speed. The average power ⟨P⟩ is the average energy per unit time: ⟨P⟩ e average total energy of segment length Δx is μ Δx A²ω² × ½. Dividing by Δt = Δx/v gives:
功率是能量通过某一点的传递速率。对于弦，长度为 Δx 的一小段能量在时间 Δt = Δx / v 内通过某点，其中 v 为波速。平均功率 ⟨P⟩ 等于单位时间的平均能量：⟨P⟩ = (长度 Δx 内的总能量) / Δt。由于 cos²θ 在一个完整周期内的平均值为 ½，因此长度为 Δx 的质元的平均总能量为 μ Δx A²ω² × ½。除以 Δt = Δx/v 得到：
Pavg = ½ μ ω² A² v
This equation shows that the power carried by a sinusoidal wave on a string is proportional to the square of the amplitude and the square of the angular frequency, as well as to the linear density and wave speed. For a given string and frequency, doubling the amplitude quadruples the power.
该公式表明，弦上正弦波传递的功率与振幅的平方和角频率的平方成正比，也与线密度和波速成正比。对于给定的弦和频率，振幅加倍会使功率变为原来的四倍。
4. Intensity of a Wave | 波的强度
Intensity I is defined as the power transmitted per unit area perpendicular to the direction of wave propagation. For a one-dimensional wave on a string, intensity is simply the power. However, for waves spreading in two or three dimensions, the area becomes important. In three dimensions, the wavefront is a sphere, and the same total power passes through successively larger spheres. Hence, intensity at a distance r from a point source is:
强度 I 定义为单位垂直面积上传递的功率。对于一维的弦上波，强度就是功率。但对于在二维或三维中传播的波，面积变得重要。在三维空间中，波前是一个球面，同样的总功率通过逐渐增大的球面。因此，距离点源 r 处的强度为：
I = P / (4π r²)
This is the inverse square law: intensity decreases with the square of the distance from the source. It applies to sound waves from a small source and to light from a star, provided there is no absorption.
这就是平方反比定律：强度随距离源的平方衰减。适用于小尺寸声源发出的声波和恒星发出的光，只要没有吸收。
5. Relationship Between Amplitude and Energy | 振幅与能量的关系
For all linear waves, the energy carried by the wave is proportional to the square of its amplitude. This can be seen directly from the power equation for a string: P ∝ A². In sound waves, the pressure amplitude Δp is proportional to the displacement amplitude; since energy ∝ (Δp)², once again energy ∝ A². For light and other electromagnetic waves, the intensity (energy per unit area per unit time) is proportional to the square of the electric field amplitude E₀.
对于所有线性波，波携带的能量与振幅的平方成正比。这可以直接从弦的功率方程看出：P ∝ A²。在声波中，压强振幅 Δp 与位移振幅成正比；由于能量 ∝ (Δp)²，因此能量 ∝ A²。对于光和其他电磁波，强度（单位面积单位时间的能量）与电场振幅 E₀ 的平方成正比。
This relationship explains why a louder sound has greater pressure variations, and why brighter light corresponds to stronger electric fields. If the amplitude of a wave doubles, the energy transported per unit time quadruples, provided all other factors remain constant.
这一关系解释了为什么更响的声音具有更大的压强变化，以及为什么更亮的光对应于更强的电场。若波的振幅加倍，则单位时间传输的能量变为四倍（前提是其他因素保持不变）。
6. Energy Density in Waves | 波的能量密度
Energy density u is the energy stored per unit volume of the medium. For the string wave, the instantaneous energy per unit length is μ A²ω² cos²(kx – ωt). Dividing by the cross-sectional area (if any) gives volume energy density for a medium. In a sound wave, the average kinetic and potential energy densities are equal, and the total average energy density is:
能量密度 u 是介质单位体积中储存的能量。对于弦波，单位长度的瞬时能量为 μ A²ω² cos²(kx – ωt)。除以横截面积（如果有）就得到介质的体积能量密度。在声波中，平均动能密度和平均势能密度相等，总平均能量密度为：
⟨u⟩ = ½ ρ ω² s₀²
he intensity of a sound wave can then be related to the energy density and wave speed: I = ⟨u⟩ v.
其中 ρ 是介质密度，s₀ 是位移振幅。声波的强度与能量密度和波速的关系为 I = ⟨u⟩ v。
7. Energy in Electromagnetic Waves | 电磁波中的能量
Electromagnetic waves carry energy in oscillating electric and magnetic fields. The instantaneous energy density in an EM wave is shared equally between the electric and magnetic fields. The magnitude of the Poynting vector describes the power per unit area (instantaneous intensity). For a sinusoidal plane wave, the time-averaged intensity is:
电磁波通过振荡的电场和磁场携带能量。电磁波中的瞬时能量密度平均分配给电场和磁场。坡印廷矢量的大小描述了单位面积的功率（瞬时强度）。对于正弦平面波，时间平均强度为：
I = ½ c ε₀ E₀²
where c is the speed of light, ε₀ the permittivity of free space, and E₀ the peak electric field strength. Once again, the intensity is proportional to the square of the amplitude (E₀). This is consistent with the general principle that wave energy scales as amplitude squared.
其中 c 为光速，ε₀ 为真空介电常数，E₀ 为电场强度的峰值。同样，强度与振幅（E₀）的平方成正比。这与波能量随振幅平方变化的一般原理一致。
8. Worked Example: Energy of a Sound Wave | 例题：声波的能量
Example: A spherical sound wave is emitted from a loudspeaker of power 0.50 W. Assuming no energy is absorbed, calculate (a) the intensity at a distance of 2.0 m, and (b) the displacement amplitude of the wave at that point given that the frequency is 1.0 kHz, the density of air is 1.2 kg m⁻³, and the speed of sound is 340 m s⁻¹.
例题：一个扬声器发出功率为 0.50 W 的球面声波。假设没有能量被吸收，计算 (a) 在距离 2.0 m 处的强度；(b) 已知频率为 1.0 kHz，空气密度为 1.2 kg m⁻³，声速为 340 m s⁻¹，求该点的位移振幅。
ion (a): Intensity I = P / (4π r²) = 0.50 / (4π × 2.0²) = 0.50 / (50.27) ≈ 9.95 × 10⁻³ W m
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