IB Physics HL · 鼎睿学苑

Unit C.2: Wave model单元 C.2：波模型

The foundational unit of Theme C "Wave behaviour". A travelling (progressive) wave transfers energy without transferring matter. This guide covers transverse vs longitudinal waves, the core wave quantities (wavelength, frequency, period, amplitude) tied together by the wave equation $v = f\lambda$, wavefronts and rays, reading $\lambda$ and $T$ off displacement graphs, and the electromagnetic spectrum from radio to gamma. Every later wave unit (wave phenomena, standing waves, Doppler) builds on the vocabulary fixed here.主题 C"波的行为"的奠基单元。行波（progressive wave）传递能量而不传递物质。本指南涵盖横波与纵波、由波动方程 $v = f\lambda$ 串联的核心波动量（波长、频率、周期、振幅）、波前与射线、从位移图像读出 $\lambda$ 与 $T$，以及从无线电到伽马的电磁波谱。后续所有波动单元（波现象、驻波、多普勒效应）都建立在本单元固定的术语之上。

IB Physics · Theme C.2 · First Assessment 2025 Papers 1 · 2 6 Topics · SL + HL core6 个核心专题 · SL + HL 核心
Read me first先读这里

How to use this guide本指南使用说明

C.2 is the vocabulary unit of waves. The single algebraic relation is $v = f\lambda$; the marks come from precise definitions (a wave transfers energy, not matter), from correctly classifying transverse vs longitudinal, and from reading wavelength off a displacement-distance graph but period off a displacement-time graph. Train the language alongside the one equation.C.2 是波动的词汇单元。唯一的代数关系是 $v = f\lambda$；分数来自精确的定义（波传递能量，不传递物质）、正确区分横波与纵波，以及在位移-距离图上读波长、在位移-时间图上读周期。术语与这一个方程一起练。

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If you are cramming如果你在临阵磨枪

Memorise $v = f\lambda$ and $T = 1/f$. A wave carries energy, not matter. Transverse: oscillation $\perp$ wave direction; longitudinal: oscillation $\parallel$ wave direction. Wavelength $\lambda$ is read off a displacement-distance graph; period $T$ is read off a displacement-time graph. All EM waves travel at $c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$ in vacuum.

背熟 $v = f\lambda$ 与 $T = 1/f$。波传递能量，不传递物质。横波：振动方向 $\perp$ 波传播方向；纵波：振动方向 $\parallel$ 波传播方向。波长 $\lambda$ 从位移-距离图读取；周期 $T$ 从位移-时间图读取。所有电磁波在真空中以 $c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$ 传播。

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If you are going for a 7如果你目标是 7 分

Be able to argue why $v = f\lambda$ follows from "the wave advances one wavelength every period". State that in a mechanical wave each particle executes simple harmonic motion about a fixed equilibrium while the disturbance propagates. Distinguish a wavefront (line of constant phase) from a ray (direction of energy flow, $\perp$ wavefront). Know the EM spectrum order and that EM waves are transverse while sound is a longitudinal mechanical wave.

能论证为何 $v = f\lambda$ 来自"波每经过一个周期前进一个波长"。说明在机械波中每个质点绕固定平衡位置做简谐运动，而扰动在传播。区分波前（等相位线）与射线（能量流动方向，$\perp$ 波前）。掌握电磁波谱顺序，并知道电磁波是横波而声波是纵向机械波。

HL flagHL 标记说明 C.2 "Wave model" is shared SL + HL core content; there is no HL-only sub-topic in this super-topic, so no block here carries the HL tag. The deeper wave mathematics (phase, the travelling-wave function) lives in C.3 Wave phenomena and C.4 Standing waves. SL and HL students study all of C.2.C.2"波模型"是 SL 与 HL 共同的核心内容；本超级专题没有 HL 专属子专题，因此本页没有任何段落带 HL 标记。更深的波动数学（相位、行波函数）位于 C.3 波现象与 C.4 驻波。SL 与 HL 学生都学习全部 C.2。
Topic C2.1 · Travelling Waves and Energy TransferTopic C2.1 · 行波与能量传递

Travelling (Progressive) Waves行波（progressive wave） C.2 SL+HL

What a travelling wave is.
  • A travelling (progressive) wave is a disturbance that propagates through a medium (or vacuum) transferring energy and momentum from one place to another.
  • No net transfer of matter. The particles of the medium oscillate about fixed equilibrium positions; they do not travel with the wave.
  • A pulse is a single, short disturbance; a continuous wave is a repeating train of pulses driven by a periodic source.
Source. A wave is produced by an oscillating source; the source's frequency sets the wave's frequency.
行波是什么。
  • 行波（progressive wave）是在介质（或真空）中传播的扰动，将能量与动量从一处传到另一处。
  • 没有物质的净传递。介质质点绕固定平衡位置振动；它们并不随波一起前进。
  • 脉冲（pulse）是单个的、短促的扰动；连续波是由周期性波源驱动的、重复的脉冲列。
波源。波由振动的波源产生；波源的频率决定波的频率。
Worked Example C2.1 (energy, not matter)C2.1 例题（传能不传质）

A cork floats on a pond. A water wave of amplitude $4.0\ \mathrm{cm}$ passes the cork, which bobs up and down once every $2.0\ \mathrm{s}$. (a) Describe the net horizontal displacement of the cork after many waves pass. (b) State what is transported across the pond.一软木塞浮在池塘上。一列振幅 $4.0\ \mathrm{cm}$ 的水波经过软木塞，使它每 $2.0\ \mathrm{s}$ 上下起伏一次。(a) 描述许多波经过后软木塞的净水平位移。(b) 说明什么被横跨池塘传输。

Identify. The cork is a marker for a particle of the medium (water surface).

识别。软木塞标记介质（水面）中的一个质点。

(a) Set up. In a travelling wave the medium particles oscillate about a fixed equilibrium. The cork moves up and down (and, for a water wave, slightly in a small loop) but does not migrate with the wave.

(a) 列式。行波中介质质点绕固定平衡位置振动。软木塞上下起伏（水波情形还略沿小圈运动），但不会随波迁移。

Evaluate (a). The net horizontal displacement is (essentially) zero: the cork returns to its mean position.

评估 (a)。净水平位移（本质上）为零：软木塞回到平均位置。

(b) Evaluate. Energy (and momentum) is transported across the pond, not water. This is the defining property of a travelling wave.

(b) 评估。横跨池塘传输的是能量（与动量），不是水。这正是行波的定义性质。

Going deeper: why energy moves but particles do not深入：为何能量前进而质点不前进

Each particle is coupled to its neighbours by intermolecular (or, for EM waves, electromagnetic) restoring forces. When one particle is displaced, it does work on the next, passing on kinetic and potential energy. The disturbance therefore advances through the chain while each particle merely oscillates about its own equilibrium. The energy carried per unit time (the wave's power) scales with the square of the amplitude, $P \propto A^{2}$, which is why a louder sound or brighter light corresponds to a larger amplitude.

每个质点通过分子间（电磁波则为电磁）回复力与相邻质点耦合。当一个质点被位移时，它对下一个质点做功，传递动能与势能。因此扰动沿链条前进，而每个质点只绕自身平衡位置振动。单位时间传递的能量（波的功率）与振幅平方成正比，$P \propto A^{2}$，这正是更响的声音或更亮的光对应更大振幅的原因。

A travelling wave moves along a rope. Which quantity is transferred along the rope by the wave?一列行波沿绳传播。波沿绳传递的是哪个量？
C2.1 · Q1
The rope material itself绳子材料本身
Energy and momentum能量与动量
Mass of the rope绳子的质量
Nothing is transferred什么都不传递
A travelling wave transfers energy (and momentum) along the medium while each particle oscillates about a fixed equilibrium. The rope material is not carried along.行波沿介质传递能量（与动量），而每个质点绕固定平衡位置振动。绳子材料并不被带走。
The defining feature of a travelling wave is transfer of energy without net transfer of matter. The rope particles stay near their equilibrium positions.行波的定义性特征是传递能量而无物质净传递。绳子质点停留在其平衡位置附近。
Doubling the amplitude of a wave (frequency unchanged) changes the power it carries by a factor of about:将波的振幅加倍（频率不变），其传递的功率约变为原来的几倍：
C2.1 · Q2
$1$
$2$
$4$
$8$
Wave power scales with the square of amplitude, $P \propto A^{2}$. Doubling $A$ gives $2^{2} = 4$ times the power.波功率与振幅平方成正比，$P \propto A^{2}$。$A$ 加倍即功率变为 $2^{2} = 4$ 倍。
Power transported by a wave is proportional to amplitude squared, not amplitude. So a factor-2 amplitude gives a factor-4 power.波传输的功率与振幅的平方成正比，而非与振幅成正比。故振幅 2 倍对应功率 4 倍。
Topic C2.2 · Transverse vs Longitudinal WavesTopic C2.2 · 横波与纵波

Transverse and Longitudinal Waves横波与纵波 C.2 SL+HL

The two wave types (classified by particle motion).
  • Transverse wave: particle oscillation is perpendicular to the direction of energy transfer. Examples: waves on a string, water-surface ripples, all electromagnetic waves. Features: crests and troughs.
  • Longitudinal wave: particle oscillation is parallel to the direction of energy transfer. Example: sound in air. Features: compressions (particles bunched, high pressure) and rarefactions (particles spread out, low pressure).
Key contrast. Transverse waves can be polarised; longitudinal waves cannot.
两类波（按质点运动分类）。
  • 横波（transverse wave）：质点振动方向与能量传递方向垂直。例：弦上的波、水面波纹、所有电磁波。特征：波峰与波谷。
  • 纵波（longitudinal wave）：质点振动方向与能量传递方向平行。例：空气中的声音。特征：压缩（compression，质点密集、高压）与稀疏（rarefaction，质点稀疏、低压）。
关键对比。横波可被偏振；纵波不能。
Property性质Transverse横波Longitudinal纵波
Particle motion vs wave direction质点运动相对波向Perpendicular ($\perp$)垂直（$\perp$）Parallel ($\parallel$)平行（$\parallel$）
Visible features可见特征Crests, troughs波峰、波谷Compressions, rarefactions压缩、稀疏
Example例子Light, string waves光、弦波Sound in air空气中的声波
Can be polarised?可偏振？Yes可以No不可
Worked Example C2.2 (classify and describe sound)C2.2 例题（分类并描述声波）

A loudspeaker cone vibrates back and forth along the line toward a listener. (a) Classify the sound wave produced. (b) Describe what happens to the air molecules at a compression and at a rarefaction.扬声器纸盆沿指向听者的方向来回振动。(a) 对所产生的声波进行分类。(b) 描述压缩处与稀疏处的空气分子发生了什么。

Identify. The cone oscillates along the same line that the sound travels.

识别。纸盆沿声音传播的同一方向振动。

(a) Set up. Particle (air molecule) oscillation is parallel to the direction of energy transfer.

(a) 列式。质点（空气分子）振动方向与能量传递方向平行。

Evaluate (a). The wave is longitudinal — a mechanical longitudinal wave that needs a medium.

评估 (a)。该波为纵波——一种需要介质的机械纵波。

(b) Evaluate. At a compression the air molecules are bunched together (locally high pressure and density). At a rarefaction they are spread apart (locally low pressure and density). Each molecule simply oscillates back and forth about its own mean position.

(b) 评估。在压缩处空气分子密集聚集（局部高压、高密度）。在稀疏处它们彼此散开（局部低压、低密度）。每个分子只是绕自身平均位置来回振动。

Going deeper: why sound needs a medium but light does not深入：为何声波需要介质而光不需要

A longitudinal mechanical wave such as sound propagates by particles colliding with and pushing their neighbours; with no particles (a vacuum) there is nothing to compress, so sound cannot travel. An electromagnetic wave, by contrast, is a self-sustaining oscillation of electric and magnetic fields that propagate through a vacuum; it needs no medium. This is the deepest distinction between a mechanical wave (sound, water, string) and an electromagnetic wave (light, radio), and it is a favourite Paper 1 discriminator.

声波这类机械纵波依靠质点碰撞并推动相邻质点而传播；没有质点（真空）就无物可压缩，故声音无法传播。相反，电磁波是电场与磁场自持的振荡，可在真空中传播，无需介质。这是机械波（声、水、弦）与电磁波（光、无线电）之间最根本的区别，也是 Paper 1 常用的判别点。

In a longitudinal wave, the direction of particle oscillation relative to the direction of energy transfer is:在纵波中，质点振动方向相对能量传递方向是：
C2.2 · Q1
Perpendicular垂直
At $45^{\circ}$成 $45^{\circ}$
Random随机
Parallel平行
By definition, longitudinal-wave particles oscillate parallel to the direction the wave travels, producing compressions and rarefactions.按定义，纵波质点平行于波传播方向振动，形成压缩与稀疏。
Longitudinal = parallel oscillation (compressions and rarefactions). Perpendicular oscillation defines a transverse wave.纵波 = 平行振动（压缩与稀疏）。垂直振动定义的是横波。
Which observation provides direct evidence that a wave is transverse rather than longitudinal?哪一观察直接证明某波是横波而非纵波？
C2.2 · Q2
It can be polarised它可以被偏振
It can be reflected它可以被反射
It can be refracted它可以被折射
It transfers energy它传递能量
Polarisation restricts oscillation to one plane, which is only possible when the oscillation is perpendicular to the wave direction. Only transverse waves can be polarised; reflection, refraction and energy transfer occur for both types.偏振将振动限制在一个平面内，只有当振动垂直于波向时才可能。只有横波可被偏振；反射、折射与能量传递两类波都有。
Reflection, refraction and energy transfer happen for both wave types and so cannot distinguish them. Polarisation is unique to transverse waves.反射、折射与能量传递两类波都有，无法区分。偏振是横波独有的。
Topic C2.3 · Wave Quantities and v = fλTopic C2.3 · 波动量与 v = fλ

Wave Quantities and the Wave Equation波动量与波动方程 C.2 SL+HL

The four quantities.
  • Wavelength $\lambda$ ($\mathrm{m}$): shortest distance between two points in phase (e.g. crest to next crest).
  • Frequency $f$ ($\mathrm{Hz} = \mathrm{s^{-1}}$): number of complete oscillations per second; set by the source.
  • Period $T$ ($\mathrm{s}$): time for one complete oscillation, $T = 1/f$.
  • Amplitude $A$ ($\mathrm{m}$): maximum displacement from equilibrium; sets the energy carried.
The wave equation (data booklet). $$ v = f\lambda, \qquad f = \frac{1}{T}. $$ A wave advances one wavelength $\lambda$ in one period $T$, so $v = \lambda / T = f\lambda$.
四个量。
  • 波长（wavelength）$\lambda$（$\mathrm{m}$）：两个同相点之间的最短距离（如相邻两波峰）。
  • 频率（frequency）$f$（$\mathrm{Hz} = \mathrm{s^{-1}}$）：每秒完成的振动次数；由波源决定。
  • 周期（period）$T$（$\mathrm{s}$）：完成一次振动所需时间，$T = 1/f$。
  • 振幅（amplitude）$A$（$\mathrm{m}$）：偏离平衡的最大位移；决定所携带的能量。
波动方程（数据手册）。 $$ v = f\lambda, \qquad f = \frac{1}{T}. $$ 波在一个周期 $T$ 内前进一个波长 $\lambda$，故 $v = \lambda / T = f\lambda$。
Worked Example C2.3 (apply v = fλ)C2.3 例题（应用 v = fλ）

A sound wave in air has frequency $440\ \mathrm{Hz}$ and travels at $340\ \mathrm{m\,s^{-1}}$. (a) Find its wavelength. (b) Find its period.空气中一列声波频率为 $440\ \mathrm{Hz}$，传播速度 $340\ \mathrm{m\,s^{-1}}$。(a) 求其波长。(b) 求其周期。

Identify. Known: $f = 440\ \mathrm{Hz}$, $v = 340\ \mathrm{m\,s^{-1}}$.

识别。已知：$f = 440\ \mathrm{Hz}$、$v = 340\ \mathrm{m\,s^{-1}}$。

(a) Set up. Rearrange the wave equation $v = f\lambda$ for $\lambda$.

(a) 列式。由波动方程 $v = f\lambda$ 解出 $\lambda$。

$$ \lambda = \frac{v}{f} = \frac{340}{440} \approx 0.773\ \mathrm{m}. $$

(b) Set up. Period from frequency: $T = 1/f$.

(b) 列式。由频率求周期：$T = 1/f$。

$$ T = \frac{1}{f} = \frac{1}{440} \approx 2.27\times10^{-3}\ \mathrm{s}. $$

Evaluate. A sub-metre wavelength and a millisecond period are typical for audible sound. Cross-check: $v = f\lambda = 440 \times 0.773 \approx 340\ \mathrm{m\,s^{-1}}$. Consistent.

评估。亚米级波长与毫秒级周期对可闻声波是典型值。互校：$v = f\lambda = 440 \times 0.773 \approx 340\ \mathrm{m\,s^{-1}}$，一致。

Going deeper: deriving v = fλ from "one wavelength per period"深入：由"每周期前进一个波长"推导 v = fλ

In exactly one period $T$, the source completes one full oscillation and the wave pattern advances by exactly one wavelength $\lambda$. Speed is distance over time:

在恰好一个周期 $T$ 内，波源完成一次完整振动，波形恰好前进一个波长 $\lambda$。速度等于距离除以时间：

$$ v = \frac{\text{distance}}{\text{time}} = \frac{\lambda}{T}. $$

Since $f = 1/T$, substitute to obtain the data-booklet form:

由于 $f = 1/T$，代入得数据手册形式：

$$ v = \frac{\lambda}{T} = f\lambda. $$

Note: for a given medium $v$ is fixed (it depends on the medium's properties, not on the source). So if a wave passes from one medium to another, $f$ stays constant (set by the source) while $v$ and $\lambda$ change together.

注意：对给定介质 $v$ 是固定的（取决于介质性质，而非波源）。故当波从一种介质进入另一种介质时，$f$ 保持不变（由波源决定），而 $v$ 与 $\lambda$ 一起改变。

A wave has wavelength $0.50\ \mathrm{m}$ and frequency $200\ \mathrm{Hz}$. Its speed is:一列波波长 $0.50\ \mathrm{m}$，频率 $200\ \mathrm{Hz}$。其波速为：
C2.3 · Q1
$0.0025\ \mathrm{m\,s^{-1}}$
$100\ \mathrm{m\,s^{-1}}$
$400\ \mathrm{m\,s^{-1}}$
$200\ \mathrm{m\,s^{-1}}$
Apply $v = f\lambda = 200 \times 0.50 = 100\ \mathrm{m\,s^{-1}}$.用 $v = f\lambda = 200 \times 0.50 = 100\ \mathrm{m\,s^{-1}}$。
Use the wave equation $v = f\lambda$: multiply frequency by wavelength, do not divide.用波动方程 $v = f\lambda$：频率乘波长，不要相除。
A wave passes from medium 1 into medium 2, where its speed halves. What happens to its frequency and wavelength?一列波由介质 1 进入介质 2，速度减半。其频率与波长如何变化？
C2.3 · Q2
Both halve两者都减半
Frequency halves, wavelength unchanged频率减半，波长不变
Frequency unchanged, wavelength halves频率不变，波长减半
Frequency doubles, wavelength halves频率加倍，波长减半
Frequency is fixed by the source and does not change when the medium changes. From $v = f\lambda$, if $v$ halves and $f$ is constant, $\lambda$ must halve.频率由波源决定，介质改变时不变。由 $v = f\lambda$，若 $v$ 减半且 $f$ 不变，则 $\lambda$ 必减半。
Frequency is set by the source and stays constant across a boundary. With $f$ fixed, $v = f\lambda$ forces $\lambda$ to follow $v$, so halving $v$ halves $\lambda$.频率由波源决定，跨越边界时保持不变。$f$ 固定时，$v = f\lambda$ 使 $\lambda$ 随 $v$ 变化，故 $v$ 减半则 $\lambda$ 减半。
Topic C2.4 · Wavefronts and RaysTopic C2.4 · 波前与射线

Wavefronts and Rays波前与射线 C.2 SL+HL

Two ways to represent a wave.
  • Wavefront: a surface (or line in 2D) joining points that are all in phase, e.g. all the crests at one instant. Adjacent wavefronts are one wavelength $\lambda$ apart.
  • Ray: a line showing the direction of energy propagation. Rays are always perpendicular to wavefronts.
Geometry. A point source gives circular/spherical wavefronts; far from the source, or from a distant source, wavefronts become straight/plane and rays become parallel.
表示波的两种方式。
  • 波前（wavefront）：连接所有同相点的面（二维中为线），如某一瞬时所有波峰的连线。相邻波前相距一个波长 $\lambda$。
  • 射线（ray）：表示能量传播方向的线。射线总是垂直于波前。
几何关系。点波源给出圆形/球形波前；远离波源处或来自远处波源时，波前趋于直线/平面，射线趋于平行。
Worked Example C2.4 (count wavefronts)C2.4 例题（数波前）

A ripple-tank diagram shows circular wavefronts from a point source. Two adjacent crest wavefronts are $2.5\ \mathrm{cm}$ apart, and the dipper vibrates at $8.0\ \mathrm{Hz}$. (a) State the wavelength. (b) Find the wave speed. (c) State the angle between a ray and the wavefront it crosses.水波槽示意图显示来自点波源的圆形波前。相邻两个波峰波前相距 $2.5\ \mathrm{cm}$，振子以 $8.0\ \mathrm{Hz}$ 振动。(a) 写出波长。(b) 求波速。(c) 写出射线与其穿过的波前之间的夹角。

Identify. Adjacent crest wavefronts are separated by exactly one wavelength.

识别。相邻波峰波前相距恰好一个波长。

(a) Evaluate. $\lambda = 2.5\ \mathrm{cm} = 0.025\ \mathrm{m}$.

(a) 评估。$\lambda = 2.5\ \mathrm{cm} = 0.025\ \mathrm{m}$。

(b) Set up. Apply $v = f\lambda$.

(b) 列式。用 $v = f\lambda$。

$$ v = f\lambda = 8.0 \times 0.025 = 0.20\ \mathrm{m\,s^{-1}}. $$

(c) Evaluate. A ray points along the direction of energy flow, which is perpendicular to the wavefront, so the angle is $90^{\circ}$.

(c) 评估。射线沿能量流动方向，垂直于波前，故夹角为 $90^{\circ}$。

Going deeper: when to draw rays, when to draw wavefronts深入：何时画射线、何时画波前

Wavefronts are the natural picture for diffraction and interference, where you track how crests line up. Rays are the natural picture for reflection and refraction (geometric optics), where you track the path of energy and apply the law of reflection or Snell's law. The two pictures are equivalent: a ray is just the perpendicular to the local wavefront. When a plane wave hits a boundary at an angle, the wavefronts change spacing (because $\lambda$ changes with $v$) and the rays bend — both descriptions give the same refraction. The IB expects you to convert fluently between a wavefront diagram and the corresponding ray diagram.

波前是描述衍射与干涉的自然图像，用于追踪波峰如何对齐。射线是描述反射与折射（几何光学）的自然图像，用于追踪能量路径并应用反射定律或斯涅尔定律。两种图像等价：射线就是局部波前的垂线。当平面波斜射到边界时，波前间距改变（因 $\lambda$ 随 $v$ 改变）、射线弯折——两种描述给出同一折射结果。IB 要求你能在波前图与对应射线图之间流畅转换。

The angle between a ray and the wavefront it crosses is always:射线与其穿过的波前之间的夹角总是：
C2.4 · Q1
$0^{\circ}$
$30^{\circ}$
$45^{\circ}$
$90^{\circ}$
A ray shows the direction of energy propagation and is by definition perpendicular to the wavefront, so the angle is $90^{\circ}$.射线表示能量传播方向，按定义垂直于波前，故夹角为 $90^{\circ}$。
Rays are perpendicular to wavefronts by definition. The angle between them is always $90^{\circ}$.射线按定义垂直于波前。两者夹角总是 $90^{\circ}$。
On a wavefront diagram of a wave of wavelength $\lambda$, the distance between two adjacent wavefronts (both crests) is:在波长为 $\lambda$ 的波前图中，相邻两个波前（均为波峰）之间的距离为：
C2.4 · Q2
$\lambda$
$\tfrac{1}{2}\lambda$
$2\lambda$
Depends on the speed取决于波速
Adjacent wavefronts join points one full cycle apart and so are separated by exactly one wavelength $\lambda$.相邻波前连接相差一个完整周期的点，因此相距恰好一个波长 $\lambda$。
Consecutive crest wavefronts are one wavelength apart by definition of $\lambda$ (shortest distance between in-phase points).相邻波峰波前按 $\lambda$ 定义（同相点间最短距离）相距一个波长。
Topic C2.5 · Displacement GraphsTopic C2.5 · 位移图像

Displacement–Distance vs Displacement–Time Graphs位移-距离图与位移-时间图 C.2 SL+HL

The two graphs — and what each reads off.
  • Displacement–distance graph: a snapshot of the whole wave at one instant. The horizontal axis is position $x$. Read the wavelength $\lambda$ off it (crest to crest).
  • Displacement–time graph: the motion of one single particle over time. The horizontal axis is time $t$. Read the period $T$ off it (one full cycle), then $f = 1/T$.
Both graphs give the amplitude $A$ (the peak displacement). Only the displacement–distance graph gives $\lambda$; only the displacement–time graph gives $T$.
两种图像——各自读什么。
  • 位移-距离图：某一瞬时整列波的快照。横轴为位置 $x$。从中读波长 $\lambda$（峰到峰）。
  • 位移-时间图：单个质点随时间的运动。横轴为时间 $t$。从中读周期 $T$（一个完整循环），再由 $f = 1/T$ 求频率。
两种图都能给出振幅 $A$（峰值位移）。只有位移-距离图给出 $\lambda$；只有位移-时间图给出 $T$。
The classic trap经典陷阱 The two graphs look identical (both are sinusoids). The only way to tell them apart is the horizontal-axis label. If it says distance/position, the repeat length is $\lambda$. If it says time, the repeat length is $T$. Reading $T$ off a distance graph (or $\lambda$ off a time graph) is the single most common error in this unit.两种图看起来完全相同（都是正弦曲线）。唯一区分方法是横轴标签。若标"距离/位置"，则重复长度是 $\lambda$。若标"时间"，则重复长度是 $T$。在距离图上读 $T$（或在时间图上读 $\lambda$）是本单元最常见的错误。
Worked Example C2.5 (read both graphs)C2.5 例题（读两种图）

For one wave, a displacement–distance graph shows a repeat every $0.60\ \mathrm{m}$, and a displacement–time graph of one of its particles shows a repeat every $0.020\ \mathrm{s}$. Both graphs have peak displacement $3.0\ \mathrm{cm}$. Find (a) the amplitude, (b) the wavelength, (c) the frequency, (d) the wave speed.对同一列波，位移-距离图每 $0.60\ \mathrm{m}$ 重复一次，其某质点的位移-时间图每 $0.020\ \mathrm{s}$ 重复一次。两图峰值位移均为 $3.0\ \mathrm{cm}$。求 (a) 振幅，(b) 波长，(c) 频率，(d) 波速。

Identify. Distance graph gives $\lambda$; time graph gives $T$; peak displacement gives $A$.

识别。距离图给出 $\lambda$；时间图给出 $T$；峰值位移给出 $A$。

(a) Evaluate. Amplitude $A = 3.0\ \mathrm{cm} = 0.030\ \mathrm{m}$.

(a) 评估。振幅 $A = 3.0\ \mathrm{cm} = 0.030\ \mathrm{m}$。

(b) Evaluate. Wavelength from the distance graph: $\lambda = 0.60\ \mathrm{m}$.

(b) 评估。由距离图得波长：$\lambda = 0.60\ \mathrm{m}$。

(c) Set up. Period from the time graph is $T = 0.020\ \mathrm{s}$, so $f = 1/T$.

(c) 列式。由时间图得周期 $T = 0.020\ \mathrm{s}$，故 $f = 1/T$。

$$ f = \frac{1}{T} = \frac{1}{0.020} = 50\ \mathrm{Hz}. $$

(d) Set up. Wave speed from $v = f\lambda$.

(d) 列式。由 $v = f\lambda$ 求波速。

$$ v = f\lambda = 50 \times 0.60 = 30\ \mathrm{m\,s^{-1}}. $$

Evaluate. The two graphs are read separately for $\lambda$ and $T$, then combined through $v = f\lambda$. Mixing up the axes would give a wildly wrong speed.

评估。分别从两图读出 $\lambda$ 与 $T$，再用 $v = f\lambda$ 合并。混淆横轴会得到完全错误的波速。

Going deeper: reading the direction of motion of a particle深入：判断质点的运动方向

A displacement–distance snapshot plus the direction the wave travels lets you deduce which way any particle is moving next. Trick: imagine the whole waveform shifting a tiny amount in the wave's direction of travel; whatever displacement the waveform brings to a particle's $x$-position is where that particle moves to next. Equivalently, a particle copies the motion of the particle just "behind" it (on the source side) a moment earlier. This is a standard Paper 1 multiple-choice item: given a snapshot and "the wave moves to the right", state whether a marked particle is moving up or down.

一张位移-距离快照加上波的传播方向，就能推断任一质点下一刻向哪运动。技巧：想象整个波形沿传播方向微移一点；波形带到该质点 $x$ 位置处的位移，就是该质点下一刻要到达的位置。等价地，质点重复其"后方"（靠波源一侧）质点稍早的运动。这是 Paper 1 标准选择题：给出快照与"波向右传播"，判断标记质点正向上还是向下运动。

From which graph do you read the period $T$ of a wave?从哪种图读取波的周期 $T$？
C2.5 · Q1
Displacement–distance graph位移-距离图
Either graph gives $T$ directly两种图都能直接给出 $T$
Displacement–time graph位移-时间图
Neither graph gives $T$两种图都不能给出 $T$
The period is a time, so it is read from the displacement–time graph (one full cycle along the time axis). The distance graph gives the wavelength instead.周期是时间量，故从位移-时间图读取（时间轴上一个完整循环）。距离图给出的是波长。
Period is a duration. Read it off the time axis of a displacement–time graph. The distance graph yields $\lambda$, not $T$.周期是时长。从位移-时间图的时间轴读取。距离图给出 $\lambda$，不是 $T$。
A displacement–distance graph repeats every $0.40\ \mathrm{m}$; a displacement–time graph of a particle repeats every $0.010\ \mathrm{s}$. The wave speed is:某波位移-距离图每 $0.40\ \mathrm{m}$ 重复，质点位移-时间图每 $0.010\ \mathrm{s}$ 重复。波速为：
C2.5 · Q2
$40\ \mathrm{m\,s^{-1}}$
$0.040\ \mathrm{m\,s^{-1}}$
$4.0\ \mathrm{m\,s^{-1}}$
$400\ \mathrm{m\,s^{-1}}$
$\lambda = 0.40\ \mathrm{m}$ (distance graph), $T = 0.010\ \mathrm{s}$ so $f = 100\ \mathrm{Hz}$. Then $v = f\lambda = 100 \times 0.40 = 40\ \mathrm{m\,s^{-1}}$.$\lambda = 0.40\ \mathrm{m}$（距离图），$T = 0.010\ \mathrm{s}$ 故 $f = 100\ \mathrm{Hz}$。则 $v = f\lambda = 100 \times 0.40 = 40\ \mathrm{m\,s^{-1}}$。
Read $\lambda$ from the distance graph and $T$ from the time graph, get $f = 1/T$, then $v = f\lambda$. Do not interchange the axes.从距离图读 $\lambda$、从时间图读 $T$，求 $f = 1/T$，再 $v = f\lambda$。不要互换横轴。
Topic C2.6 · The Electromagnetic SpectrumTopic C2.6 · 电磁波谱

The Electromagnetic Spectrum电磁波谱 C.2 SL+HL

The electromagnetic spectrum (low $f$ / long $\lambda$ → high $f$ / short $\lambda$).

radio → microwave → infrared → visible → ultraviolet → X-ray → gamma

  • All EM waves are transverse (oscillating electric and magnetic fields, mutually $\perp$ and $\perp$ to propagation) and can be polarised.
  • All EM waves travel at the same speed $c$ in a vacuum: c = 3.00×10⁸ m s⁻¹. Differences between the bands are differences in $f$ and $\lambda$ only, related by $c = f\lambda$.
  • Sound is NOT part of this spectrum: it is a mechanical longitudinal wave needing a medium.
电磁波谱（低 $f$ / 长 $\lambda$ → 高 $f$ / 短 $\lambda$）。

无线电 → 微波 → 红外 → 可见光 → 紫外 → X 射线 → 伽马射线

  • 所有电磁波都是横波（振荡的电场与磁场互相 $\perp$ 且 $\perp$ 于传播方向），且可被偏振。
  • 所有电磁波在真空中以相同速度 $c$ 传播：c = 3.00×10⁸ m s⁻¹。各波段之间的差别仅在 $f$ 与 $\lambda$，由 $c = f\lambda$ 关联。
  • 声波不属于电磁波谱：它是需要介质的机械纵波。
Band波段Typical $\lambda$典型 $\lambda$Use / source用途 / 来源
Radio无线电$> 1\ \mathrm{m}$Broadcasting, communications广播、通信
Microwave微波$\sim 10^{-2}\ \mathrm{m}$Radar, ovens, mobile/WiFi雷达、微波炉、手机/WiFi
Infrared红外$\sim 10^{-5}\ \mathrm{m}$Heat, thermal imaging, remotes热、热成像、遥控器
Visible可见光$\sim (4\text{–}7)\times10^{-7}\ \mathrm{m}$Sight, optics视觉、光学
Ultraviolet紫外$\sim 10^{-8}\ \mathrm{m}$Sterilisation, fluorescence消毒、荧光
X-rayX 射线$\sim 10^{-10}\ \mathrm{m}$Medical imaging医学成像
Gamma伽马射线$< 10^{-12}\ \mathrm{m}$Nuclear decay, radiotherapy核衰变、放射治疗
Worked Example C2.6 (EM frequency from c = fλ)C2.6 例题（由 c = fλ 求电磁波频率）

Red light has wavelength $\lambda = 7.0\times10^{-7}\ \mathrm{m}$ in vacuum. (a) Find its frequency. (b) State how its speed compares with that of an X-ray in vacuum.红光在真空中的波长 $\lambda = 7.0\times10^{-7}\ \mathrm{m}$。(a) 求其频率。(b) 说明其速度与真空中 X 射线速度的比较。

Identify. In vacuum every EM wave travels at $c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$, so use $c = f\lambda$.

识别。真空中每列电磁波都以 $c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$ 传播，故用 $c = f\lambda$。

(a) Set up. Rearrange $c = f\lambda$ for $f$.

(a) 列式。由 $c = f\lambda$ 解出 $f$。

$$ f = \frac{c}{\lambda} = \frac{3.00\times10^{8}}{7.0\times10^{-7}} \approx 4.3\times10^{14}\ \mathrm{Hz}. $$

(b) Evaluate. X-rays and red light both travel at exactly $c$ in vacuum — identical speed. They differ only in frequency and wavelength (X-rays have far higher $f$ and shorter $\lambda$).

(b) 评估。X 射线与红光在真空中都恰以 $c$ 传播——速度完全相同。它们只在频率与波长上不同（X 射线 $f$ 高得多、$\lambda$ 短得多）。

Going deeper: EM waves vs sound — the defining contrasts深入：电磁波与声波——定义性对比

EM waves and sound are the two waves the IB contrasts most often. EM waves are transverse, need no medium, travel at $c$ in vacuum, and can be polarised. Sound is a mechanical longitudinal wave, needs a medium, travels much slower (about $340\ \mathrm{m\,s^{-1}}$ in air), and cannot be polarised. A common exam item asks why you see a distant lightning flash before you hear the thunder: light (EM) reaches you almost instantly at $c$, whereas sound crawls at $\sim 340\ \mathrm{m\,s^{-1}}$, so the delay (about $3\ \mathrm{s}$ per kilometre) measures the distance to the strike.

电磁波与声波是 IB 最常对比的两种波。电磁波是横波，不需要介质，真空中以 $c$ 传播，可被偏振。声波是机械纵波，需要介质，速度慢得多（空气中约 $340\ \mathrm{m\,s^{-1}}$），不可被偏振。常见考题问：为何先看到远处闪电再听到雷声？光（电磁波）以 $c$ 几乎瞬间到达，而声音以约 $340\ \mathrm{m\,s^{-1}}$ 缓慢前进，故时间差（每千米约 $3\ \mathrm{s}$）可量出到雷击处的距离。

Which of the following correctly orders three EM bands from longest to shortest wavelength?下列哪项将三个电磁波段按波长从长到短正确排序？
C2.6 · Q1
Gamma → visible → radio伽马 → 可见光 → 无线电
Visible → radio → gamma可见光 → 无线电 → 伽马
Radio → infrared → ultraviolet无线电 → 红外 → 紫外
X-ray → microwave → radioX 射线 → 微波 → 无线电
Wavelength decreases (frequency increases) along radio → microwave → infrared → visible → ultraviolet → X-ray → gamma. Radio → infrared → ultraviolet is correctly long-to-short.沿无线电 → 微波 → 红外 → 可见光 → 紫外 → X 射线 → 伽马，波长递减（频率递增）。无线电 → 红外 → 紫外 即从长到短正确。
Order the spectrum radio (longest) → gamma (shortest). Only radio → infrared → ultraviolet is monotonically decreasing in wavelength.按无线电（最长）→ 伽马（最短）排序。只有无线电 → 红外 → 紫外 的波长单调递减。
A radio wave and a gamma ray both travel through a vacuum. Which statement is correct?无线电波与伽马射线都在真空中传播。哪项陈述正确？
C2.6 · Q2
The gamma ray travels faster伽马射线传播更快
The radio wave travels faster无线电波传播更快
Both are longitudinal waves两者都是纵波
Both travel at $c$ and are transverse两者都以 $c$ 传播且为横波
All electromagnetic waves travel at $c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$ in vacuum, regardless of band, and all are transverse. They differ only in $f$ and $\lambda$.所有电磁波在真空中都以 $c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$ 传播，与波段无关，且都是横波。它们只在 $f$ 与 $\lambda$ 上不同。
All EM waves share the same vacuum speed $c$ and are transverse; only frequency and wavelength differ between bands.所有电磁波在真空中速度都是 $c$ 且为横波；波段间只有频率与波长不同。
Exam Tactics考试技巧

Exam Strategy and Common Pitfalls考试策略与常见陷阱

Definitions earn marks (every paper)定义就是分数（每张试卷）
  • "A wave transfers energy without transferring matter." Markschemes reward this exact phrasing for "describe a travelling wave".
  • "波传递能量而不传递物质。""描述行波"题的评分认可这一精确表述。
  • Transverse = oscillation perpendicular; longitudinal = oscillation parallel to energy transfer. State the reference direction.
  • 横波 = 振动垂直；纵波 = 振动平行于能量传递方向。要写明参考方向。
Using v = fλ使用 v = fλ
  • Convert to SI first. $\mathrm{cm} \to \mathrm{m}$, $\mathrm{kHz}/\mathrm{MHz} \to \mathrm{Hz}$ before substituting, or your speed is off by powers of ten.
  • 先换算成 SI 单位。代入前把 $\mathrm{cm} \to \mathrm{m}$、$\mathrm{kHz}/\mathrm{MHz} \to \mathrm{Hz}$，否则波速会差几个数量级。
  • For EM waves in vacuum, $v = c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$ is given. Use it directly via $c = f\lambda$.
  • 真空中电磁波 $v = c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$ 是已知量。直接由 $c = f\lambda$ 使用。
Reading graphs (data-response Section B)读图（数据题 B 部分）
  • Check the horizontal-axis label before reading the repeat length. Distance axis → $\lambda$. Time axis → $T$.
  • 读重复长度前先看横轴标签。距离轴 → $\lambda$。时间轴 → $T$。
  • Amplitude is the peak displacement on either graph; do not confuse it with the peak-to-peak distance (that is $2A$).
  • 振幅是任一图上的峰值位移；不要与峰-峰距离混淆（那是 $2A$）。
EM spectrum and wave types电磁波谱与波的类型
  • Memorise the band order radio → gamma and that all EM waves are transverse and travel at $c$ in vacuum.
  • 记住波段顺序无线电 → 伽马，并记住所有电磁波都是横波、真空中以 $c$ 传播。
  • Sound is a mechanical longitudinal wave; it needs a medium and is NOT on the EM spectrum. A favourite trap is "sound travels in a vacuum" — it does not.
  • 声波是机械纵波；需要介质，不在电磁波谱上。常见陷阱是"声音能在真空中传播"——不能。
Active Recall主动回忆

Flashcards闪卡

0 / 14 flipped已翻 0 / 14
A travelling wave transfers?行波传递什么？
Energy (and momentum); no net matter transfer.能量（与动量）；无物质净传递。
Transverse wave?横波？
Oscillation $\perp$ to energy transfer; e.g. light, string waves.振动 $\perp$ 能量传递；如光、弦波。
Longitudinal wave?纵波？
Oscillation $\parallel$ to energy transfer; e.g. sound (compressions/rarefactions).振动 $\parallel$ 能量传递；如声波（压缩/稀疏）。
Wave equation?波动方程？
$$v = f\lambda$$
Period from frequency?由频率求周期？
$$T = \frac{1}{f}$$
Wavelength $\lambda$?波长 $\lambda$？
Shortest distance between two in-phase points (crest to crest).两个同相点之间的最短距离（峰到峰）。
Amplitude $A$?振幅 $A$？
Maximum displacement from equilibrium; sets energy carried.偏离平衡的最大位移；决定所携带的能量。
Wavefront?波前？
Line/surface joining in-phase points; adjacent ones $\lambda$ apart.连接同相点的线/面；相邻者相距 $\lambda$。
Ray, and its angle to a wavefront?射线，及其与波前的夹角？
Direction of energy flow; always $\perp$ (90°) to wavefronts.能量流动方向；总是 $\perp$（90°）于波前。
Read $\lambda$ off which graph?$\lambda$ 从哪种图读？
Displacement–distance graph (snapshot).位移-距离图（快照）。
Read $T$ off which graph?$T$ 从哪种图读？
Displacement–time graph (one particle).位移-时间图（单个质点）。
EM spectrum order (long→short $\lambda$)?电磁波谱顺序（$\lambda$ 长→短）？
Radio, microwave, infrared, visible, UV, X-ray, gamma.无线电、微波、红外、可见光、紫外、X 射线、伽马。
Speed of all EM waves in vacuum?所有电磁波在真空中的速度？
$$c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$$
Is sound on the EM spectrum?声波在电磁波谱上吗？
No — sound is a mechanical longitudinal wave that needs a medium.否——声波是需要介质的机械纵波。
Mixed Practice综合练习

Unit C.2 Practice Quiz单元 C.2 练习测验

An FM radio station broadcasts at $96\ \mathrm{MHz}$. The wavelength of the wave in air ($v \approx c$) is closest to:某调频电台以 $96\ \mathrm{MHz}$ 广播。该波在空气中（$v \approx c$）的波长最接近：
Q1
$0.31\ \mathrm{m}$
$96\ \mathrm{m}$
$3.1\ \mathrm{m}$
$0.96\ \mathrm{m}$
$\lambda = c/f = (3.00\times10^{8})/(96\times10^{6}) \approx 3.1\ \mathrm{m}$. Note the $\mathrm{MHz}\to\mathrm{Hz}$ conversion.$\lambda = c/f = (3.00\times10^{8})/(96\times10^{6}) \approx 3.1\ \mathrm{m}$。注意 $\mathrm{MHz}\to\mathrm{Hz}$ 换算。
Use $\lambda = c/f$ with $c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$ and $f = 96\times10^{6}\ \mathrm{Hz}$. Convert MHz to Hz first.用 $\lambda = c/f$，$c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$，$f = 96\times10^{6}\ \mathrm{Hz}$。先把 MHz 换成 Hz。
A displacement–time graph for a particle in a wave shows one complete cycle in $0.025\ \mathrm{s}$. The frequency of the wave is:某波中一质点的位移-时间图在 $0.025\ \mathrm{s}$ 内完成一个完整循环。该波频率为：
Q2
$25\ \mathrm{Hz}$
$40\ \mathrm{Hz}$
$400\ \mathrm{Hz}$
$0.025\ \mathrm{Hz}$
One cycle in $0.025\ \mathrm{s}$ means $T = 0.025\ \mathrm{s}$, so $f = 1/T = 1/0.025 = 40\ \mathrm{Hz}$.$0.025\ \mathrm{s}$ 完成一个循环即 $T = 0.025\ \mathrm{s}$，故 $f = 1/T = 40\ \mathrm{Hz}$。
The time for one cycle on a displacement–time graph is the period $T$. Then $f = 1/T$.位移-时间图上一个循环的时间是周期 $T$。再由 $f = 1/T$ 求频率。
Which single statement best describes the difference between sound waves and light waves?哪一句话最能描述声波与光波的区别？
Q3
Sound is longitudinal and needs a medium; light is transverse and does not声波是纵波且需介质；光是横波且不需介质
Both are transverse, but light needs a medium两者都是横波，但光需要介质
Sound is transverse; light is longitudinal声波是横波；光是纵波
Both travel at $c$ in air两者在空气中都以 $c$ 传播
Sound is a mechanical longitudinal wave requiring a medium; light is a transverse EM wave that travels through a vacuum at $c$. The other options reverse or confuse these facts.声波是需要介质的机械纵波；光是可在真空中以 $c$ 传播的横向电磁波。其余选项颠倒或混淆了这些事实。
Sound: longitudinal, needs a medium, slow. Light: transverse EM, no medium, travels at $c$. Keep the two contrasts straight.声波：纵波、需介质、速度慢。光：横向电磁波、不需介质、以 $c$ 传播。两组对比不要弄混。
A water wave slows down as it enters a shallower region. Assuming its frequency is unchanged, its wavelength:一列水波进入较浅区域时减速。设频率不变，其波长：
Q4
Increases增大
Stays the same不变
Becomes zero变为零
Decreases减小
From $v = f\lambda$ with $f$ constant, $\lambda \propto v$. If $v$ decreases, $\lambda$ decreases in proportion. This is why wavefronts bunch up in shallow water.由 $v = f\lambda$ 且 $f$ 不变，$\lambda \propto v$。$v$ 减小则 $\lambda$ 成比例减小。这正是浅水中波前密集的原因。
Frequency is set by the source and stays fixed. With $f$ constant in $v = f\lambda$, wavelength tracks speed: slower wave → shorter wavelength.频率由波源决定、保持不变。$v = f\lambda$ 中 $f$ 不变时波长随波速变化：波速变慢 → 波长变短。
A wave of speed $1500\ \mathrm{m\,s^{-1}}$ (ultrasound in water) has a frequency of $3.0\ \mathrm{MHz}$. Its wavelength is closest to:速度 $1500\ \mathrm{m\,s^{-1}}$ 的波（水中超声）频率为 $3.0\ \mathrm{MHz}$。其波长最接近：
Q5
$5.0\ \mathrm{mm}$
$0.50\ \mathrm{mm}$
$2.0\ \mathrm{mm}$
$50\ \mathrm{mm}$
$\lambda = v/f = 1500/(3.0\times10^{6}) = 5.0\times10^{-4}\ \mathrm{m} = 0.50\ \mathrm{mm}$. The fine wavelength is why medical ultrasound resolves small structures.$\lambda = v/f = 1500/(3.0\times10^{6}) = 5.0\times10^{-4}\ \mathrm{m} = 0.50\ \mathrm{mm}$。波长很短正是医用超声能分辨细小结构的原因。
Use $\lambda = v/f$ with $f = 3.0\times10^{6}\ \mathrm{Hz}$. The result is $5.0\times10^{-4}\ \mathrm{m}$, i.e. $0.50\ \mathrm{mm}$.用 $\lambda = v/f$，$f = 3.0\times10^{6}\ \mathrm{Hz}$。结果为 $5.0\times10^{-4}\ \mathrm{m}$，即 $0.50\ \mathrm{mm}$。
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IB Physics HL · Unit C.2 Wave model · v1 · First Assessment 2025IB 物理 HL · 单元 C.2 波模型 · v1 · 首次考核 2025