📚 Case Study Practice: Coupled Oscillations – Integrating Matrices, DEs and Complex Numbers | 案例分析实战演练：耦合振动——矩阵、微分方程与复数的融合
In CIE Year 13 Further Mathematics, the ability to connect seemingly separate topics is a key skill. This case study walks you through a complete worked example of coupled oscillations, where matrices, differential equations, eigenvalues, eigenvectors and complex numbers all come together. By following the reasoning step by step, you will see how these tools interact to produce clean, physical solutions, and you will gain confidence in tackling the multi‑stage problems that frequently appear in Paper 2 and Paper 3.
在 CIE 13 年级进阶数学中，将看似独立的知识点联系起来是一项关键能力。本案例带你完整地走一遍耦合振动的求解过程，矩阵、微分方程、特征值、特征向量和复数在此交汇。通过逐步跟随推理，你将看到这些工具如何相互作用产生简洁的物理解，并增强处理 Paper 2 和 Paper 3 中常见多阶段问题的信心。
1. Introduction to the Case Study | 案例介绍
We consider a mechanical system consisting of two identical masses connected by springs. Such a model appears not only in physics but also in engineering and applied mathematics modules. The challenge is to describe the motion of both masses over time when they are disturbed from equilibrium. Because the masses are coupled, their equations of motion must be solved simultaneously. This leads naturally to a system of second‑order linear differential equations that can be handled elegantly with matrix methods and complex exponentials.
我们考虑一个由两根相同质量块和弹簧连接而成的力学系统。这种模型不仅出现在物理中，也出现在工程和应用数学模块里。挑战在于描述两个质量块偏离平衡后随时间的运动。由于质量块相互耦合，它们的运动方程必须联立求解。这自然地导出一个二阶线性微分方程组，可以用矩阵方法和复指数优雅地处理。
2. Setting up the Physical Model | 建立物理模型
Imagine two identical carts, each of mass 1 kg, placed on a smooth horizontal track. They are attached to fixed walls and to each other by three light springs. The outer springs have stiffness 2 N/m, while the central coupling spring has stiffness 1 N/m. Let x₁(t) and x₂(t) be the displacements of the two carts from their equilibrium positions, measured to the right. The forces acting on each mass follow Hooke's law, taking into account the relative extension of the coupling spring.
想象两个相同的小车，质量均为 1 kg，放置在光滑水平轨道上。它们分别通过轻质弹簧固定在两端墙壁上，彼此之间再用一根弹簧连接。外侧弹簧的劲度系数为 2 N/m，中间耦合弹簧的劲度系数为 1 N/m。设 x₁(t) 和 x₂(t) 分别为两小车偏离平衡位置向右的位移。作用在每个质量上的力遵循胡克定律，并考虑耦合弹簧的相对伸长。
3. Formulating the Differential Equations | 构建微分方程
For mass 1, the left spring exerts a restoring force –2x₁, while the middle spring exerts a force equal to –1·(x₁ – x₂) = –x₁ + x₂. Thus the net force results in the equation x₁'' = –3x₁ + x₂. Similarly, mass 2 feels –2x₂ from the right spring and –1·(x₂ – x₁) = –x₂ + x₁ from the coupling spring, giving x₂'' = x₁ – ions hold in the absence of damping, making the system conservative and solvable by energy‑based or algebraic methods.
对质量块 1，左弹簧产生 –2x₁ 的恢复力，中间弹簧产生的力为 –1·(x₁ – x₂) = –x₁ + x₂。因此净力给出方程 x₁'' = –3x₁ + x₂。类似地，质量块 2 受到右弹簧的 –2x₂ 以及耦合弹簧的 –1·(x₂ – x₁) = –x₂ + x₁，得到 x₂'' = x₁ – 3x₂。两个方程在无阻尼时成立，系统是保守的，可用能量法或代数方法求解。
4. Matrix Representation | 矩阵表示
The system can be written compactly in vector form. Let x be the column vector [x₁, x₂]ᵀ. Then the coupled equations become x'' icients. Representing the problem as a matrix differential equation immediately suggests the use of eigenvalues, because the motions are coupled and we wish to find independent normal modes.
系统可以用向量形式简洁地写出。令 x 为列向量 [x₁, x₂]ᵀ。则耦合方程变为 x'' = A x，其中 A 是 2×2 的系数矩阵。将问题表示为矩阵微分方程后，立刻提示我们使用特征值，因为运动是耦合的，而我们希望找到独立的简正模式。
A =
| -3 | 1 |
| 1 | -3 |
5. Solving the System: Eigenvalues and Eigenvectors | 求解系统：特征值与特征向量
To solve x'' = A x, we seek solutions of the form x = v eλt, where v is a constant vector. Substitution yields the algebraic condition (λ²I – A)v = 0. For non‑ tions, λ² must be an eigenvalue of A. We therefore compute the eigenvalues of A from det(A – µI) = 0, giving µ = –2 and µ = –4. Hence the allowable values of λ² are –2 and –4, so λ = ±i√2 and λ = ±2i. The corresponding eigenvectors are v₁ = [1, 1]ᵀ for µ = –2, and v₂ = [1, –1]ᵀ for µ = –4.
为求解 x'' = A x，我们寻求形如 x = v eλt 的解，其中 v 是常向量。代入后得到代数条件 (λ²I – A)v = 0。为了有非平凡解，λ² 必须是 A 的特征值。因此我们通过 det(A – µI) = 0 计算 A 的特征值，得到 µ = –2 和 µ = –4。于是 λ² 允许的值为 –2 和 –4，因此 λ = ±i√2 和 λ = ±2i。对应的特征向量为 µ = –2 时的 v₁ = [1, 1]ᵀ，以及 µ = –4 时的 v₂ = [1, –1]ᵀ。
6. Interpreting Eigenvalues as Frequencies | 解读特征值与频率
Each eigenvalue λ² = –ω² gives an angular frequency ω. The first mode has ω₁ = √2 . The eigenvectors reveal the shapes of these normal modes: in mode 1 the masses move in phase with equal amplitude (symmetric stretching of the coupling spring), whereas in mode 2 they move exactly out of phase (antisymmetric motion). This geometric interpretation is invaluable for checking the physical plausibility of the solution.
每个特征值 λ² = –ω² 对应一个角频率 ω。第一模式 ω₁ = √2 rad/s；第二模式 ω₂ = 2 rad/s。特征向量揭示了这些简正模式的形态：模式 1 中质量块同相等幅运动（耦合弹簧对称拉伸），模式 2 中它们反相运动（反对称运动）。这种几何解释对于检查解的物理合理性非常有价值。
7. Constructing the General Solution Using Complex Exponentials | 用复指数构造通解
Because the eigenvalues of λ are purely imaginary, the solutions are oscillatory. We can write the contribution from the first mode as v₁ [C₁ ei√2 t + D₁ e–i√2 t]. Using Euler's formula, this combination is equivalent to v₁ [A₁ cos(√2 t) + B₁ sin(√2 t)], or more compactly v₁ R₁ cos(√2 t + φ₁). Applying the same procedure to the second mode, the most general real solution is a superposition of the two normal modes.
由于 λ 的特征值为纯虚数，解是振荡的。我们可以将第一模式的贡献写作 v₁ [C₁ ei√2 t + D₁ e–i√2 t]。借助欧拉公式，这个组合等价于 v₁ [A₁ cos(√2 t) + B₁ sin(√2 t)]，或更简洁地表示为 v₁ R₁ cos(√2 t + φ₁)。对第二模式应用同样的步骤，最一般的实解就是两个简正模式的叠加。
x(t) = c₁ [1, 1]ᵀ cos(√2 t + φ₁) + c₂ [1, –1]ᵀ cos(2t + φ₂)
8. Applying Initial Conditions | 应用初始条件
the right and cart 2 held at its equilibrium position. This means x₁(0) = 1, x₂(0) = 0, x₁'(0) = 0, x₂' phases. Solving yields φ₁ = φ₂ = 0, c₁ = 0.5, c₂ = 0.5, leading to the explicit solution.
假设两小车从静止释放，其中小车 1 向右偏离 1 单位，小车 2 保持在平衡位置。这意味着 x₁(0) = 1, x₂(0) = 0, x₁'(0) = 0, x₂'(0) = 0。将 t = 0 代入通解及其导数，我们得到关于未知振幅和相位的线性方程组。求解得 φ₁ = φ₂ = 0，c₁ = 0.5，c₂ = 0.5，从而给出显式解。
x₁(t) = 0.5 cos(√2 t) + 0.5 cos(2t)
x₂(t) = 0.5 cos(√2 t) – 0.5 cos(2t)
9. Visualising the Normal Modes | 可视化简正模式
The expressions above show that the motion is a beat‑like superposition of two frequencies. When √2 t and 2t are nearly in phase, the masses move together; when they are out of phase, one mass moves while the other opposes. Plotting x₁ and x₂ against time (either by hand or with technology) reveals a periodic exchange of energy between the two masses, a hallmark of coupled oscillators. The normal mode vectors [1,1]ᵀ and [1,–1]ᵀ form an orthogonal basis, which confirms that any initial condition can be decomposed uniquely.
上述表达式表明运动是两个频率的类拍频叠加。当 √2 t 和 2t 接近同相时，两质量块同向运动；当它们反相时，一个质量块运动而另一个反向。对 x₁ 和 x₂ 作图（手绘或借助技术）可展现两质量块之间能量的周期性交换，这正是耦合振子的特征。简正模式向量 [1,1]ᵀ 和 [1,–1]ᵀ 构成正交基，这确认了任何初始条件均可唯一分解。
10. Verification and Insights | 验证与洞见
We can verify the solution by substituting back into the original differential equations and checking that the initial conditions are met. The process also deepens understanding: the matrix approach is far more efficient than eliminating variables, and the complex exponential method reveals why the frequencies are simply the square roots of the negative eigenvalues. Moreover, if damping were added, the eigenvalues would gain real parts, turning oscillatory modes into damped ones – a natural extension within the Year 13 syllabus.
我们可以将解代回原微分方程并检查初始条件以验证正确性。这个过程还加深了理解：矩阵方法比消元法高效得多，而复数指数方法揭示了频率正是负特征值的平方根的原因。此外，如果加入阻尼，特征值将获得实部，使振荡模式变为衰减模式——这是 13 年级大纲内的自然延伸。
11. Summary and Exam Tips | 总结与考试技巧
of linear ODEs, always form the matrix A and study its eigenvalues. Use the ansatz x = v eλt to convert the differential problem into an algebraic eigenvalue problem. Remember that complex λ values correspond to oscillatory solutions, and Euler’s formula bridges the gap to real trigonometric forms. In CIE exams, clearly state the general solution in both complex and real forms, and show all steps of eigenvector calculation – marks are awarded for method, not just the final answer.
面对耦合线性常微分方程组时，首先要构造矩阵 A 并研究其特征值。使用预设解 x = v eλt 把微分问题转化为代数特征值问题。记住，复数 λ 对应振荡解，而欧拉公式衔接了复数形式与实三角函数形式。在 CIE 考试中，应明确写出复数形式和实数形式的通解，并展示特征向量计算的所有步骤——评分依据方法，而不仅仅是最终答案。
Published by TutorHao | Further Mathematics Revision Series | aleveler.com
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