WJEC ALevel Physics Waves 横波驻波 考点解析
波动是 WJEC A-Level 物理大纲中的核心章节，覆盖横波与纵波、波的描述、驻波、叠加与干涉、衍射等一连串考点。它既出现在选择题里考概念辨析，也出现在计算题中要求熟练运用 v = fλ。本文用中英双语把这一章的知识点与解题套路完整梳理一遍，帮助你从"看得懂公式"进阶到"拿得到分数"。
Waves are one of the most heavily examined topics in WJEC A-Level Physics. From classifying transverse and longitudinal waves to solving stationary wave problems, examiners test both conceptual understanding and calculation fluency. This bilingual guide walks through every core idea in order, with worked examples and exam tips, so you can turn formula familiarity into real marks.
1. 波的分类：横波与纵波
按质点振动方向与波的传播方向的关系，机械波分为两类。横波中质点振动方向与传播方向垂直，例如绳波、水波、以及电磁波；横波的明显特征是能够发生偏振。纵波中质点振动方向与传播方向平行，例如声波；纵波表现出疏部和密部，即压缩区与稀疏区交替前进。
In a transverse wave, particles vibrate at right angles to the direction of energy transfer, as seen in a wave on a string, on water, or in electromagnetic radiation. Polarisation is only possible for transverse waves, which is a key distinction to remember. In a longitudinal wave, particles vibrate parallel to the direction of travel, as in sound, producing alternating compressions and rarefactions.
考试中常见的一种出法，是给一幅波形图或一段文字描述，让你判断它是横波还是纵波，并说明判断依据。回答时要点出"振动方向与传播方向是否垂直或平行"，必要时补充"只有横波可以偏振"。
A common exam question shows a diagram or description and asks whether the wave is transverse or longitudinal. State clearly whether particle vibration is perpendicular or parallel to the direction of travel, and add the polarisation point when a transverse wave is involved.
2. 描述波的五个量：振幅 波长 周期 频率 相位
振幅 A 是质点偏离平衡位置的最大距离，单位米，它决定波携带的能量，波的能量与振幅的平方成正比。波长 λ 是相邻两个同相点之间的距离，例如相邻两个波峰或波谷之间的距离。周期 T 是质点完成一次完整振动所需的时间，单位秒。频率 f 是每秒振动的次数，单位赫兹，f 与 T 互为倒数：f = 1/T。相位描述质点在振动周期中所处的位置，相位差对应两个质点振动步调的差异，通常用弧度或角度表示。
Amplitude A is the maximum displacement from equilibrium; wave energy is proportional to A squared. Wavelength λ is the distance between two adjacent points in phase, such as two successive crests. The period T is the time for one complete oscillation, and the frequency f = 1/T is the number of oscillations per second. Phase describes where a particle is within its cycle, and phase difference compares how two particles, or two waves, are out of step, usually measured in radians or degrees.
图像题是这一节的主要题型。位移-时间图描述的是某一个质点随时间如何振动，读出的横轴间隔是周期 T；位移-距离图是某一瞬间沿波传播方向各质点的位置快照，读出的横轴间隔是波长 λ。把这两种图混淆是失分最多的原因之一。
Graph questions dominate this section. A displacement-time graph tracks one particle over time, so the horizontal spacing gives the period T. A displacement-distance graph is a snapshot of the whole wave at one instant, so the horizontal spacing gives the wavelength λ. Mixing up these two graphs is one of the most common mark-losers in the exam.
关于相位，常考"某两点间的相位差是多少"。如果两点的间距是 n 个波长，则相位差为 2πn 弧度或 360n 度；相距半个波长的两点相位差为 π 弧度，振动步调完全相反，称为反相。
For phase difference questions, remember that a separation of one wavelength corresponds to a phase difference of 2π radians, or 360 degrees. Two points half a wavelength apart oscillate exactly out of phase, with a phase difference of π radians, and are said to be in antiphase.
3. 核心公式 v = fλ 与波速的决定因素
所有波的传播速度、频率与波长之间都满足波速公式：v = fλ，也可以写成 v = λ/T。频率由波源决定，波从一个介质进入另一个介质时频率不变；波速由介质本身的性质决定，因此波长会随介质改变而改变。例如光从空气进入水中，速度变小，频率不变，于是波长变短。
Every wave obeys the wave equation v = fλ, equivalently v = λ/T. The frequency is fixed by the source and never changes when the wave enters a new medium, while the speed is determined by the medium itself. This is why light entering water slows down and its wavelength shortens while its frequency stays the same.
典型计算：一列声波在空气中传播，频率为 425 Hz，测得相邻波峰之间的距离为 0.80 m，求波速。解：v = fλ = 425 × 0.80 = 340 m/s。这个结果与声波在空气中的常见速度吻合，说明计算正确。注意计算前先把所有单位换算成国际单位，厘米要除以 100 变成米。
Worked example: a sound wave in air has frequency 425 Hz and adjacent crests are 0.80 m apart. Using v = fλ gives v = 425 × 0.80 = 340 m/s, which matches the typical speed of sound in air, a good sanity check. Always convert every quantity to SI units before calculating, for example dividing centimetres by 100 to obtain metres.
进阶考点：绳上横波的波速由张力与线密度决定，张力越大波速越快，线密度越大波速越慢。这类定性判断在 WJEC 简答题中很常见，回答时把"波速由介质性质决定"这句话写出来即可得分。
For waves on a string, the speed increases with tension and decreases with linear density. These qualitative points appear regularly in WJEC short-answer questions, so the sentence "wave speed depends on the properties of the medium" is often enough to secure the mark.
4. 驻波：节点 腹点 与弦上的谐波
两列频率相同、振幅相同、传播方向相反的波叠加，会形成驻波。驻波的外形看起来"定住不动"，波腹处振幅最大，波节处振幅始终为零，相邻波节之间的距离等于半个波长。驻波与行波的最大区别是：驻波不传递能量，能量被束缚在波节与波腹之间来回振荡。理解这一点，是概念题拿分的关键。
A stationary wave forms when two waves of the same frequency and amplitude travel in opposite directions and superpose. It appears frozen in place: antinodes vibrate with maximum amplitude while nodes stay permanently at zero displacement, and the distance between adjacent nodes is half a wavelength. Crucially, a stationary wave transfers no energy, which is the single most examined conceptual point.
两端固定的弦上可以形成一系列驻波模式，称为谐波。基频对应波长 λ = 2L，即弦长等于半个波长；第 n 次谐波的波长满足 λ = 2L/n，其中 L 是弦长。开口管与闭口管的驻波边界条件不同：闭口管的末端必须是波节，因此只能产生奇数次谐波。
A string fixed at both ends supports a family of standing wave patterns called harmonics. The fundamental has wavelength λ = 2L, so the string holds half a wavelength; the nth harmonic satisfies λ = 2L/n, where L is the string length. Pipes behave differently at each end: a closed end must be a node, so a pipe closed at one end can only produce odd harmonics.
典型计算：一根长 1.5 m 的弦两端固定，以第三谐波振动，波速为 60 m/s，求振动频率。解：第三谐波 λ = 2L/3 = 2 × 1.5/3 = 1.0 m；f = v/λ = 60/1.0 = 60 Hz。这类题目的套路是先由谐波次数写出波长，再用 v = fλ 求频率，两步之间不要跳步。
Worked example: a string of length 1.5 m is fixed at both ends and vibrates in its third harmonic with wave speed 60 m/s. For the third harmonic, λ = 2L/3 = 2 × 1.5/3 = 1.0 m, so f = v/λ = 60/1.0 = 60 Hz. The routine is always the same: write the wavelength from the harmonic number first, then apply v = fλ. Do not skip the first step.
5. 叠加原理与干涉
叠加原理指出：当两列或多列波在空间某点相遇时，该点的合位移等于各列波单独存在时在该点位移的矢量和。干涉是叠加的结果：当两列相干波在一点同相相遇，振幅相加，形成相长干涉，合振幅最大；当它们反相相遇，振幅相减，形成相消干涉。相干波必须频率相同且相位差恒定。
The principle of superposition states that when waves meet at a point, the resultant displacement is the vector sum of the individual displacements. Interference follows directly: coherent waves arriving in phase interfere constructively, giving maximum amplitude, while waves arriving in antiphase interfere destructively. Coherent sources must have the same frequency and a constant phase difference.
用光程差判断干涉类型是最常见的计算题：如果两列波到达某点的路程差等于波长的整数倍 nλ，则该点相长；如果路程差等于半波长的奇数倍 (n + 1/2)λ，则该点相消。典型计算：两相干波源发出波长为 0.25 m 的波，某点到两波源的路程差为 0.50 m，问该点振动加强还是减弱。解：0.50/0.25 = 2，是波长的整数倍，因此该点是加强点。
Interference questions are usually solved by comparing path difference with wavelength. A path difference of nλ gives constructive interference, while (n + 1/2)λ gives destructive interference. Example: two coherent sources emit waves of wavelength 0.25 m, and a point is 0.50 m further from one source than the other. Since 0.50/0.25 = 2 is a whole number of wavelengths, the point vibrates constructively and the amplitude is a maximum.
6. 衍射与衍射光栅
衍射是波绕过障碍物或通过狭缝后发生弯曲传播的现象。衍射是否明显，取决于障碍物或缝的尺寸与波长的相对大小：缝越窄、波长越长，衍射越明显。光通过单缝发生衍射后，屏幕上出现明暗相间的条纹，中央明纹最宽最亮。声波波长约为几分米到几米，容易绕过门窗，所以隔墙能听到声音，这正是波长较长导致的明显衍射。
Diffraction is the spreading of waves as they pass an obstacle or through a gap. It is noticeable only when the gap or obstacle size is comparable to the wavelength. Light diffracted by a narrow single slit produces bright and dark fringes, with the central maximum the widest and brightest. Sound waves have wavelengths of decimetres to metres, which is why you can hear through a doorway: the wavelength is long enough to diffract strongly.
A2 阶段的衍射光栅公式是 d sinθ = nλ，其中 d 是相邻狭缝的间距，θ 是第 n 级明纹的衍射角。典型计算：一块光栅每毫米有 500 条刻线，用波长为 550 nm 的光垂直照射，求第一级明纹的衍射角。解：d = 1/500 mm = 2.0 × 10-6 m，即 2.0 μm；sinθ = λ/d = 550 × 10-9 / (2.0 × 10-6) = 0.275，因此 θ 约等于 16 度。注意刻线密度的单位换算，每毫米 N 条线时 d = 10-3/N 米。
At A2 level the diffraction grating equation d sinθ = nλ relates slit spacing d to the angle θ of the nth order maximum. Example: a grating has 500 lines per millimetre and is illuminated normally by light of wavelength 550 nm. Then d = 1/500 mm = 2.0 μm, and sinθ = λ/d = 550 × 10-9 divided by 2.0 × 10-6 = 0.275, so θ is about 16 degrees. Watch the unit conversion: with N lines per millimetre, d = 10-3/N, in metres.
7. 考试技巧与常见错误
总结 WJEC 波动章节的高频失分点。第一，混淆位移-时间图和位移-距离图，导致把周期读成波长；应对方法是先看横轴单位，时间是秒就找 T，距离是米就找 λ。第二，忘记 f = 1/T 或换算单位出错，尤其是厘米与米的换算、以及波长纳米与米的换算，10-9 这个因子经常被漏掉。第三，驻波概念题答不出"不传递能量"这个要点。第四，干涉判断只写结论不写过程，题目要求 show or explain 时，必须写出路程差与波长比较的完整推导。
The most common mark-losers in WJEC wave questions are: confusing displacement-time graphs with displacement-distance graphs, so a period is misread as a wavelength; forgetting f = 1/T or mishandling unit conversions, especially the factor of 10-9 when converting nanometres to metres; missing the point that stationary waves transfer no energy; and writing only a conclusion for interference questions without showing the path difference comparison when the command word is explain or show.
答题时还要注意 WJEC 的命令词。State 只需要一句话结论；Calculate 需要公式、代入数值与带单位的答案；Explain 必须给出物理原因，不能只重复现象；Derive 要求从基础关系一步步推出公式。计算题最后检查两件事：答案是否有单位，有效数字是否与题目数据一致。
Read the command words carefully. State needs a one-sentence answer; Calculate requires a formula, substitution and an answer with units; Explain demands a physical reason rather than a restatement; Derive means building the result step by step from first principles. Before finishing any calculation, check that your answer has units and that the significant figures match the data given in the question.
波动章节的知识点环环相扣：先分清横波纵波，再掌握描述波的五个量，然后以 v = fλ 为桥梁连接驻波、干涉与衍射。建议把本文中的四道例题独立重算一遍，再配合真题练习，正确率会明显提升。需要系统辅导或者对某个考点有疑问，都可以联系我们，一对一帮你把薄弱环节补起来。
Every idea in this topic connects: classify the wave, describe it with the five quantities, then use v = fλ as the bridge into stationary waves, interference and diffraction. Recompute the four worked examples in this guide by yourself, then practise with past paper questions, and your accuracy will improve quickly. If you want structured help or are stuck on a particular point, contact us and we can work through your weak areas one on one.
Cambridge International AS ALevel Physics Coursebook 正版剑桥国际AS ALevel考试物理教材带线上账号 英文原版¥557.0📱 去京东购买京东精选需要一对一辅导？Need One-on-One Tutoring?
A-Level / GCSE / IB / KS2-KS3 全科辅导
或截图后用「扫一扫 → 相册」选择本图