WJEC物理 简谐运动 共振 引力场 电磁感应
WJEC 的 A Level 物理把振动放在 A2 的 Unit 3，把场放在 Unit 4，两卷加起来占整科 300 分里的 100 分。这两块内容看着分散，其实共用同一套工具：位移、速度与加速度之间的微积分关系，以及场强与势之间的斜率关系。下面按 WJEC 官方 spec 的单元顺序拆成 12 个板块，每个板块给出公式、可以手动核对的数值例子，以及阅卷时真正扣分的位置。文中出现的分值、时长与单元名称都取自 WJEC 官方 spec 与历年真题卷，可以自己回查。
WJEC puts oscillations in the A2 Unit 3 paper and fields in Unit 4, and the two papers together carry 100 of the 300 marks in the A Level; they also share one toolkit, namely the calculus linking displacement, velocity and acceleration and the slope that links field strength to potential. The twelve blocks below follow the official unit order and give the formulas, the numbers that can be checked by hand, and the places where marks are actually lost. Every mark tariff, duration and unit name quoted here comes from the WJEC specification and the published past papers.
卷面位置：Unit 3 与 Unit 4 的分值分布 / Where the Marks Sit
WJEC 的 A Level 物理课程代码是 1420QS，A2 单元在 2017 年夏季首次认证。Unit 3 是 Oscillations and Nuclei，2 小时 15 分钟、100 分，占整科 25%，主题为圆周运动、振动、气体动理论、热学、核衰变与核能。Unit 4 是 Fields and Options，2 小时、100 分、25%，四类场的内容加最后 20 分的选项题。两卷都是 Section A 80 分必答、Section B 20 分，June 2024 的 Unit 3 卷（1420U30-1）Section A 七道题分值为 10、10、11、23、7、7、12。振动题量通常在 10 到 20 分之间，那道 23 分长题常把气体动理论与热学串起来考。很多学生以为背熟公式就够了，实际上这两卷一半以上的分值写在推导过程、单位与有效数字上。
WJEC's A Level Physics carries the code 1420QS and first awarded its A2 units in summer 2017; Unit 3 is Oscillations and Nuclei at 2 hours 15 minutes and 100 marks, which is 25% of the A Level. Unit 4 is Fields and Options at 2 hours and 100 marks, built from four field topics plus a 20-mark option section, and both papers open with an 80-mark Section A; the June 2024 Unit 3 paper (1420U30-1) split that Section A as 10, 10, 11, 23, 7, 7 and 12. The official Summer 2024 examiner report notes that recall questions keep scoring below data-driven ones, so the working and the units matter as much as the formulas.
简谐运动的判据与三个极值 / The Test for Simple Harmonic Motion
简谐运动的判据只有一句话：加速度与离平衡位置的位移成正比，方向相反，写成 a = -ω^2 x，其中 ω 是角频率，单位 rad/s。很多学生以为来回摆动就是简谐运动，判断时只看图像像不像正弦曲线，反而漏掉题目要的那句线性关系与负号。角频率与周期的关系是 ω = 2π/T，周期 0.50 s 的振子 ω 等于 2π 除以 0.50，也就是 12.6 rad/s。位移取 x = A cos(ωt) 时速度是它的导数，v = -ωA sin(ωt)，振幅 0.05 m 时最大速度 v_max = ωA = 0.63 m/s。加速度是速度的导数，a_max = ω^2 A = 158 乘以 0.05，等于 7.9 m/s^2。位移最大处速度为零而加速度最大，位移为零处速度最大而加速度为零，这组对应关系在图像题里反复出现。
Simple harmonic motion is defined by one condition, that the acceleration is proportional to the displacement from equilibrium and directed towards it, written a = -ω^2 x, where ω is the angular frequency in rad/s. Angular frequency links to the period through ω = 2π/T, so a 0.50 s oscillator runs at 12.6 rad/s, and differentiating x = A cos(ωt) gives v = -ωA sin(ωt) with a maximum speed of 0.63 m/s at an amplitude of 0.05 m. Acceleration peaks at 7.9 m/s^2, and pairing zero speed with maximum acceleration at the extremes is the pattern that graph questions keep testing.
位移、速度、加速度图像的相位关系 / Phase in the Three Graphs
三条曲线的相位关系是固定考点：速度比位移超前四分之一个周期，加速度与位移反相。位移图从正向最大值开始时，位移用余弦描述，速度用负正弦，加速度用负余弦，写完三条式子图像就能直接画出来。t = 0 时振子在正向最大位移，四分之一周期后回到平衡位置且速度最大，半个周期后到达负向最大位移。判断方向看斜率比看数值快，位移图斜率为正说明速度为正，斜率为零的位置速度取极值。阅卷时最常见的失分是只写最大值不写方向，题目问 velocity 时答案必须带正负号或者"向左"这样的方向词。另一种常见的错误是把加速度图画成与位移同相，实际上两者的符号永远相反，负号就写在这个判据里。
Speed leads displacement by a quarter of a cycle and acceleration is in antiphase with displacement, so a graph starting at the positive extreme can be written as a cosine, a negative sine and a negative cosine in turn. At t = 0 the oscillator sits at maximum positive displacement, reaches equilibrium with maximum speed a quarter of a cycle later, and arrives at the negative extreme after half a period. Marks are lost when a value is quoted without a direction, and the commonest graph error is drawing acceleration in phase with displacement rather than in antiphase.
弹簧振子与单摆的两个周期公式 / Spring and Pendulum Periods
两个周期公式要背到不用想：弹簧振子 T = 2π√(m/k)，单摆 T = 2π√(l/g)。弹簧的周期只取决于质量与劲度系数，竖直悬挂时重力只是把平衡位置下移，并不改变周期。0.25 kg 配上 40 N/m 的弹簧，T 约 0.50 s，与上一段的 12.6 rad/s 正好对应。单摆 0.80 m，T = 2π√(0.80/9.81)，约 1.79 s，频率 0.56 Hz。单摆公式只在小角度下成立，10 度以内误差不到百分之一，摆角越大测出的周期越长。学生最常把周期当成与摆球质量有关，或者把摆长量成悬线长度漏掉小球半径，两个错在实验题里都直接扣分。
The spring period is T = 2π√(m/k) and the pendulum period is T = 2π√(l/g), and the spring value depends only on mass and stiffness because gravity merely shifts the equilibrium position downwards. A 0.25 kg mass on a 40 N/m spring gives about 0.50 s, while a 0.80 m pendulum gives about 1.79 s, or 0.56 Hz. The pendulum formula holds within about 1% up to 10 degrees, and plotting period squared against length gives a gradient of 4π^2/g, which removes the need to square-root every reading.
振动中的能量转换 / Energy in an Oscillator
无阻尼振子的总能量恒定，动能是 ½mω^2(A^2 - x^2)，势能是 ½mω^2x^2，两者相加等于 ½mω^2A^2。用 0.25 kg、ω = 12.6 rad/s、振幅 0.05 m 的振子算一遍，总能量约 0.049 J。位移等于半振幅 0.025 m 时，势能占总量的四分之一约 0.012 J，动能占四分之三约 0.037 J，速度是 0.54 m/s。这组比例是固定结论，半振幅处动能一定是总量的四分之三，不用重新推。很多学生以为最大动能要另列公式，其实就是把总能量里的 x 取零，结果同样是 ½mω^2A^2。错得最多的是在势能公式里漏掉平方，或者把 ½kx^2 与 ½mω^2x^2 混用而忘了 k 与 ω 的换算。
An undamped oscillator keeps a constant total energy of ½mω^2A^2, split into kinetic energy ½mω^2(A^2 - x^2) and potential energy ½mω^2x^2. For a 0.25 kg mass with ω = 12.6 rad/s and an amplitude of 0.05 m the total is about 0.049 J, and at half amplitude the potential energy holds a quarter of that while the kinetic energy holds three quarters, giving a speed of 0.54 m/s. The maximum kinetic energy is the same expression with x set to zero, so no separate formula is needed.
阻尼的三种形态 / Light, Heavy and Critical Damping
阻尼来自摩擦与空气阻力，能量被移出振动系统，振幅按指数规律衰减，而周期基本不变。轻微阻尼的系统还在振荡，只是每个周期的振幅都小一点；重阻尼的系统缓慢回到平衡位置，根本不越过平衡点。临界阻尼介于两者之间，在最短时间内回到平衡位置且不振荡。三种曲线的区分点是过零次数，临界阻尼与重阻尼都不过零，差别只在回到平衡所用的时间。汽车减震器与闭门器都按临界阻尼设计，指针式电表也让读数尽快稳定下来。常见的错误是只看第一条曲线的振幅大小下结论，正确做法是看整条曲线是否穿过平衡线。
Damping comes from friction and air resistance, so energy leaves the system and the amplitude decays exponentially while the period stays essentially constant. A lightly damped oscillator still crosses equilibrium each cycle with a shrinking amplitude, a heavily damped one crawls back without crossing at all, and critical damping returns to equilibrium in the shortest time without overshooting. Car shock absorbers and door closers are designed for critical damping, and a curve comparison question is settled by counting crossings rather than by comparing first amplitudes.
受迫振动与共振 / Forced Oscillation and Resonance
受迫振动由周期性驱动力维持，稳态下的振动频率等于驱动力频率，而不是系统的固有频率，这一点是整节的核心。振幅随驱动频率变化，驱动频率接近固有频率时振幅达到最大，这个现象叫共振。阻尼越小，共振峰越高也越窄；阻尼越大，峰值越低而且稍微偏向低频一侧。2000 年伦敦千禧桥开桥当天因行人的侧向步伐与桥的侧向固有频率接近而出现大幅晃动，随后关闭两天并加装阻尼器。收音机的调谐回路、微波炉对水分子的作用、乐器共鸣箱都用同一套口径解释。阅卷时常见的问题是只写振幅变大，没写出前提是驱动频率接近固有频率，这类答案通常只给一半的分。
In a forced oscillation the steady state frequency equals the driving frequency rather than the natural frequency of the system, and the amplitude peaks when the two frequencies approach each other. Smaller damping gives a taller and narrower resonance peak, while larger damping lowers the peak and shifts it to a slightly lower frequency. London's Millennium Bridge closed for two days in 2000 after pedestrian footsteps matched its lateral natural frequency, and tuning circuits, microwave heating and instrument soundboxes are explained with the same language.
引力场与平方反比定律 / Gravitational Fields and the Inverse Square Law
引力场强定义为单位质量受到的引力，g = GM/r^2，方向指向场源，所以场强是矢量。地球表面取 r = 6.37 x 10^6 m、GM = 3.986 x 10^14 m^3 s^-2，算得 g 约 9.81 N/kg。到了 400 km 高空，r 变成 6.77 x 10^6 m，g 降到 8.7 N/kg，约是地面的 89%，国际空间站就在这个高度附近运行。场线从质量指向内部而且不会相交，均匀场只在小范围内近似成立。很多学生以为高处完全没有重力，实际上 400 km 处仍有 8.7 N/kg，失重来自自由落体状态而不是引力消失。解题先判断研究点在场源外部还是内部，外部才能直接用平方反比，这一句判断经常决定整题能否起手。
Gravitational field strength is the force per unit mass, g = GM/r^2, directed towards the source mass, which makes it a vector. With r = 6.37 x 10^6 m and GM = 3.986 x 10^14 m^3 s^-2 the surface value comes out at 9.81 N/kg, and at an altitude of 400 km it falls to 8.7 N/kg, about 89% of the surface figure. Field lines run inward and never cross, and the inverse square law applies outside the source mass, so the first step in any field question is deciding whether the point of interest is outside or inside.
场强与势的关系 / Field Strength, Potential and Equipotentials
引力势定义为把单位质量从无穷远移到该点所做的功，V = -GM/r，负号说明质量落进场里会释放能量。地球表面的引力势约 -6.26 x 10^7 J/kg，这个值本身不用记，但量级与负号要写对。场强等于势的负梯度，在径向场里就是 g = -dV/dr，在图像上体现为势-距离曲线的斜率。等势面与场线垂直，把质量沿等势面移动不做功，判断四条路径哪条做功为零就是靠这一条。脱离地球引力所需的最小速度由能量守恒得到，v = √(2GM/r)，代入地球数据约 11.2 km/s。阅卷时常见的失分是把势当成矢量处理，势是标量，相加时直接代数相加减。
Gravitational potential is the work done per unit mass in bringing a mass from infinity, V = -GM/r, and the minus sign records that work is released as a mass falls into the field. At the Earth's surface the potential is about -6.26 x 10^7 J/kg, field strength equals the negative gradient of potential, and equipotential surfaces are perpendicular to the field lines, so no work is done along them. The escape speed follows from energy conservation as v = sqrt(2GM/r), which comes to about 11.2 km/s, and potential stays a scalar so the values simply add.
匀强电场与带电粒子的加速 / Uniform Fields and Accelerating Charges
两块平行金属板之间的场强由电势差与板间距决定，E = V/d，方向从正板指向负板。板上电势差 500 V、间距 25 mm 时场强是 2.0 x 10^4 V/m，场线平行等距，等势面是与板平行的平面。带电粒子在板间受到的力 F = QE 大小恒定，正电荷从正板移向负板时电场力做正功，动能增加量等于 QV。一个质子经过 500 V 的电势差获得 8.0 x 10^-17 J 的动能，也就是 500 eV；电子伏特在粒子与核物理里很常用，1 eV 等于 1.6 x 10^-19 J。反接电源时场强方向相反，受力方向也随之相反，粒子被减速而不是加速，这一步判断最容易被忽略。学生最常把 E = V/d 里的 d 当成板的长度，实际上 d 是两板之间的垂直间距。
Between two parallel plates the field strength is set by the potential difference and the separation, E = V/d, pointing from the positive plate to the negative one, so 500 V across 25 mm gives 2.0 x 10^4 V/m. The field is uniform, with parallel and equally spaced lines and plane equipotentials, and a charge feels a constant force of QE, so a proton crossing 500 V gains 8.0 x 10^-17 J, which is 500 eV. Reversing the supply reverses both the field and the force, and the separation d is the perpendicular gap between the plates rather than their length.
磁场力与带电粒子的圆周运动 / Magnetic Force and Circular Motion
磁场对运动电荷的力写成 F = Bqv，方向与速度垂直，用左手定则判断，对通电导线则写成 F = BIl。力与速度垂直意味着磁场力不做功，粒子速率与动能都不变，只改变方向，所以轨迹是圆。圆周运动需要向心力，令 Bqv = mv^2/r 得到半径 r = mv/(Bq)，周期 T = 2πm/(Bq) 与速率无关，这是回旋加速器能工作的前提。电子以 3.0 x 10^6 m/s 垂直进入 0.10 T 的磁场，半径是 1.7 x 10^-4 m，也就是 0.17 mm，周期约 3.6 x 10^-10 s。很多学生以为磁场会让粒子加速，实际上磁场力永远垂直于速度，只有电场才能改变动能。定则方向判断出错是最常见的失分点，把电流、磁场与力的方向都标在图上可以减少这类错误。
A magnetic field exerts a force of Bqv on a moving charge, perpendicular to the velocity, and BIl on a current-carrying wire; because the force stays perpendicular it does no work, so the speed and kinetic energy are unchanged while the direction turns, producing a circular path. Setting Bqv equal to mv^2/r gives r = mv/(Bq) and a period of 2πm/(Bq) that is independent of speed, which is what makes a cyclotron work. An electron entering 0.10 T at 3.0 x 10^6 m/s turns on a radius of 1.7 x 10^-4 m, with a period near 3.6 x 10^-10 s.
电磁感应：磁通、法拉第定律与楞次定律 / Flux, Faraday and Lenz
磁通描述穿过某个面积的磁场总量，写成 Φ = BA，单位是韦伯，只有 B 与面积垂直时才能直接相乘。法拉第定律说感应电动势等于磁通变化率的负值乘以匝数，emf = -N dΦ/dt，负号来自楞次定律。感应电流的方向总是反抗引起它的变化，所以磁通增加时感应电流产生的磁场与原磁场方向相反。200 匝线圈、面积 2.0 x 10^-3 m^2，磁通密度 0.20 s 内从零升到 0.15 T，每匝磁通变化 3.0 x 10^-4 Wb，感应电动势是 0.30 V。速率翻倍则电动势翻倍，线圈停下来磁通不再变化，电动势立刻回到零。错得最多的是把楞次定律的负号丢掉，物理上它决定了电流方向，方向写反整题的分都会掉。
Magnetic flux measures the field passing through an area, Φ = BA in webers, and it multiplies directly only when the field is perpendicular to the surface. Faraday's law scales the rate of change of flux by the number of turns, emf = -N dΦ/dt, and the minus sign carries Lenz's law, which says the induced current opposes the change that produced it. A 200-turn coil of area 2.0 x 10^-3 m^2 whose field rises from zero to 0.15 T in 0.20 s has 3.0 x 10^-4 Wb per turn changing, and the emf comes out at 0.30 V.
四周复习排期与作答口径 / A Four-Week Plan and Answer Technique
复习排期可以这样定：一周圆周运动与振动，重点是简谐运动判据、两个周期公式与能量分配；一周气体动理论与热学，把状态方程与分子平均动能连起来；一周核衰变与核能，练活度、半衰期与质量亏损；最后一周每天一类场，收尾用完整的两小时做一套 Unit 4 真题。每周留半小时专门写 Section B 的阅读题，答案要引用题目材料里的数据，只写结论不给分。真题按年份倒着做，2024 与 2023 的卷子留到最后当模考，做完对照 WJEC 官网的 mark scheme 逐点核对自己的表述。口径上记住三件事：非精确答案保留三位有效数字，中途计算不要过早四舍五入，带单位的答案必须写单位。阅卷时过程完整但最后一步算术出错的卷面通常给方法分，只有一个数字的答案拿不到过程分。
A four-week plan can run as circular motion and oscillations first, then kinetic theory and thermal physics, then nuclear decay and nuclear energy, and finally one field topic per day in the last week with two full Unit 4 papers under timed conditions. Half an hour each week goes to Section B extended writing, where answers have to quote the data given in the passage rather than state conclusions alone. Non-exact answers are given to three significant figures, intermediate values are not rounded early, and every numerical answer carries its unit.
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