📚 Year 12 WJEC Further Mathematics: Case Analysis Practical Drills | Year 12 WJEC 进阶数学：案例分析实战演练
In Year 12 WJEC Further Mathematics, mastering pure and applied concepts goes beyond routine exercises. This article presents a series of case studies that require you to apply complex numbers, matrices, and differential equations to real-world problems. These practical drills will sharpen your analytical skills and prepare you for exam-style modelling questions.
在 WJEC 12 年级进阶数学中，掌握纯数学与应用概念不能仅靠常规练习。本文通过一系列案例分析，要求你将复数、矩阵和微分方程应用于实际问题。这些实战演练将提升你的分析能力，并帮助你为考试中的建模题型做好准备。
1. Introduction to Case Analysis | 案例分析概述
Case analyses in WJEC Further Mathematics are designed to blend theory with application. A typical problem will describe a physical, engineering, or computational scenario, ask you to formulate a mathematical model, and then find a solution using techniques from the pure core. The key is not just calculating answers but interpreting results in context.
WJEC 进阶数学中的案例分析旨在将理论与应用相结合。典型问题会描述一个物理、工程或计算场景，要求你建立数学模型，然后运用纯数学知识求解。关键在于不仅要算出答案，还要在情境中解释结果。
Throughout this article, we will work through three carefully chosen cases: complex impedance in AC circuits, matrix encryption, and damped mechanical vibrations. Each case integrates multiple syllabus topics, emphasising calculation, verification, and the communication of mathematical reasoning.
在本文中，我们将详细演练三个精选案例：交流电路中的复阻抗、矩阵加密和阻尼机械振动。每个案例整合了多个考纲知识点，强调计算、验证以及数学推理的表达。
2. Case Study 1: Complex Impedance in an RLC Circuit | 案例一：RLC 电路中的复阻抗
s RLC circuit has a resistance R = 50 Ω, an inductance L = 0.2 H, and a capacitance C = 50 µ ne the total impedance Z and the phase angle φ between the supply voltage and the current.
一个串联 RLC 电路中，电阻 R = 50 Ω，电感 L = 0.2 H，电容 C = 50 µF，且连接至频率 f = 50 Hz 的交流电源。试求总阻抗 Z 以及电源电压与电流之间的相位角 φ。
compute the angular frequency ω = 2πf = 100π ≈ 314.16 rad/s. The inductive reactance is XL = ωL ≈ 314.16 × 0.2 = 62.83 Ω, and the capacitive reactance is XC = 1/(ωC) = 1/(314.16 × 50 × 10⁻⁶) ≈ 63.66 Ω.
首先计算角频率 ω = 2πf = 100π ≈ 314.16 rad/s。感抗为 XL = ωL ≈ 314.16 × 0.2 = 62.83 Ω，容抗为 XC = 1/(ωC) = 1/(314.16 × 50 × 10⁻⁶) ≈ 63.66 Ω。
The total impedance is a complex number: Z = R + j(XL - XC eal part is purely resistive, and the imaginary part is small and negative, indicating a slightly capacitive overall behaviour.
总阻抗为复数：Z = R + j(XL - XC)。代入数值得 Z = 50 + j(62.83 - 63.66) = 50 - j0.83 Ω。实部为纯电阻，虚部为很小的负值，这表明电路整体呈轻微容性。
3. Solving and Interpreting Impedance | 求解与解释阻抗
The magnitude of the impedance is |Z| = √(R² + (XL - XC)²) = √50² + 0.83² ≈ 50.007 Ω. The phase angle is φ = arctan((XL - XC e negative sign means the current leads the voltage by about 0.95°, as expected when capacitive reactance dominates slightly.
阻抗的模为 |Z| = √(R² + (XL - XC)²) = √50² + 0.83² ≈ 50.007 Ω。相位角 φ = arctan((XL - XC)/R) = arctan(-0.83/50) ≈ -0.95°。负号表示电流超前电压约 0.95°，这与容抗略占优势的情形吻合。
Suppose the supply voltage is V = 230∠ . The current phasor is I = V / Z = 230 / (50 - j0.83). Converting Z to polar form: 50.007∠-0.95°. Then I ≈ 4.60∠ rrent leads the voltage by 0.95°. The power factor is cos φ ≈ 0.9999, very close to unity.
假设电源电压为 V = 230∠0° V，则电流相量为 I = V / Z = 230 / (50 - j0.83)。将 Z 化为极坐标形式：50.007∠-0.95°，得到 I ≈ 4.60∠0.95° A。电流大小约为 4.60 A，且超前电压 0.95°。功率因数 cos φ ≈ 0.9999，十分接近 1。
table using j² = -1, rationalising complex denominators, and interpreting Argand diagrams.
该案例强化了在直角坐标和极坐标形式之间转换复数的能力，这是 WJEC FP1 的核心技能。你需要熟练运用 j² = -1、复数分母有理化以及解释 Argand 图。
4. Case Study 2: Encryption with Matrices | 案例二：矩阵加密
A simple encryption scheme uses the 2×2 matrix K = [[2, 3], [1, 2]]. A plaintext message 'HELP' 5, L=12, P=16. The message is split into two column vectors p₁ = [8; 5] and p₂ = [12; 16].
一种简单的加密方案使用 2×2 矩阵 K = [[2, 3], [1, 2]]。明文 'HELP' 按 A=1, B=2, …, Z=26 转换为数字 H=8, E=5, L=12, P=16，并分成两个列向量 p₁ = [8; 5] 与 p₂ = [12; 16]。
16] = [2×12+3×16; 1×12+2×16] = [72; 44]. The ciphertext vectors are [31; 18] and [72; 44].
密文向量通过将每个明文向量乘以 K 得到。对 p₁：K × [8; 5] = [2×8+3×5; 1×8+2×5] = [31; 18]；对 p₂：K × [12; 16] = [2×12+3×16; 1×12+2×16] = [72; 44]。得到的密文向量为 [31; 18] 与 [72; 44]。
non-zero, K is invertible, and K⁻¹ = (1/det(K)) × [[2, -3], [-1, 2]] = [[2, -3], [-1, 2]].
解密时，接收方需要求出逆矩阵 K⁻¹。计算行列式 det(K) = 2×2 - 3×1 = 1。由于行列式非零，K 可逆，且 K⁻¹ = (1/det(K)) × [[2, -3], [-1, 2]] = [[2, -3], [-1, 2]]。
5. Matrix Inversion and Decoding | 矩阵求逆与解码
+ 2×18] = [62 - 54; -31 + 36] = [8; 5], which corresponds to H and E. For c₂ = [72; 44]: K⁻¹ × [72; 44] = [2×72 - 3×44; -1×72 + 2×44] = [144 - 132; -72 + 88] = [12; 16], giving L and P. The message is correctly retrieved.
将每个密文向量乘以 K⁻¹ 即可恢复明文。对 c₁ = [31; 18]： , -3], [-1, 2]] × [31; 18] = [2×31 + (-3)×18; -1×31 + 2×18] = [62 - 54; -31 + 36] = [8; 5]，对应 H 和 E。对 c₂ = [72; 44]：K⁻¹ × [72; 44] = [2×72 - 3×44; -1×72 + 2×44] = [144 - 132; -72 + 88] = [12; 16]，得到 L 和 P。消息被正确还原。
ix. In WJEC Further Pure, you may encounter questions that ask you to show why a matrix is suitable for encryption (it must have an integer inverse) or to crack a message when the inverse is given.
这个案例巩固了矩阵乘法、求行列式以及求 2×2 逆矩阵等知识点。在 WJEC 进阶纯数中，你可能会遇到要求说明为何某个矩阵适合加密（必须具有整数逆矩阵）的题目，或给你逆矩阵后破解消息的题目。
6. Case Study 3: Mechanical Vibrations – A Damped Oscillator | 案例三：机械振动——阻尼振子
mass is released from rest from an initial displacement of 0.1 m. Formulate and solve the differential equation for the displacement x(t).
质量 m = 2 kg 连接于刚度 k = 10 N/m 的弹簧和阻尼系数 c = 4 N·s/m 的阻尼器上。该质量从静止状态、在初始位移 0.1 m 处释放。试建立并求解位移 x(t) 的微分方程。
0, or dividing by 2: d²x/dt² + 2 dx/dt + 5 x = 0. The initial conditions are x(0) = 0.1, x'(0) = 0.
控制方程为 m d²x/dt² + c dx/dt + kx = 0，即 2 d²x/dt² + 4 dx/dt + 10 x = 0，两边除以 2 得 d²x/dt² + 2 dx/dt + 5 x = 0。初始条件为 x(0) = 0.1，x'(0) = 0。
= e⁻ᵗ (A cos 2t + B sin 2t). Applying x(0)=0.1 gives A = 0.1. Differentiating and using x'(0)=0 yields B = 0.05. Thus, x(t) = e⁻ᵗ (0.1 cos 2t + 0.05 sin 2t).
辅助方程为 r² + 2r + 5 = 0，解得根 r = -1 ± 2i，表明系统为欠阻尼状态。通解为 x(t) = e⁻ᵗ (A cos 2t + B sin 2t)。代入 x(0)=0.1 得 A = 0.1，求导并利用 x'(0)=0 可得 B = 0.05。因此 x(t) = e⁻ᵗ (0.1 cos 2t + 0.05 sin 2t)。
7. Interpreting the Vibration Solution | 振动解的解释
lation is ωₑ = 2 rad/s, so the period is T = π ≈ 3.14 s. The time constant of decay is 1 second, meaning the amplitude reduces by about 63% every second.
因子 e⁻ᵗ 作为指数衰减包络，使振荡逐渐衰减。阻尼振荡的角频率为 ωₑ = 2 rad/s，因此周期为 T = π ≈ 3.14 s。衰减时间常数为 1 秒，意味着振幅每秒约减小 63%。
had intended critical damping, we would set the discriminant of the auxiliary equation to zero, which occurs when c = 2√mk The given c = 4 produces the under-damped case, which is a typical exam analysis scenario.
从初始位移 0.1 m 开始，质量以逐渐减小的振幅振荡。几秒钟后
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