Year 12 Cambridge Further Mathematics: Case Study Practical Exercise | Year 12 剑桥进阶数学：案例分析实战演练

📚 Year 12 Cambridge Further Mathematics: Case Study Practical Exercise | Year 12 剑桥进阶数学：案例分析实战演练

In this article we work through a realistic engineering case study that integrates several core topics from the Cambridge Year 12 Further Mathematics syllabus. By modelling the vibration of a bridge deck under wind loading, we apply complex numbers, second‑order differential equations, matrix methods and hyperbolic functions – all within a single coherent problem. The step‑by‑step approach helps you see how different branches of further mathematics support one another in practical analysis.

本文通过一个真实的工程案例，串联起剑桥 Year 12 进阶数学课程中的多个核心知识点。我们模拟风荷载作用下桥面的振动，综合运用复数、二阶微分方程、矩阵方法和双曲函数，在一个完整问题中展示它们的内在联系。逐层递进的分析能帮助大家理解进阶数学各分支如何在实践中相互支撑。

1. Case Study Overview | 案例概览

The structure under investigation is a simplified suspension footbridge. A main suspension cable hangs between two towers, and the deck is modelled as a rigid beam supported by hangers from that cable. We first determine the static shape of the cable using hyperbolic functions. Then a dynamic model of the deck is built, treating it as a single degree‑of‑freedom mass‑spring‑damper system excited by a periodic wind gust. The resulting second‑order ODE is solved with complex numbers. Finally, a two‑segment deck model is written in matrix state‑space form, and eigenvalue analysis reveals stability conditions.

研究对象是一座简化的悬索人行桥。主缆悬挂在两座塔柱之间，桥面假定为由缆索吊挂的刚性梁。我们先用双曲函数求出缆绳的静态线形，随后将桥面建模为单自由度质量‑弹簧‑阻尼系统，承受阵风的周期性激励。产生的二阶常微分方程用复数法求解。最后，两节段桥面模型写成矩阵状态空间形式，通过特征值分析揭示稳定条件。

2. Modelling the Static Cable: Hyperbolic Functions | 静态缆绳建模：双曲函数

ve can be written as y = a cosh(x/a). The parameter a is the ratio of the horizontal tension to the weight per unit length. For a bridge cable we are interested in the length and the end slope.

均匀柔软的缆绳在自重下形成悬链线。若最低点在 (0, a) 且下垂量由此量起，曲线可写为 y = a cosh(x/a)。参数 a 是水平拉力与单位长度重量之比。对桥缆而言，我们关心缆长和端点斜率。

s = a sinh(L/a). The tension at any point has vertical component proportional to sinh, leading to the familiar result that the maximum tension occurs at the supports.

从 x=0 到 x=L 的缆长 s 由 √(1+(dy/dx)²) 积分得到。因 dy/dx = sinh(x/a)，被积函数化为 cosh(x/a)，故 s = a sinh(L/a)。任一点张力的竖直分量与 sinh 成正比，因此最大张力出现在支座处，这是经典结论。

3. Equation of Motion for the Bridge Deck | 桥面运动方程

the hangers and cable is represented by a spring constant k, and the inherent damping is modelled by a dashpot with coefficient c. Newton’s second law gives m d²x/dt² + c dx/dt + k x = F₀ sin(ωt).

考虑阵风力 F₀ sin(ωt) 作用下桥面的竖向振动，记 x(t) 为偏离平衡的位移。吊杆和缆绳的组合刚度用弹簧常数 k 表示，固有阻尼用阻尼系数 c 的阻尼器模拟。牛顿第二定律给出 m d²x/dt² + c dx/dt + k x = F₀ sin(ωt)。

Divide through by the mass m to obtain the standard form

d²x/dt² + 2ζωₙ dx/dt + ωₙ² x = (F₀/m) sin(ωt)

where ωₙ = √k/m is the natural frequency and ζ = c / (2√mk) is the damping ratio.

各项除以质量 m 得到标准形式

d²x/dt² + 2ζωₙ dx/dt + ωₙ² x = (F₀/m) sin(ωt)

其中 ωₙ = √k/m 为无阻尼固有频率，ζ = c/(2√mk) 为阻尼比。

4. Homogeneous Solution via Characteristic Equation | 齐次方程的特征根求解

eneous equation is d²x/dt² + 2ζωₙ dx/dt + ωₙ² x = 0. Assuming a solution of the form x = e^λt, we obtain the characteristic equation

λ² + 2ζωₙ λ + ωₙ² = 0

齐次方程为 d²x/dt² + 2ζωₙ dx/dt + ωₙ² x = 0。设解为 x=e^λt，得到特征方程

λ² + 2ζωₙ λ + ωₙ² = 0

Its roots are λ = –ζωₙ ± ωₙ√ζ² – 1. For an under‑damped bridge, ζ < 1, so the roots are complex conjugates: λ = –ζωₙ ± i ωₚ, where ωₚ = ωₙ√1 – ζ² is the damped natural frequency. Thus the homogeneous solution is x_h = e^{–ζωₙ t} (A cos(ωₚ t) + B sin(ωₚ t)). This transient response decays exponentially.

其根为 λ = –ζωₙ ± ωₙ√ζ² – 1。对欠阻尼桥面 ζ<1，根是一对共轭复数：λ = –ζωₙ ± i ωₚ，其中 ωₚ = ωₙ√1 – ζ² 为有阻尼固有频率。因此齐次解为 x_h = e^{–ζωₙ t}(A cos(ωₚ t) + B sin(ωₚ t))。这一瞬态响应按指数衰减。

5. Particular Integral for Periodic Forcing | 周期强迫项的特解

To find a particular solution for the sinusoidal right‑hand side, it is efficient to use complex exponentials. Replace sin(ωt) by the imaginary part of e^iωt. Consider the auxiliary equation d²z/dt² + 2ζωₙ dz/dt + ωₙ² z = (F₀/m) e^iωt. Assume a trial solution z = X e^iωt. Substituting gives X ( –ω² + i 2ζωₙω + ωₙ² ) = F₀/m. Hence

X = (F₀/m) / ( (ωₙ² – ω²) + i (2ζωₙω) )

对正弦型右端项，运用复指数可高效求出特解。将 sin(ωt) 替换为 e^iωt 的虚部。考虑辅助方程 d²z/dt² + 2ζωₙ dz/dt + ωₙ² z = (F₀/m) e^iωt。设试解 z = X e^iωt，代入得 X (–ω² + i 2ζωₙω + ωₙ²) = F₀/m，因此

X = (F₀/m) / ( (ωₙ² – ω²) + i (2ζωₙω) )

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The particular integral for the original equation is x_p = Im( X e^iωt ). Expressing X in polar form, X = |X| e^{–i φ}, where the modulus is |X| = (F₀/m) / √[ (ωₙ² – ω²)² + (2ζωₙω)² ] and the phase lag φ satisfies tan φ = (2ζωₙω) / (ωₙ² – ω²). Therefore x_p = |X| sin(ωt – φ). This steady‑state response oscillates at the driving frequency.

原方程的特解为 x_p = Im(X e^iωt)。将 X 写成极坐标形式 X = |X| e^{–i φ}，其中模 |X| = (F₀/m) / √[ (ωₙ² – ω²)² + (2ζωₙω)² ]，相位滞后 φ 满足 tan φ = (2ζωₙω)/(ωₙ² – ω²)。因此 x_p = |X| sin(ωt – φ)。这个稳态响应以外加频率作简谐振荡。

6. Full Response and Damping Analysis | 全响应与阻尼分析

The complete solution is x(t) = x_h(t) + x_p(t). The transient part x_h dies out because of the e^{–ζωₙ t} factor, leaving the steady‑state forced vibration x_p. Resonance occurs when the forcing frequency ω is close to ωₙ, causing the amplitude |X| to peak. The sharpness of the peak is controlled by the damping ratio ζ: small ζ gives a large amplification near resonance.

完整解为 x(t) = x_h(t) + x_p(t)。瞬态部分 x_h 因含 e^{–ζωₙ t} 因子而消逝，最终留下稳态受迫振动 x_p。当激励频率 ω 接近 ωₙ 时，振幅 |X| 急剧增大，即发生共振。共振峰的尖锐程度由阻尼比 ζ 控制：ζ 越小，近共振区的放大倍数越大。

Engineers set the design so that the bridge' m typical wind gust frequencies, or they increase damping artificially. The maximum steady‑te amplitude is (F₀/(2mζωₙ²)) when ω = ωₙ, showing the critical role of the damping ratio.

工程师通过设计使桥梁固有频率远离阵风典型频率，或人为增大阻尼。当 ω=ωₙ 时，最大稳态振幅为 F₀/(2mζωₙ²)，突显了阻尼比的关键作用。

7. Matrix Formulation for Coupled Deck Sections | 耦合段的矩阵表示

Real bridges have many coupled modes. As a next step, consider two identical deck segments, each with mass m, connected by springs of stiffness k and dampers c. Let x₁ and x₂ be their displacements. The coupled equations are

m d²x₁/dt² + c dx₁/dt + k x₁ + k_c (x₁ – x₂) = F₁(t)

m d²x₂/dt² + c dx₂/dt + k x₂ + k_c (x₂ – x₁) = F₂(t)

实际桥梁存在众多耦合模态。作为进阶，考虑两个相同桥面节段，质量各为 m，通过刚度 k 的弹簧和阻尼器 c 连接。记 x₁、x₂ 为位移，耦合方程为

m d²x₁/dt² + c dx₁/dt + k x₁ + k_c (x₁ – x₂) = F₁(t)

m d²x₂/dt² + c dx₂/dt + k x₂ + k_c (x₂ – x₁) = F₂(t)

oduce velocity variables v₁ = dx₁/dt, v₂ = dx₂/dt. Then the system can be written in first‑order state‑space form d/dt [x₁, x₂, v₁, v₂]ᵀ = A [x₁, x₂, v₁, v₂]ᵀ mass, damping and stiffness matrices. This linear system is ideal for eigenvalue analysis.

引入速度变量 v₁=dx₁/dt, v₂=dx₂/dt，可将系统写成一阶状态空间形式 d/dt [x₁, x₂, v₁, v₂]ᵀ = A [x₁, x₂, v₁, v₂]ᵀ + B u(t)。4×4 矩阵 A 由质量、阻尼和刚度矩阵构成。该线性系统非常适合特征值分析。

8. Eigenvalue Analysis and Stability | 特征值与稳定性分析

The eigenvalues of the state matrix A dictate the system' haviour depends on the real parts of the eigenvalues: if all eigenvalues have negative real parts, the equilibrium is stable and vibrations die out. A purely imaginary pair implies undamped oscillation, while any positive real part signals instability.

状态矩阵 A 的特征值决定系统的动力学行为。在齐次情形（F₁=F₂=0），解的行为取决于特征值的实部：若所有特征值具有负实部，平衡点稳定，振动衰减；纯虚数对表示无阻尼振荡；出现正实部则系统失稳。

The characteristic equation for A is a quartic. Using the parameters m, k, k_c, c we can show that a sufficient condition for stability is that all damping coefficients are positive. This links back to the scalar damping ratio ζ derived earlier. The eigenvalues come in complex‑conjugate pairs, reflecting the dual‑oscillator nature. Matrix diagonalisation (if A is diagonalisable) then decouples the system into independent modes, each governed by its own eigenvalue.

A 的特征方程为四次方程。代入参数 m, k, k_c, c 可以证明系统稳定的一个充分条件是所有阻尼系数为正，这与前面标量阻尼比 ζ 呼应。特征值成对共轭出现，反映双振子本质。若 A 可对角化，矩阵对角化可将系统解耦为独立模态，每一模态由其对应的特征值支配。

9. Practical Insights and Mathematical Connections | 工程洞察与数学联系

This case study demonstrates how relatively abstract Further Mathematics techniques become powerful tools in structural dynamics. Complex numbers turn the tedious search for a particular integral into algebraic manipulation of phasors. Hyperbolic functions describe the static cable shape compactly. Matrix eigenvalue theory reveals stability margins that are invisible from a single‑equation analysis. Extending the idea, one could apply series expansions (e.g. Taylor series for cosh) to approximate the catenary by a parabola for small sags, leading to simpler design formulas.

本案例展示了一些看似抽象的进阶数学技术如何在结构动力学中成为有力工具。复数把繁琐的特解搜寻转化为相量的代数操作；双曲函数简洁描述缆绳静态线形；矩阵特征值理论揭示单一方程分析看不到的稳定裕度。进一步延伸，可利用级数展开（如 cosh 的泰勒级数）在小垂度时将悬链线近似为抛物线，从而导出更简化的设计公式。

The unified application of FP1 and FP2 content – roots of polynomial equations, complex numbers, hyperbolic functions, differential equations, and matrices – underlines the interconnected nature of the syllabus. Regular practice with such integrated case studies builds the fluency needed for both examinations and for further studies in engineering or physics.

FP1 与 FP2 中各知识点的统一运用——多项式求根、复数、双曲函数、微分方程和矩阵——凸显了课程各部分的相互关联。经常进行这类综合案例训练，既能培养应对考试的熟练度，也为工程或物理方向的深造打下扎实的数学基础。

Published by TutorHao | Further Mathematics Revision Series | aleveler.com

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