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IB OPH SB4 Student Cards: Waves Revision Essentials | IB OPH SB4 学习卡片：波动复习要点

2026-08-22 00:43 · 本文部分内容由AI辅助创作

📚 IB OPH SB4 Student Cards: Waves Revision Essentials | IB OPH SB4 学习卡片：波动复习要点

These IB OPH SB4 student cards cover the wave topic as a set of quick-fire revision prompts. Use them to test definitions, equations, diagrams, and the standard problem types - reflection, refraction, interference and standing waves - so that every mark in the waves section becomes predictable rather than hopeful.

这套 IB OPH SB4 学习卡片把波动专题整理成一组快速复习提示。用它自测定义、方程、图示和标准题型——反射、折射、干涉与驻波——让波动部分的每一分都变得可预期，而不是靠运气。

1. What Are SB4 Student Cards? | 什么是 SB4 学习卡片？

an self-quiz in a queue, on a bus, or five minutes before a lesson without opening a 600-page textbook. For the waves topic the deck should cover four families of content: definitions and units, wave graphs, the equation v = f λ together with its rearrangements, and the standard problem types on reflection, refraction, interference and standing waves. A realistic target is 40 to 60 cards for the topic. Keep each answer under about twenty words; if you cannot compress it, the card is really two cards.

学习卡片是一种便携式复习工具。每张卡片正面写一个清晰的问题，背面写一个简短、可直接用于考试的答案，这样你可以在排队、坐车或上课前五分钟自测，而不必翻开六百页的教材。就波动专题而言，卡片组应覆盖四类内容：定义与单位、波形图、方程 v = f λ 及其各种变形，以及反射、折射、干涉和驻波的标准题型。整个专题合理的数量是 40 到 60 张。每张答案尽量压缩在二十个词以内；如果压不下去，说明这其实是两张卡片。

2. Wave Terminology You Must Know | 必须掌握的波动术语

Before attempting any calculation you must be able to define each wave quantity and match it to its symbol and SI unit. Examiners regularly award a mark for the unit alone, so learn the quantities as a set rather than one by one.

在做任何计算之前，你必须能定义每一个波动量，并把它与符号和 SI 单位对应起来。考官经常单独给单位一分，所以要把这些量作为一个整体来记，而不是零散地背。

Quantity | 物理量 Symbol | 符号 Definition | 定义 Unit | 单位
Amplitude | 振幅 A Maximum displacement from the equilibrium position | 离开平衡位置的最大位移 m
Wavelength | 波长 λ Distance between two successive crests, or between any two points in phase | 相邻两个波峰之间、或任意两个同相点之间的距离 m
Frequency | 频率 f Number of complete cycles passing a point per second | 每秒通过某点的完整周期数 Hz (s⁻¹)
Period | 周期 T Time for one complete cycle | 完成一个完整周期所需的时间 s
Wave speed | 波速 v Distance travelled by a wavefront per second | 波前每秒传播的距离 m s⁻¹
Displacement | 位移 x or y Distance of a particle from equilibrium in a stated direction | 质点沿指定方向离开平衡位置的距离 m
Phase difference | 相位差 Δφ Fraction of a cycle by which one oscillation leads another, in radians or degrees | 一个振动相对另一个振动超前的周期比例，用弧度或度表示 rad or °
Wavefront | 波前 - A line joining points that are in phase | 连接同相各点的线 -
Crest / trough | 波峰 / 波谷 - Point of maximum positive / negative displacement | 正向 / 负向最大位移点 -
Compression / rarefaction | 密部 / 疏部 - Region of maximum / minimum pressure in a longitudinal wave | 纵波中压强最大 / 最小的区域 -

3. Transverse and Longitudinal Waves | 横波与纵波

A transverse wave has oscillations perpendicular to the direction in which energy is transferred; a longitudinal wave has oscillations parallel to that direction. Light and all other electromagnetic waves are transverse and can therefore be polarised, whereas sound in air is longitudinal and cannot. Water surface waves are a mixture of both motions but are treated as transverse in almost every IB question. A favourite multiple-choice comparison is seismic waves: S-waves are transverse, P-waves are longitudinal, which is why S-waves cannot travel through the liquid outer core.

横波的振动方向与能量传播方向垂直；纵波的振动方向与能量传播方向平行。光和所有其他电磁波都是横波，因此可以被偏振；而空气中的声波是纵波，不能偏振。水面波是两种运动的混合，但在几乎所有的 IB 题目中都被当作横波处理。选择题里最常出现的一个对比是地震波：S 波是横波，P 波是纵波，这正是 S 波无法穿过液态外核的原因。

Write on your card that polarisation is the definitive test: only transverse waves can be polarised, so a wave that can be blocked completely by two crossed polarising filters must be transverse.

在卡片上写下这句话：偏振是决定性的判据——只有横波才能被偏振，所以能被两块正交偏振片完全挡住的波必定是横波。

4. The Wave Equation and Its Rearrangements | 波动方程及其变形

The wave equation links the three quantities that appear in nearly every waves calculation. You must be able to use it in all three rearrangements without hesitation, because exam questions rarely ask for speed directly.

波动方程把几乎每道波动计算题都会用到的三个量联系起来。你必须在三种变形之间毫不迟疑地切换，因为考题很少直接问波速。

v = f λ = λ / T     f = 1 / T     λ = v / f

00 × 10⁸ m s⁻¹, and sound in air at room temperature is about 340 m s⁻¹. For a stretched string the wave speed depends on tension and mass per unit length:

频率 50 Hz、波速 340 m s⁻¹ 的波，波长为 340 ÷ 50 = 6.8 m。要记住参考值：真空中光速 c = 3.00 × 10⁸ m s⁻¹，室温下空气中的声速约为 340 m s⁻¹。对于张紧的弦，波速取决于张力和线密度：

v = √T / μ

where T is the tension in newtons and μ is the mass per unit length in kg m⁻¹. This equation explains why tightening a guitar string raises its pitch.

其中 T 是张力（牛顿），μ 是线密度（kg m⁻¹）。这个方程解释了为什么拧紧吉他弦会提高音调。

5. Reading Wave Graphs Correctly | 正确读取波形图

Two graphs look almost identical but mean completely different things. A displacement-time graph has time on the horizontal axis: the repeat distance is the period T, and the gradient gives the velocity of one particle. A displacement-distance graph has position along the wave on the horizontal axis: the repeat distance is the wavelength λ, and the graph is effectively a photograph of the whole wave at one instant. Confusing the two is one of the most expensive errors in the topic.

两种图看起来几乎一样，含义却完全不同。位移—时间图横轴是时间：重复距离是周期 T，斜率给出单个质点的速度。位移—距离图横轴是波上的位置：重复距离是波长 λ，这张图实际上相当于某一瞬间整列波的"照片"。把两者混淆是这一专题里代价最高的错误之一。

now mark on a d-x graph. Also note that the wave speed and the particle speed are never the same thing: the wave moves forward at constant speed while each particle oscillates about a fixed equilibrium point.

在这两种图之间转换需要波速。如果位移—时间图给出 T = 0.02 s，波速为 340 m s⁻¹，则 λ = v T = 340 × 0.02 = 6.8 m，可以据此标注到位移—距离图上。还要注意，波速与质点速度永远不是一回事：波以恒定速度向前传播，而每个质点只是围绕固定平衡位置振动。

6. Reflection and Refraction | 反射与折射

Reflection obeys two laws: the incident ray, the reflected ray and the normal lie in the same plane, and the angle of incidence equals the angle of reflection, both measured from the normal. Refraction happens because the wave speed changes when it crosses a boundary. The frequency stays the same - the source has not changed - so if the speed drops, the wavelength must drop too.

反射遵守两条定律：入射光线、反射光线和法线在同一平面内；入射角等于反射角，两者都从法线量起。折射发生的原因是波越过界面时波速改变。频率保持不变——因为波源没有改变——所以波速减小时波长必然减小。

n₁ sin θ₁ = n₂ sin θ₂     n = c / v

Snell' medium (larger n, smaller v), the ray bends towards the normal; leaving it, the ray bends away. Air has n ≈ 1.00, water n ≈ 1.33, glass n ≈ 1.50, so light entering glass slows to about 2.0 × 10⁸ m s⁻¹.

斯涅尔定律会出现在公式表中，但你必须会用 n₁ sin θ₁ = n₂ sin θ₂ 这一形式，并能解释偏折方向：进入光密介质（n 更大、v 更小）时，光线向法线偏折；离开时则远离法线。空气 n ≈ 1.00，水 n ≈ 1.33，玻璃 n ≈ 1.50，因此光进入玻璃后速度降为约 2.0 × 10⁸ m s⁻¹。

7. Total Internal Reflection and the Critical Angle | 全反射与临界角

Total internal reflection occurs only when two conditions are met together: light travels from a denser medium towards a less dense one, and the angle of incidence exceeds the critical angle. At exactly the critical angle the refracted ray emerges along the boundary at 90°, which gives the working equation.

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全反射只有在两个条件同时满足时才会发生：光从光密介质射向光疏介质，且入射角大于临界角。恰好在临界角时，折射光线沿界面以 90° 射出，由此得到计算公式。

sin C = n₂ / n₁     (n₁ > n₂)

tions are optical fibres, where the signal stays trapped in the core, right-angled prisms used in periscopes and binoculars, and mirages above hot roads where light bends in layers of air of different density. Do not forget the word "total" acted at all beyond C, which is why optical fibres lose almost nothing along their length.

对玻璃到空气，sin C = 1.00 ÷ 1.50 = 0.667，所以 C ≈ 41.8°。经典应用包括光纤（信号被限制在纤芯中）、潜望镜和双筒望远镜中的直角棱镜，以及热路面上的海市蜃楼（光在不同密度空气层中弯曲）。不要忘记"全"字：超过临界角后完全没有折射光，这正是光纤长距离传输几乎不损耗的原因。

8. Superposition, Interference and Path Difference | 叠加、干涉与路程差

The principle of superposition states that when two waves meet, the resultant displacement at any point is the vector sum of the individual displacements. Constructive interference requires a path difference of a whole number of wavelengths; destructive interference requires an odd number of half wavelengths.

叠加原理指出：两列波相遇时，任意点的合位移等于各波单独位移的矢量和。相长干涉要求路程差为波长的整数倍；相消干涉要求路程差为半波长的奇数倍。

Constructive: Δx = n λ     Destructive: Δx = (n + ½) λ

spacing is given by λ = a x / D, where a is the slit separation, x the spacing between adjacent bright fringes and D the distance from the slits to the screen. Increasing a decreases x - a favourite "what happens if" question.

只有波源相干时干涉才是稳定的：频率相同、相位差恒定。在双缝实验中，条纹间距由 λ = a x / D 给出，其中 a 是缝间距，x 是相邻亮纹间距，D 是缝到屏的距离。增大 a 会使 x 减小——这是最常被问到的"若……会怎样"题型。

9. Standing Waves and Resonance | 驻波与共振

A standing wave forms when two identical waves of the same frequency travel in opposite directions through the same medium, usually an incident wave and its reflection. The result is a pattern of nodes, where displacement is always zero, and antinodes, where displacement is maximum. Nodes are spaced half a wavelength apart, and so are antinodes.

当两列频率相同的相同波在同一介质中沿相反方向传播（通常是一列入射波和它的反射波）时，就形成驻波。其结果是节点（位移始终为零）和腹点（位移最大）组成的图案。相邻节点间距为半个波长，腹点也是如此。

String or open pipe: L = n λ / 2     fₙ = n v / (2 L)

Closed pipe: L = (2n − 1) λ / 4

A string fixed at both ends and a pipe open at both ends share the same condition and produce all harmonics. A pipe closed at one end always has a node at the closed end and an antinode at the open end, so it only produces odd harmonics, and its fundamental frequency is half that of an open pipe of the same length. Resonance occurs when a driving frequency matches one of these natural frequencies, producing a large amplitude vibration - the physics behind musical instruments and the Tacoma Narrows bridge collapse.

两端固定的弦和两端开口的管遵循同一条件，能产生所有谐波。一端封闭的管在封闭端必为节点、开口端必为腹点，因此只产生奇次谐波，而且其基频是同长度开管的一半。当驱动频率与这些固有频率之一相同时就发生共振，产生大幅度振动——这正是乐器发声和塔科马海峡大桥坍塌背后的物理原理。

10. The Doppler Effect | 多普勒效应

The Doppler effect is the apparent change in observed frequency when the source, the observer, or both move relative to the medium. The observed frequency rises as source and observer approach, and falls as they separate. Crucially, the actual frequency emitted by the source never changes - only what is received changes.

多普勒效应是指当波源、观察者或两者相对介质运动时，观察到的频率发生表观变化。当波源与观察者相互靠近时观测频率升高，相互远离时降低。关键在于：波源实际发出的频率从不改变——改变的只是接收到的频率。

f′ = f (v ± vo) / (v ∓ vs)

Use the upper signs when source and observer approach each other, the lower signs when they separate. Everyday examples include the drop in pitch of an ambulance siren as it passes you and the redshift of light from receding galaxies, which is direct evidence for the expansion of the universe. In a card answer, always state which signs you chose and why.

相互靠近时取上面的符号，相互远离时取下面的符号。日常例子包括救护车驶过时警笛音调骤降，以及退行星系光谱的红移——这是宇宙膨胀的直接证据。在卡片答案中，一定要说明你选了哪组符号以及理由。

11. Common Exam Traps | 常见考试陷阱

Most lost marks in this topic come from a small number of repeated mistakes. Put each of these on its own card, with the correction written underneath in a different colour.

这一专题丢分大多来自少数几个反复出现的错误。把每一条单独做成一张卡片，下面用另一种颜色写上改正方法。