Edexcel物理 Waves 波长 频率 折射 地震波

Edexcel物理 Waves 波长 频率 折射 地震波

Edexcel GCSE Physics 的 Topic 4 讲 Waves，考纲 Issue 4（2024 年 3 月）把这一章拆成 4.1 到 4.17 十几个条目。前几条讲波怎么描述，中间几条讲波速怎么算、怎么测，后面几条讲折射、声波、超声与地震波。很多学生以为这一章只要背下 v = f x lambda 就够了，结果在解释题上丢分。物理单科还要额外考标 P 的条目，Combined Science 的学生不会碰到这些条目。Paper 1 覆盖 Topic 1 到 Topic 7，Waves 是其中第四块。下面按考纲顺序把每个条目落到具体题目和算例上。

Topic 4 of the Pearson Edexcel GCSE Physics specification covers waves in a sequence of numbered points, running from the vocabulary of wave motion to refraction, sound, ultrasound and seismic waves. The paper rewards students who can move between a description and a calculation without pausing to look anything up; the physics-only points marked P are examined in the separate physics course only. What follows works through each point with the readings and numbers that actually turn up in questions.

波传递能量和信息，不搬运物质

考纲 4.1 的说法是波传递能量与信息，但不传递物质。水波经过时，水面上的一片叶子上下浮动，位置几乎不变，说明移动的是波形而不是水本身。声波经过时空气分子只在平衡位置附近来回振动，也没有整团空气从说话的人那里跑到你耳朵里。地震发生时地面起伏传得很远，土壤本身并没有被搬到几百公里外。这些都是 4.2 要求写出的证据，答题时要讲清振动与传播是两件事。很多学生以为波就是物质在往前跑，写解释题时把这两件事混成一句。阅卷时看到只写能量传递而不提振动方向的答案，通常拿不到满分。

A wave is a disturbance that carries energy and information from place to place while the medium itself stays where it is. A floating leaf on a pond rises and falls as ripples pass, yet it does not surf towards the bank, and that observation is the evidence 4.2 asks you to describe. Sound behaves in the same way: air molecules oscillate about fixed positions, so no parcel of air travels from a speaker to a listener. Examiners want that distinction spelled out rather than the bare phrase energy transfer.

横波与纵波：看振动方向与传播方向

4.5 要求用质点振动方向和波传播方向的关系区分横波与纵波。横波里振动方向与传播方向垂直，绳子上抖出的一道波峰就是典型例子。纵波里质点沿传播方向来回压缩和拉伸，形成疏密相间的区域。声波是纵波，考试里最常拿来比较的就是它。电磁波是横波，光、无线电、X 射线都归这一类。地震波两种都有，P 波是纵波，S 波是横波，这也解释了它们到达时间不同。水面波看起来像横波，严格说水质点在画圆周，GCSE 阶段按横波处理即可。这一步最容易出错的地方是把横波说成上下移动的波，而题目问的是振动方向。

The classification in 4.5 rests on one comparison: the direction in which particles oscillate against the direction in which the wave travels. Where the two are perpendicular the wave is transverse, as with a pulse sent along a rope or a beam of light from a torch. Where particles oscillate along the direction of travel the wave is longitudinal, producing compressions and rarefactions, which is exactly what sound does. Seismic waves appear in both forms: P waves are longitudinal and arrive first, while S waves are transverse and follow behind.

波长、频率、振幅、周期与波前

4.3 和 4.4 把这些术语的用法定死了。频率是一定时间内通过某点的完整波数，单位赫兹，等于每秒的波数。波长是相邻两个同相点之间的距离，比如两个波峰之间，单位是米。振幅是质点离开平衡位置的最大距离，它决定波携带的能量大小，与波长无关。周期是一个完整波通过所用的时间，所以周期与频率互为倒数。波前是连接同一时刻所有同相点的假想线，画图题里用来表示传播方向。常见的错误是把振幅写成两个波峰之间的距离，这等于把振幅和波长混成一个。单位也要跟着查，厘米和毫米换成米之后数值才对。

Frequency counts complete waves passing a point each second and is measured in hertz, while wavelength measures the distance between one crest and the next, in metres. Amplitude is the maximum displacement of a particle from its rest position and controls how much energy the wave carries, so it has nothing to do with wavelength. Period is the time taken for one complete wave to pass, which makes it the reciprocal of frequency. Wavefronts are drawn as lines joining points that are in phase, and in ray diagrams they mark the direction of travel.

两个公式：v = f x lambda 与 v = x / t

4.6 要求同时掌握两个方程，v = f x lambda 与 v = x / t。前者把波速、频率、波长连在一起，后者用距离除以时间求速度。两条式子说的是同一个物理量，用哪一条取决于题目给了什么。比如水波频率 25 Hz，波长 12 mm，先把 12 mm 换成 0.012 m，乘得 0.30 m/s。反过来给波速 340 m/s 和频率 170 Hz，波长就是 2.0 m。单位不统一是这一节最常见的失分点，答案写 0.3 而不写单位同样扣分。计算过程建议先写公式再代数字，阅卷按公式给方法分。频率与周期互换算用 f = 1 / T，别把倒数算反。

Two equations cover every wave calculation on this paper, and choosing between them is the first decision: v equals f times lambda links speed to frequency and wavelength, while v equals x over t gives speed from a measured distance and time. Convert units before substituting, because millimetres and centimetres are the usual trap; a wavelength of 12 mm is 0.012 m, so 25 Hz gives 0.30 m/s. Working the other way, 340 m/s divided by 170 Hz leaves a wavelength of 2.0 m. Writing the equation first and the numbers second keeps the method marks safe even when the arithmetic slips.

核心实验 4.17：水波槽里量出波速

4.17 是八个必做核心实验之一，要求判断设备是否适合测量固体与流体中的波速、频率和波长。水波槽里沿槽边放一把直尺，用电机带动振子产生平行波前，先数 10 秒内通过的波数再除以 10，就得到频率。波长用相邻波峰的距离读出，拍照冻结波形读数更稳。另有一种测速办法，在槽边标两个点，量出距离，用秒表记一个波峰走过这段距离的时间，按 v = x / t 算。两种方法给出的速度互相校核，接近说明读数可靠。常见的错误是只用秒表追一个波峰，反应时间占的比例太大，读数就不可信了。固体一侧用振动发生器带动绳子或金属杆，频率直接从信号发生器读，比肉眼数波靠谱得多。

The core practical asks you to judge how suitable each set of equipment is for measuring speed, frequency and wavelength in a solid and in a fluid, not simply to take one reading. In a ripple tank a ruler gives crest spacing, and counting the waves passing a mark in ten seconds gives frequency once you divide by ten, while a photograph freezes the pattern so the spacing can be read without blur. Speed can also come from timing one crest between two marks a measured distance apart, and the two estimates should agree. Chasing a single crest with a hand-held stopwatch is the weak link, since reaction time is a large fraction of such a short interval, which is why a signal generator driving a string gives a far better frequency reading in the solid case.

声速在空气中的两种测法

4.7 要求会描述测量空气中声速与水面波速的做法。空气里常用两个学生配合，相隔一段已知距离，一个拍手或敲击木块，另一个看着动作开始计时，听到声音时停表。距离除以时间就得到声速，实验里常用 100 m 距离，得到 330 m/s 左右的数值。这种办法的误差来自人的反应时间，距离越短误差占比越大。另一种做法用回声，对着远处的墙或山崖发声，记下往返时间，距离要按往返算。超声测距是同一思路，4.8P 就要求用时间和波速求深度或距离。很多学生以为声速在空气里永远是 340 m/s，于是不看题干给什么值，阅卷时这类答案一律按题给数据处理。

Sound speed in air is usually measured by two students standing a known distance apart: one claps or strikes two blocks, the other starts a stopwatch on seeing the action and stops it on hearing the sound. Dividing the distance by the time gives a value near 330 m/s when the gap is about 100 m, and shortening that gap makes reaction time a larger share of the error. An echo method works the same way, except the distance must be doubled because the sound travels out to the reflector and back. Questions on 4.8P apply this to depth and distance, and the value to use is always the one printed in the question.

折射的成因：速度改变导致方向改变

4.10 要求用速度变化解释折射。波从一种介质进入另一种介质时速度改变，如果斜着射入界面，方向就会偏折。光从空气进入玻璃速度变小，折射光线向法线靠拢。光从玻璃射入空气速度变大，折射光线远离法线。垂直射入时不偏折，因为入射角为零，速度变化只改变波长。声波从空气进入水中速度变大，同样会偏离法线，水下判断声音方向容易错就是这个原因。常见的错误是只写光被玻璃弯曲，不写速度变化，这类答案通常只拿一分。4.11 还要求记住不同物质对不同波长的吸收、透射、折射与反射会不一样。

Refraction happens because speed changes at a boundary, and the change of direction follows from that rather than from the material bending the wave. Light slows on entering glass, so a ray arriving at an angle bends towards the normal; leaving glass it speeds up and bends away. A ray arriving along the normal is not deviated at all, since there is no angle to change, although its wavelength still alters. Point 4.11 adds that absorption, transmission, reflection and refraction all depend on wavelength, which is why a prism spreads white light into a spectrum.

界面上的四种结果：反射、折射、透射、吸收

4.9P 把界面上的结果分成反射、折射、透射、吸收四种，答题时要挑对词。反射是波返回原介质，角度关系与光相同。折射是进入新介质并改变方向。透射是穿过介质继续传播，方向可能不变也可能变。吸收是波的能量转成介质内能，比如黑色表面吸收可见光后温度升高。同一个界面可以同时发生几种结果，一块玻璃既反射一部分光又让大部分光透过。这几条只在物理单科考，标 P 的条目在 Combined Science 里不出现。写解释题时把能量去了哪里写清，比堆术语更容易拿分。

At any interface a wave can be reflected, refracted, transmitted or absorbed, and the mark often depends on naming the right one. Reflection sends the wave back into the original medium, refraction turns it as it enters the new one, transmission lets it carry on through, and absorption converts its energy into internal energy in the material. A sheet of glass does several of these at once, reflecting a little light while transmitting most of it. Points marked P belong to the physics-only course, so candidates sitting Combined Science will not meet them in the exam.

声波跨介质：频率不变，速度和波长变

4.16P 要求说明声波从一种介质传到另一种介质时速度、频率、波长怎么变。频率由声源决定，穿过界面时不变。速度由介质决定，固体里最快，液体次之，气体最慢。波长由 v = f x lambda 决定，速度变了波长跟着变。这三点连起来能解释隔墙听到的声音为什么发闷，固体传声效率为什么更高。假如频率也跟着变，音调就会改变，实际听不到这种现象，这条推理常用来验证频率不变。常见的错误是三种量一起放大，忘了频率是声源属性。答题时先把不变量抓住，另外两个按公式推。

When sound crosses into a new medium its frequency stays the same because the source fixes it, while speed is set by the medium: fastest in solids, slower in liquids, slowest in gases. Wavelength therefore has to change, since v equals f times lambda and only two of the three quantities can be independent. That single rule explains why a voice heard through a wall sounds muffled rather than higher or lower. Students who change all three quantities at once lose the mark, because pitch would then shift, and it does not.

超声与次声：20000 Hz 与 20 Hz 两条线

4.13P 把高于 20000 Hz 的声波叫超声，4.14P 把低于 20 Hz 的叫次声。人耳能听到的范围是 20 Hz 到 20000 Hz，超出这个范围的声波听不到但依然存在。4.15P 要求会解释用途，声呐用超声反射测距离和鱼群位置，胎儿扫描用超声成像，探测地球内部结构靠地震波这类低频振动。超声方向性强、波长短，能分辨很小的结构，这是它用于成像的原因。次声传播远、衰减慢，可用来监测火山和地震。这两条只在物理单科考，写题时把频率范围和用途一一对上。

Sound above 20000 Hz is ultrasound and sound below 20 Hz is infrasound, with human hearing filling the band between those two limits. Ultrasound is useful because short wavelengths resolve small structures and the beam can be aimed, which underlies sonar ranging and foetal scanning. Infrasound travels long distances with little loss, so it is used to monitor volcanic activity and earthquakes, and seismic waves of this kind also reveal the structure of the Earth's interior. Both limits are physics-only content, and questions usually pair a frequency with a use.

地震波 P 波与 S 波：为什么 S 波过不了液态外核

4.5 要求用地震波举例说明横波与纵波的区别。P 波是纵波，速度大约每秒 6 到 8 公里，能穿过固体和液体。S 波是横波，速度慢一些，只能穿过固体。地震台站记录到 P 波先到、S 波后到，时间差可以用来估算震中距离，这就是 4.8P 的用武之地。地球上有些区域收不到 S 波，因为外核是液态，横波无法在其中传播，这条证据支持了外核为液态的结论。4.15P 提到的地球内部探测就建立在这种观测上。算距离时把 v = x / t 变形，注意时间要用单个波的传播时间，别把两个波的到达时间差直接代进去。

Earthquake waves illustrate both types at once: P waves are longitudinal, travel at roughly 6 to 8 km per second and pass through solids and liquids, while S waves are transverse, slower, and confined to solids. Stations record the P arrival first and the S arrival later, and that gap feeds the distance calculation in 4.8P. The shadow zones where S waves never arrive are the evidence that the outer core is liquid, since a transverse wave cannot cross a fluid. Rearranging v equals x over t gives the distance, but the time to use is the travel time of one wave, not the gap between the two arrivals.

计算题规范：单位换算、回声除以二、有效数字

Waves 的计算题失分很少因为不会公式，多是因为步骤不完整。单位先统一到米、秒、赫兹，答案后面写单位。回声和超声测距要记住波走的是往返，深度等于速度乘时间再除以二。超声在水中速度约 1500 m/s，往返用时 0.08 s，总路程 120 m，深度就是 60 m。有效数字按题干给的数据定，给两位就写两位，中间不要提前四舍五入。Paper 1 是 1 小时 45 分钟、100 分、10 道题，Waves 通常落在前面的短答和后面的解释题里。考纲没有公布每个 Topic 的分值表，按预测题量安排复习并不靠谱。

Marks on wave calculations are usually lost in the method rather than the formula: convert to metres, seconds and hertz first, then attach the unit to the answer. Echo and ultrasound questions double the distance, so depth is speed times time divided by two; at 1500 m/s in water with a 0.08 s round trip the path is 120 m and the depth 60 m. Significant figures should match the data given, and rounding halfway through a calculation is a common way to lose the final mark. Paper 1 runs for 1 hour 45 minutes with 100 marks across ten questions, and the specification publishes no marks quota per topic, so predicting question counts is a poor way to plan revision.

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