Case Study Practical Exercises | 案例分析实战演练

📚 Case Study Practical Exercises | 案例分析实战演练

In Advanced Higher Mathematics, case study exercises challenge you to apply your knowledge of calculus, differential equations, and numerical methods to model real-world phenomena. This article walks through a complete worked example—modelling the motion of a damped pendulum—demonstrating the step-by-step approach required in SQA exam questions.

在进阶数学中，案例分析题要求你运用微积分、微分方程和数值方法的知识对现实世界现象进行建模。本文通过一个完整的工作示例——阻尼摆运动的建模——展示 SQA 考试题目所需的逐步分析方法。

1. Introduction to the Case Study | 案例简介

We consider a classic mechanical system: a simple pendulum of length L, bob mass m, and a damping force proportional to the angular velocity caused by a viscous medium. The goal is to derive the equation of motion, solve it under small-angle approximation, classify the damping regimes, and then explore a numerical solution to the original nonlinear model.

我们考虑一个经典的力学系统：长度为 L、摆锤质量为 m 的单摆，受到由粘性介质引起的、与角速度成正比的阻尼力。目标是推导运动方程，在小角度近似下求解方程，对阻尼区域进行分类，然后探索对原非线性模型的数值解。

2. Setting Up the Physical Model | 建立物理模型

The pendulum consists of a light rod of length L and a bob of mass m. The air or fluid resistance exerts a force proportional to the bob's speed, F_drag cient. The angular displacement θ(t) is measured from the downward vertical, with the positive direction anticlockwise.

单摆由一根长度为 L 的轻杆和质量为 m 的摆锤组成。空气或流体阻力施加一个与摆锤速度成正比的力，F_drag = −b v，其中 b > 0 是阻尼系数。角位移 θ(t) 从向下的竖直线量起，逆时针方向为正。

dθ/dt and the moment arm is L). Rotational inertia of the bob about the pivot is I = mL².

关于悬挂点的相关力矩有重力矩 −mgL sinθ 和阻尼力矩 −b L² dθ/dt（因为 v = L dθ/dt 且力臂为 L）。摆锤对悬挂点的转动惯量为 I = mL²。

3. Deriving the Nonlinear Differential Equation | 导出非线性微分方程

t. Substituting I = mL² and dividing through by mL² yields the second-order nonlinear ODE:

转动形式的牛顿第二定律给出 I d²θ/dt² = −mgL sinθ − bL² dθ/dt。代入 I = mL² 并除以 mL²，得到二阶非线性常微分方程：

d²θ/dt² + (b/m) dθ/dt + (g/L) sinθ = 0

ping factor γ = b/m and the natural frequency squared ω₀² = g/L. Then the equation becomes

定义阻尼因子 γ = b/m 和自然频率平方 ω₀² = g/L。于是方程化为

d²θ/dt² + γ dθ/dt + ω₀² sinθ = 0

This is the exact nonlinear pendulum equation with linear viscous damping.

这是带有线性粘性阻尼的精确非线性摆方程。

4. Small-Angle Linearisation | 小角度线性化

ituting this approximation removes the nonlinearity and gives the linear second-order ODE:

对于小角度位移，有 sinθ ≈ θ（当 θ 以弧度为单位）。代入此近似可消除非线性，得到线性二阶常微分方程：

d²θ/dt² + γ dθ/dt + ω₀² θ = 0

This linearised model is valid for |θ| less than about 0.2 rad and provides the foundation for analytical treatment. It is a homogeneous linear ODE with constant coefficients.

该线性化模型在 |θ| 约小于 0.2 rad 时成立，并为解析处理奠定了基础。这是一个常系数齐次线性常微分方程。

5. The Characteristic Equation | 特征方程

Assume a solution of the form θ(t) = e^rt. Differentiating and substituting into the linear ODE gives:

设解的形式为 θ(t) = e^rt。求导并代入线性常微分方程得到：

r² + γ r + ω₀² = 0

This is the characteristic (auxiliary) equation. Its roots determine the nature of the damping.

这就是特征（辅助）方程。它的根决定了阻尼的性质。

The discriminant is Δ = γ² − 4ω₀². The three cases—underdamped, critically damped, and overdamped—arise from Δ < 0, Δ = 0, and Δ > 0, respectively.

判别式为 Δ = γ² − 4ω₀²。欠阻尼、临界阻尼和过阻尼这三种情况分别对应于 Δ < 0, Δ = 0 和 Δ > 0。

6. General Solution of the Linear ODE | 线性常微分方程的通解

Case 1 (underdamped, Δ < 0): roots are complex, r = −γ/2 ± i ω, where ω = √ω₀² − γ²/4. The general solution is

情况 1（欠阻尼，Δ < 0）：根为复数，r = −γ/2 ± i ω，其中 ω = √ω₀² − γ²/4。通解为

θ(t) = e^−γt/2 [A cos(ωt) + B sin(ωt)]

(critically damped, Δ = 0): repeated real root r = −γ/2. The solution is θ(t) = (A + Bt) e^−γt/2.

情况 2（临界阻尼，Δ = 0）：重实根 r = −γ/2。解为 θ(t) = (A + Bt) e^−γt/2。

rdamped, Δ > 0): two distinct real roots r₁, r₂, both negative. The solution is θ(t) = A e^r₁t + B e^r₂t.

情况 3（过阻尼，Δ > 0）：两个相异实根 r₁, r₂，均为负值。解为 θ(t) = A e^r₁t + B e^r₂t。

The constants A and B are determined from initial conditions θ(0) and dθ/dt(0).

常数 A 和 B 由初始条件 θ(0) 和 dθ/dt(0) 确定。

7. Classifying Damping Behaviour | 阻尼行为分类

ose ω₀ = 2 rad/s (so ω₀² = 4). Then the critical damping coefficient satisfies γ_c² = 4ω₀² ⇒ γ_c = 4. Thus:

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利用判别式 Δ，我们可以对真实系统进行分类。设 ω₀ = 2 rad/s（则 ω₀² = 4）。那么临界阻尼系数满足 γ_c² = 4ω₀² ⇒ γ_c = 4。因此：

• If γ = 1, Δ = 1 − 16 = −15 < 0: underdamped, oscillations slowly decay.

  若 γ = 1，Δ = 1 − 16 = −15 < 0：欠阻尼，振荡缓慢衰减。

• ically damped, the system returns to equilibrium in the shortest time without oscillation.

  若 γ = 4，Δ = 16 − 16 = 0：临界阻尼，系统无振荡地以最短时间回到平衡位置。

• = 20 > 0: overdamped, the motion decays without oscillation but more slowly than critical.

  若 γ = 6，Δ = 36 − 16 = 20 > 0：过阻尼，运动无振荡衰减，但比临界情况慢。

In SQA questions, you are often asked to identify which regime a set of parameters corresponds to.

在 SQA 试题中，经常要求你判断一组参数对应于哪个区域。

8. Applying Initial Conditions: Worked Example | 应用初始条件：计算示例

Take the underdamped case with ω₀ = 2 rad/s, γ = 0.5. Then ω = √4 − 0.0625 = √3.9375 ≈ 1.9843 rad/s. The general solution is θ(t) = e^{−0.25 t} [A cos(ωt) + B sin(ωt)].

取欠阻尼情况，ω₀ = 2 rad/s，γ = 0.5。则 ω = √4 − 0.0625 = √3.9375 ≈ 1.9843 rad/s。通解为 θ(t) = e^{−0.25 t} [A cos(ωt) + B sin(ωt)]。

Let initial conditions be θ(0) = 0.1 rad and dθ/dt(0) = 0. From θ(0) = 0.1 ⇒ A = 0.1. Differentiating and evaluating at t=0 gives:

设初始条件为 θ(0) = 0.1 rad 且 dθ/dt(0) = 0。由 θ(0) = 0.1 得 A = 0.1。求导并代入 t=0 得到：

dθ/dt = −0.25 e^−0.25t[...] + e^−0.25t[−Aω sin(ωt) + Bω cos(ωt)]

At t=0: 0 = −0.25 A + Bω ⇒ B = 0.25 A / ω ≈ 0.25×0.1 / 1.9843 ≈ 0.0126.

t=0 时：0 = −0.25 A + Bω ⇒ B = 0.25 A / ω ≈ 0.25×0.1 / 1.9843 ≈ 0.0126。

Thus the particular solution is θ(t) = e^−0.25t [0.1 cos(1.9843t) + 0.0126 sin(1.9843t)]. The amplitude envelope decays as 0.1 e^−0.25t.

因此特解为 θ(t) = e^−0.25t [0.1 cos(1.9843t) + 0.0126 sin(1.9843t)]。振幅包络按 0.1 e^−0.25t 衰减。

9. Graphical Interpretation of the Solution | 解的图形解释

The solution exhibits damped oscillation. The exponential factor e^−γt/2 ₀. The time constant of decay is τ = 2/γ, so after t = τ the amplitude reduces to about 37% of the initial value.

解表现出衰减振荡。指数因子 e^−γt/2 控制极值的衰减。准频率 ω 略小于自然频率 ω₀。衰减时间常数为 τ = 2/γ，因此经过 t = τ 后，振幅减小到初始值的约 37%。

Sketching the graph shows successive maxima joined by the envelope ±0.1 e^−0.25t 7 s. In exam solutions, labelling the axes and indicating the envelope earns method marks.

画图时，可通过包络线 ±0.1 e^−0.25t 连接连续的最大值。振荡周期为 T = 2π/ω ≈ 3.17 秒。在考试解答中，标记坐标轴并标明包络线可获得方法分。

10. Revisiting the Nonlinear Model | 重新审视非线性模型

umerically. The nonlinear restoring force −ω₀² sinθ limits the growth of amplitude and causes the period to become amplitude-dependent. For amplitude 0.5 rad, the linear period differs by about 1.6% from the true period.

当小角度近似不成立时，必须用数值方法求解原方程 d²θ/dt² + γ dθ/dt + ω₀² sinθ = 0。非线性恢复力 −ω₀² sinθ 限制了振幅的增长，并导致周期与振幅相关。对于振幅 0.5 rad，线性周期与真实周期相差约 1.6%。

In many SQA Advanced Higher assignments, you are asked to compare the linear approximation with a numerical method such as Euler’s method or a Runge–Kutta scheme.

在许多 SQA 高级高等数学作业中，会要求你将线性近似与诸如欧拉法或龙格–库塔法等数值方法进行比较。

11. Numerical Solution Using Euler’s Method | 使用欧拉方法的数值解

We convert the second-order ODE into two first-order equations. Let ω = dθ/dt. Then:

我们将二阶常微分方程转化为两个一阶方程。令 ω = dθ/dt。则有：

dθ/dt = ω
dω/dt = −γ ω − ω₀² sinθ

Using Euler’s method with step size h, the iterative formulae are:

使用步长为 h 的欧拉方法，迭代公式为：

θₙ₊₁ = θₙ + h ωₙ
ωₙ₊₁ = ωₙ + h (−γ ωₙ − ω₀² sinθₙ)

tial θ₀ = 0.1, ω₀ = 0, and h = 0.1. The first few steps are summarised in the table below.

取相同参数 ω₀ = 2，γ = 0.5，初始 θ₀ = 0.1，ω₀ = 0，步长 h = 0.1。前几个步骤总结于下表中。

Step n	tₙ	θₙ	ωₙ
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