📚 Mastering Formula Derivations from A-Level Physics Paper 4 (June 2019) | 精通A-Level物理Paper 4 (2019年6月) 公式推导
Understanding how to derive key formulas is essential for success in A-Level Physics Paper 4. The June 2019 examiner report highlighted that many students lost marks not because they couldn’t recall equations, but because they could not demonstrate a clear logical path from fundamental principles to the final expression. Topics such as circular motion, gravitational fields, simple harmonic motion, capacitor circuits, and electromagnetic induction all require confident derivations. This article revisits those derivations step by step, pairing each English explanation with its Chinese equivalent, so you can master both the language and the logic.
理解如何推导关键公式是 A-Level 物理试卷 4 取得成功的必要条件。2019 年 6 月的考官报告指出，很多学生失分并非因为记不住方程，而是因为他们无法从基本原理出发，展示出清晰的逻辑路径得出最终表达式。圆周运动、引力场、简谐运动、电容器电路以及电磁感应等专题都要求考生能自信地进行推导。本文将逐步重温这些推导过程，每组英文解释后紧跟对应的中文，帮助你同时掌握语言与逻辑。
1. Centripetal Acceleration: From Velocity Vector to a = v²/r | 向心加速度：从速度矢量到 a = v²/r
Consider an object moving at constant speed v in a circle of radius r. In a short time Δt, its position vector sweeps an angle Δθ, and the velocity vector, always tangential, also rotates through the same angle Δθ. The change in velocity Δv is the base of an isosceles triangle formed by two velocity vectors of magnitude v separated by angle Δθ.
考虑一个物体以恒定速率 v 在半径为 r 的圆周上运动。在短时间 Δt 内，其位置矢量扫过角度 Δθ，而始终沿切线方向的速度矢量也转过相同的角度 Δθ。速度的变化量 Δv 是由两个大小为 v、夹角为 Δθ 的速度矢量构成的等腰三角形的底边。
For small Δθ, Δv ≈ vΔθ. The acceleration magnitude a is |Δv/Δt| ≈ v(Δθ/Δt). Since angular speed ω = Δθ/Δt and v = ωr, we obtain a = vω = v(v/r) = v²/r. The direction of Δv points towards the centre of the circle, giving centripetal acceleration a = v²/r along the radius inward.
对于很小的 Δθ，Δv ≈ vΔθ。加速度大小 a = |Δv/Δt| ≈ v(Δθ/Δt)。由于角速度 ω = Δθ/Δt 且 v = ωr，我们得到 a = vω = v(v/r) = v²/r。Δv 的方向指向圆心，因此向心加速度 a = v²/r，方向沿半径向内。
a = v²/r = ω²r
2. Deriving v = ωr and Periodic Time Relations | 推导 v = ωr 及周期关系
Angular speed ω is defined as the rate of change of angle: ω = Δθ/Δt. For one complete revolution, Δθ = 2π radians and the time taken is the period T, so ω = 2π/T. The linear distance covered in one revolution is the circumference 2πr, so the constant speed v = distance/time = 2πr/T.
角速度 ω 定义为角度的变化率：ω = Δθ/Δt。对于一整圈，Δθ = 2π 弧度，所用时间为周期 T，因此 ω = 2π/T。一圈内通过的直线距离是周长 2πr，所以匀速速率 v = 距离/时间 = 2πr/T。
Combining v = 2πr/T and ω = 2π/T gives v = ωr. Frequency f = 1/T, so we also have the very useful forms ω = 2πf and v = 2πfr. Students often confuse T and f in exam pressure; the June 2019 report noted many misplaced 2π factors.
结合 v = 2πr/T 和 ω = 2π/T 得到 v = ωr。频率 f = 1/T，因此我们还有非常有用的形式 ω = 2πf 和 v = 2πfr。考生常在考试压力下混淆 T 和 f；2019 年 6 月报告指出许多学生错放了 2π 因子。
v = ωr ω = 2πf T = 1/f
3. Gravitational Field Strength g = GM/r² | 引力场强度 g = GM/r²
Newton’s law of gravitation states that the force between two point masses M and m separated by distance r is F = GMm/r². The gravitational field strength g at a point is the force per unit mass on a small test mass placed at that point: g = F/m.
牛顿万有引力定律指出，两个质点 M 和 m 相距 r 时的作用力为 F = GMm/r²。某点的引力场强度 g 是放置在该点的小检验质量所受的力与其质量之比：g = F/m。
Substituting F = GMm/r² gives g = (GMm/r²)/m = GM/r². This derivation is straightforward, yet the examiner report highlights that candidates often write g = GM/r without the square, or incorrectly apply the formula for points inside a planet. Remember this applies outside a spherical mass.
代入 F = GMm/r² 得到 g = (GMm/r²)/m = GM/r²。此推导很简单，但考官报告强调，考生经常写成 g = GM/r 而没有平方，或者在行星内部的点错误地应用该公式。记住这适用于球体外部。
g = GM/r²
4. From Newton’s Law to Kepler’s Third Law | 从牛顿定律推导开普勒第三定律
For a planet of mass m orbiting a star of mass M in a circular orbit of radius r, gravitational force provides the centripetal force: GMm/r² = mv²/r. Cancel m and one power of r: GM/r = v². Replace v with 2πr/T using the periodic relation: GM/r = (2πr/T)² = 4π²r²/T².
对于一颗质量为 m 的行星在半径为 r 的圆形轨道上绕恒星 M 运行，引力提供向心力：GMm/r² = mv²/r。消去 m 和一个 r 次幂：GM/r = v²。用周期关系 v = 2πr/T 替换：GM/r = (2πr/T)² = 4π²r²/T²。
Rearrange to group r terms: T² = (4π²/GM) r³. This is Kepler’s Third Law: T² ∝ r³ with the constant of proportionality depending on the central mass M. In June 2019, some students lost marks by using r for two different quantities or failing to show clear algebraic rearrangement.
重组 r 项：T² = (4π²/GM) r³。这就是开普勒第三定律：T² ∝ r³，比例常数取决于中心质量 M。2019 年 6 月，一些学生因对两个不同量使用 r 或未展示清晰代数变换而失分。
T² = (4π²/GM) r³
5. Simple Harmonic Motion: Deriving a = – ω²x | 简谐运动：推导 a = -ω²x
An oscillation is simple harmonic if the acceleration is directly proportional to displacement from the equilibrium position and is always directed towards it. For a mass-spring system, Hooke’s law gives F = -kx, and Newton’s second law gives F = ma, so ma = -kx, thus a = -(k/m)x. Defining ω² = k/m yields a = -ω²x.
若加速度与相对于平衡位置的位移成正比且总指向平衡位置，则振动是简谐的。对质量-弹簧系统，胡克定律给出 F = -kx，牛顿第二定律给出 F = ma，因此 ma = -kx，于是 a = -(k/m)x。定义 ω² = k/m 即得 a = -ω²x。
The minus sign is crucial; the June 2019 report noted that candidates often omitted it, leading to a misunderstanding of phase. More generally, the solution of this differential equation gives the displacement x = A sin(ωt) or x = A cos(ωt), where A is amplitude.
负号至关重要；2019 年 6 月报告指出考生常遗漏负号，导致对相位的误解。一般地，这个微分方程的解给出位移 x = A sin(ωt) 或 x = A cos(ωt)，其中 A 为振幅。
a = -ω²x
6. Energy in SHM: Kinetic and Potential Expressions | 简谐运动的能量：动能与势能表达式
Starting from x = A cos(ωt), the velocity is v = dx/dt = -Aω sin(ωt). The maximum speed vₘₐₓ = ωA occurs as the oscillator passes through equilibrium. Kinetic energy Eₖ = ½mv² = ½m ω²A² sin²(ωt).
从 x = A cos(ωt) 出发，速度为 v = dx/dt = -Aω sin(ωt)。最大速度 vₘₐₓ = ωA 出现在振荡器经过平衡位置时。动能 Eₖ = ½mv² = ½m ω²A² sin²(ωt)。
Potential energy for a spring system is Eₚ = ½kx² = ½m ω²x², because k = m ω². Using x = A cos(ωt), Eₚ = ½m ω²A² cos²(ωt). The total mechanical energy Eₜₒₜ = Eₖ + Eₚ = ½m ω²A² (sin² + cos²) = ½m ω²A², which is constant.
弹簧系统的势能为 Eₚ = ½kx² = ½m ω²x²，因为 k = m ω²。利用 x = A cos(ωt)，Eₚ = ½m ω²A² cos²(ωt)。总机械能 Eₜₒₜ = Eₖ + Eₚ = ½m ω²A² (sin² + cos²) = ½m ω²A²，是守恒的。
This derivation shows that energy is proportional to the square of amplitude, a favourite exam question. Many candidates forget to link k and m ω².
该推导显示能量与振幅的平方成正比，这是考试常考题目。许多考生忘记联系 k 和 m ω²。
Eₜₒₜ = ½mω²A² vₘₐₓ = ωA
7. Capacitor Discharge: Exponential Decay Derivation | 电容器放电：指数衰减推导
When a capacitor C discharges through a resistor R, the current I = dq/dt, and the p.d. across the capacitor V = q/C. By Kirchhoff’s voltage law, V = IR, so q/C = -R dq/dt (negative because charge decreases). Rearranging gives dq/dt = -q/(RC).
当电容器 C 通过电阻 R 放电时，电流 I = dq/dt，电容器两端电压 V = q/C。由基尔霍夫电压定律，V = IR，所以 q/C = -R dq/dt（负号因为电荷减少）。整理得 dq/dt = -q/(RC)。
Separating variables and integrating from t=0 (q=Q₀) to t=t (q=q): ∫(1/q)dq = -∫(1/RC)dt → ln(q/Q₀) = -t/RC. Taking exponentials yields q = Q₀ e^(-t/RC). The product RC is the time constant τ. Similarly, I = I₀ e^(-t/RC) and V = V₀ e^(-t/RC).
分离变量并从 t=0 (q=Q₀) 积分到 t=t (q=q)：∫(1/q)dq = -∫(1/RC)dt → ln(q/Q₀) = -t/RC。两边取指数得 q = Q₀ e^(-t/RC)。乘积 RC 是时间常数 τ。类似地，I = I₀ e^(-t/RC) 以及 V = V₀ e^(-t/RC)。
q = Q₀ e^(-t/RC)
8. Time Constant τ = RC and Its Significance | 时间常数 τ = RC 及其意义
The time constant τ = RC has units: ohm × farad = (V/A) × (C/V) = C/A = second. After t = τ, the charge on the capacitor has fallen to Q₀ e⁻¹ ≈ 0.37 Q₀, i.e. about 37% of its original value. The current also decays to 37% of its initial value in one time constant.
时间常数 τ = RC 的单位为：欧姆 × 法拉 = (V/A) × (C/V) = C/A = 秒。经过 t = τ 后，电容器上的电荷已降为 Q₀ e⁻¹ ≈ 0.37 Q₀，即约原值的 37%。电流在一个时间常数内亦衰减至初始值的 37%。
In a charging circuit, the p.d. across the capacitor rises as V = V₀(1 – e^(-t/RC)). After one time constant, it reaches about 63% of the supply voltage. Examiners expect you to derive either discharge or charge equations, and to relate the gradient of ln(V) vs time graphs to -1/RC.
在充电电路中，电容器两端电压按 V = V₀(1 – e^(-t/RC)) 上升。经过一个时间常数后，电压达到约电源电压的 63%。考官期望你能推导放电或充电方程，并能将 ln(V) 对时间图的斜率与 -1/RC 联系起来。
τ = RC
9. Magnetic Flux Linkage and Induced e.m.f. | 磁链与感应电动势
Magnetic flux Φ = BA cosθ, where B is magnetic flux density, A is area, and θ is the angle between the field and the normal to the area. Flux linkage NΦ is the product of the number of turns N and the flux through each turn. Faraday’s law states that the magnitude of induced e.m.f. is equal to the rate of change of flux linkage: ε = – d(NΦ)/dt.
磁通量 Φ = BA cosθ，其中 B 为磁通量密度，A 为面积，θ 为磁场与面积法线间的夹角。磁链 NΦ 是匝数 N 与每匝磁通量的乘积。法拉第定律指出，感应电动势的大小等于磁链的变化率：ε = – d(NΦ)/dt。
For a coil rotating uniformly in a magnetic field, Φ = BA cos(ωt), where ω is angular speed. Then ε = -N d(BA cosωt)/dt = NBAω sin(ωt), giving an alternating e.m.f. with peak value ε₀ = NBAω. The June 2019 paper frequently tested this derivation with attention to the correct differentiation of cosine.
对于在磁场中匀速旋转的线圈，Φ = BA cos(ωt)，其中 ω 为角速度。那么 ε = -N d(BA cosωt)/dt = NBAω sin(ωt)，产生峰值 ε₀ = NBAω 的交变电动势。2019 年 6 月试卷频繁考查这个推导，关注对余弦函数正确求导。
ε₀ = NBAω
10. Transformer Equation Derivation (Ideal Transformer) | 理想变压器公式推导
An ideal transformer has two coils wound on a common iron core, with no flux leakage and no power loss. The alternating flux Φ in the core links both primary and secondary coils. Using Faraday’s law: εₚ = -Nₚ dΦ/dt and εₛ = -Nₛ dΦ/dt. Taking magnitudes, εₚ/εₛ = Nₚ/Nₛ.
理想变压器具有绕在同一铁芯上的两个线圈，无漏磁且无功率损耗。铁芯中交变的磁通量 Φ 同时链接初级和次级线圈。由法拉第定律：εₚ = -Nₚ dΦ/dt，εₛ = -Nₛ dΦ/dt。取大小得 εₚ/εₛ = Nₚ/Nₛ。
Since power input equals power output for an ideal transformer, Iₚ Vₚ = Iₛ Vₛ. Combining with the turns ratio, we obtain Iₛ/Iₚ = Nₚ/Nₛ. Many candidates stumble by mixing up the step-up and step-down current relationship; the report warned against blindly applying formulas without reasoning.
因理想变压器输入功率等于输出功率，Iₚ Vₚ = Iₛ Vₛ。结合匝数比，可得 Iₛ/Iₚ = Nₚ/Nₛ。许多考生因混淆升压与降压的电流关系而出错；报告警告不要盲目套用公式而不加推理。
Vₛ/Vₚ = Nₛ/Nₚ Iₛ/Iₚ = Nₚ/Nₛ
11. Deriving Root Mean Square (RMS) for Alternating Current | 交流电有效值推导
For an alternating current I = I₀ sin(ωt), the heating effect depends on I². The mean power dissipated in a resistor R is ⟨P⟩ = ⟨I²R⟩ = I₀² R ⟨sin²(ωt)⟩. Over one complete cycle, the average value of sin²(ωt) is ½, so ⟨P⟩ = ½ I₀² R.
对于交变电流 I = I₀ sin(ωt)，热效应取决于 I²。在电阻 R 上消耗的平均功率为 ⟨P⟩ = ⟨I²R⟩ = I₀² R ⟨sin²(ωt)⟩。在一个完整周期内，sin²(ωt) 的平均值为 ½，于是 ⟨P⟩ = ½ I₀² R。
The rms current I_rms is defined as the direct current that would dissipate the same average power: I_rms² R = ½ I₀² R ⇒ I_rms = I₀/√2. Similarly, V_rms = V₀/√2. The June 2019 report commented that some students incorrectly used I_rms = I₀/2, losing easy marks.
有效值电流 I_rms 定义为能产生相同平均功率的直流电流值：I_rms² R = ½ I₀² R ⇒ I_rms = I₀/√2。类似地，V_rms = V₀/√2。2019 年 6 月报告提到，部分学生错误地使用了 I_rms = I₀/2，丢掉了很容易的分数。
I_rms = I₀/√2 V_rms = V₀/√2
12. Common Pitfalls from the June 2019 Paper 4 Report | 2019年6月Paper 4报告中的常见错误
The examiner report highlighted specific mistakes that compromised formula derivations. Students often inserted numerical values before completing the algebraic derivation, making it impossible to follow the logic. In circular motion, forgetting that centripetal force is a resultant, not a separate force, led to double-counting. During SHM derivations, the phase relationship between displacement, velocity and acceleration was frequently described incorrectly; velocity leads displacement by π/2 and acceleration is π out of phase with displacement.
考官报告指出了影响公式推导的具体错误。学生常在完成代数推导前就代入数值，导致逻辑无法追踪。在圆周运动中，忘记向心力是合力而非某种单独的力，造成重复计算。进行简谐运动推导时，位移、速度和加速度之间的相位关系经常被描述错误；速度超前位移 π/2，而加速度与位移反相（π）。
Furthermore, in capacitor circuits, candidates confused the charge and voltage decay equations with charging equations, or misapplied the time constant. When dealing with flux linkage, many failed to include the factor N or misidentified the angle in BA cosθ. In AC theory, the factor √2 was often misplaced, and the idea of average power over a cycle was poorly understood. The key takeaway is that a thorough, step-by-step derivation with clear annotation of symbols is the safest route to full marks.
此外，在电容器电路中，考生混淆了放电与充电的电荷、电压方程，或错误应用了时间常数。在处理磁链时，许多人遗漏了因子 N，或未能正确识别 BA cosθ 中的角度。在交流电理论中，√2 因子常被放错位置，且对周期内平均功率的概念理解不佳。核心要点是：彻底、逐步的推导并清楚注明符号才是取得满分的最稳妥路径。
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