📚 4 Free oscillations of a linear oscillator | 4 线性振子的自由振荡
In the study of differential equations within IB Mathematics, the free oscillation of a linear oscillator provides a perfect bridge between pure calculus and physical modelling. It arises from a second-order linear homogeneous ODE, leading to simple harmonic motion and demonstrating the periodic solutions that emerge from complex roots of the characteristic equation.
在 IB 数学的微分方程学习中，线性振子的自由振荡是连接纯微积分与物理建模的完美桥梁。它由一个二阶线性齐次常微分方程产生，导出简谐运动，并展示了由特征方程复根产生的周期解。
1. The Linear Oscillator Model | 线性振子模型
A linear oscillator is any system where the restoring force is directly proportional to the displacement from equilibrium. In one dimension, if x(t) denotes displacement and the acceleration is proportional to -x, we obtain Hooke’s law in a mathematical form.
线性振子是指恢复力与偏离平衡位置的位移成正比的任何系统。在一维情况下，若用 x(t) 表示位移且加速度与 -x 成正比，我们就得到数学形式的胡克定律。
Using Newton’s second law, the net force is mass times acceleration. For an ideal spring of stiffness k, the equation m d2x/dt2 = -k x holds, which simplifies to d2x/dt2 + (k/m)x = 0.
利用牛顿第二定律，合外力等于质量乘加速度。对于劲度系数为 k 的理想弹簧，有方程 m d2x/dt2 = -k x，可化简为 d2x/dt2 + (k/m)x = 0。
Defining the constant ω2 = k/m (angular frequency squared), we obtain the standard form of the free undamped oscillator equation.
定义常数 ω2 = k/m（角频率的平方），我们得到无阻尼自由振荡方程的标准形式。
d2x/dt2 + ω2x = 0
2. Classification as a Second-Order Linear ODE | 作为二阶线性常微分方程的分类
The equation is a second-order linear homogeneous ordinary differential equation with constant coefficients. Its general form is a y” + b y’ + c y = 0. Comparing with d2x/dt2 + ω2x = 0, we have a = 1, b = 0, c = ω2.
该方程是一个常系数二阶线性齐次常微分方程。其一般形式为 a y” + b y’ + c y = 0。与 d2x/dt2 + ω2x = 0 对比，可得 a = 1, b = 0, c = ω2。
Since b = 0, the damping term is absent, which corresponds to free, undamped motion. The absence of a first derivative term is the mathematical signature of energy conservation in the oscillator.
由于 b = 0，阻尼项缺失，这对应于自由、无阻尼的运动。一阶导数项的缺失在数学上是振子能量守恒的标志。
The ODE is autonomous because the independent variable t does not explicitly appear; the system is time-invariant, meaning solutions can be shifted in time.
该常微分方程是自治的，因为自变量 t 没有显式出现；系统是时不变的，意味着解可以在时间上平移。
3. Characteristic Equation and Its Roots | 特征方程及其根
To solve the ODE, we assume a trial solution x = ert. Substituting yields the characteristic (auxiliary) equation: r2 + ω2 = 0.
为了求解该常微分方程，我们假设试探解 x = ert。代入后得到特征（辅助）方程：r2 + ω2 = 0。
The roots are purely imaginary: r = ± iω, where i is the imaginary unit. This pair of complex conjugate roots signals oscillatory behaviour.
根为纯虚数：r = ± iω，其中 i 为虚数单位。这对共轭复根预示着振荡行为。
In IB Mathematics AA and AI, students learn that complex roots α ± iβ lead to a general solution of the form eαt(C₁ cos βt + C₂ sin βt). Here α = 0, β = ω.
在 IB 数学 AA 和 AI 中，学生学到复根 α ± iβ 会导出通解形式 eαt(C₁ cos βt + C₂ sin βt)。此处 α = 0, β = ω。
4. General Solution in Trigonometric Form | 三角形式的通解
Applying the formula with α = 0 gives the general solution x(t) = C₁ cos(ωt) + C₂ sin(ωt), where C₁, C₂ are arbitrary real constants determined by initial conditions.
将 α = 0 代入公式，得通解 x(t) = C₁ cos(ωt) + C₂ sin(ωt)，其中 C₁, C₂ 是由初始条件确定的任意实常数。
An equivalent and often more insightful form is the amplitude–phase representation: x(t) = A cos(ωt + φ), where A is the amplitude and φ is the initial phase shift.
一个等价且通常更具洞察力的形式是振幅–相位表示：x(t) = A cos(ωt + φ)，其中 A 为振幅，φ 为初相位。
Using the cosine addition formula, A cos(ωt + φ) = A cos φ cos(ωt) – A sin φ sin(ωt), so C₁ = A cos φ and C₂ = -A sin φ. Both representations are useful in different contexts.
利用余弦加法公式，A cos(ωt + φ) = A cos φ cos(ωt) – A sin φ sin(ωt)，因此 C₁ = A cos φ，C₂ = -A sin φ。两种表示在不同语境中各有用途。
5. Physical Interpretation of Parameters | 参数的物理解释
The constant ω is the angular frequency, measured in rad s⁻¹. It determines how rapidly the oscillation completes a cycle. The amplitude A is the maximum displacement from equilibrium.
常数 ω 是角频率，单位为 rad s⁻¹，它决定了振荡完成一个周期的快慢。振幅 A 是偏离平衡位置的最大位移。
The phase φ describes where the oscillator is at t = 0. If φ = 0, the oscillator starts at maximum positive displacement; if φ = -π/2, it starts at equilibrium moving forward.
相位 φ 描述了在 t = 0 时振子的状态。若 φ = 0，振子从最大正向位移开始；若 φ = -π/2，它从平衡位置开始正向运动。
All solutions of the free linear oscillator are purely sinusoidal, and no energy is lost. This idealisation is crucial before adding damping or forcing.
自由线性振子的所有解都是纯正弦的，且没有能量损失。这一理想化在添加阻尼或外力之前至关重要。
6. Period and Frequency | 周期与频率
The period T of the oscillation is the time taken for one complete cycle. Since cos(ω(t+T)+φ) = cos(ωt+φ + 2π), we require ωT = 2π, giving T = 2π/ω.
振荡周期 T 是完成一个完整循环所需的时间。因为 cos(ω(t+T)+φ) = cos(ωt+φ + 2π)，需要 ωT = 2π，从而 T = 2π/ω。
The ordinary frequency f is the number of cycles per second: f = 1/T = ω/(2π). In mathematics, ω is often specified first, and T follows.
常规频率 f 是每秒的循环数：f = 1/T = ω/(2π)。在数学中，ω 通常先于 T 被指定。
These relationships are particularly useful when linking the differential equation parameter ω to observable quantities in a modelling context.
这些关系在将微分方程参数 ω 与建模情境中的可观测量相联系时尤为有用。
Table of key relationships:
| Quantity | Symbol | Expression |
|---|---|---|
| Angular frequency | ω | √(k/m) |
| Period | T | 2π/ω |
| Frequency | f | 1/T = ω/(2π) |
关键关系表:
| 量 | 符号 | 表达式 |
|---|---|---|
| 角频率 | ω | √(k/m) |
| 周期 | T | 2π/ω |
| 频率 | f | 1/T = ω/(2π) |
7. Applying Initial Conditions | 利用初始条件
To determine the constants C₁ and C₂, we need the initial displacement x(0) = x0 and initial velocity v(0) = x'(0) = v0. Substituting t = 0 into x(t) gives C₁ = x0.
要确定常数 C₁ 和 C₂，我们需要初始位移 x(0) = x0 和初始速度 v(0) = x'(0) = v0。将 t = 0 代入 x(t) 得 C₁ = x0。
Differentiating x(t) yields v(t) = -C₁ ω sin(ωt) + C₂ ω cos(ωt). Setting t = 0 gives v(0) = C₂ ω, so C₂ = v0/ω.
对 x(t) 求导得 v(t) = -C₁ ω sin(ωt) + C₂ ω cos(ωt)。令 t = 0 得 v(0) = C₂ ω，故 C₂ = v0/ω。
Therefore, the particular solution is x(t) = x0 cos(ωt) + (v0/ω) sin(ωt). The amplitude–phase form then gives A = √(x0² + (v0/ω)²) and φ = arctan(-v0/(ω x0)) with quadrant adjustments.
因此，特解为 x(t) = x0 cos(ωt) + (v0/ω) sin(ωt)。振幅–相位形式则为 A = √(x0² + (v0/ω)²)，且 φ = arctan(-v0/(ω x0))（需根据象限调整）。
These expressions are extremely common in IB exam questions, where either the constants or the amplitude and phase are requested.
这些表达式在 IB 考题中极为常见，通常要求求出常数或振幅与相位。
8. Worked Example | 示例解析
Consider an oscillator governed by d2x/dt2 + 16x = 0, with x(0) = 3 cm and v(0) = -8 cm s⁻¹. Here ω = √16 = 4 rad s⁻¹. The general solution is x(t) = C₁ cos(4t) + C₂ sin(4t).
考虑一个满足 d2x/dt2 + 16x = 0 的振子，初始条件为 x(0) = 3 cm，v(0) = -8 cm s⁻¹。这里 ω = √16 = 4 rad s⁻¹。通解为 x(t) = C₁ cos(4t) + C₂ sin(4t)。
From x(0) = C₁ = 3. From v(t) = -3·4 sin(4t) + C₂·4 cos(4t), so v(0) = 4C₂ = -8 ⇒ C₂ = -2. Hence x(t) = 3 cos(4t) – 2 sin(4t).
由 x(0) = C₁ = 3，由 v(t) = -3·4 sin(4t) + C₂·4 cos(4t)，得 v(0) = 4C₂ = -8 ⇒ C₂ = -2。因此 x(t) = 3 cos(4t) – 2 sin(4t)。
The amplitude A = √(3² + (-2)²) = √13 ≈ 3.61 cm. The phase φ satisfies cos φ = 3/A, sin φ = 2/A (since C₂ = -A sin φ, -2 = -A sin φ gives sin φ = 2/A). Thus φ ≈ arctan(2/3) ≈ 0.588 rad. So x(t) = √13 cos(4t + 0.588).
振幅 A = √(3² + (-2)²) = √13 ≈ 3.61 cm。相位 φ 满足 cos φ = 3/A，sin φ = 2/A（因为 C₂ = -A sin φ，-2 = -A sin φ 得 sin φ = 2/A）。从而 φ ≈ arctan(2/3) ≈ 0.588 rad。故 x(t) = √13 cos(4t + 0.588)。
This explicit form readily gives the position at any time and shows that the motion is bounded between -3.61 cm and 3.61 cm.
这一显式形式可直接给出任意时刻的位置，并表明运动被限制在 -3.61 cm 与 3.61 cm 之间。
9. Conservation of Energy from the ODE | 从常微分方程看能量守恒
Multiplying both sides of the ODE by dx/dt and integrating gives a first integral: ½ (dx/dt)² + ½ ω² x² = constant. This mathematical identity represents the conservation of mechanical energy.
将常微分方程两边乘以 dx/dt 并积分，得到一个首次积分：½ (dx/dt)² + ½ ω² x² = 常数。此数学恒等式代表机械能守恒。
The term ½ (dx/dt)² is proportional to kinetic energy, and ½ ω² x² to elastic potential energy. Their sum remains unchanged throughout the motion, confirming no energy dissipation.
项 ½ (dx/dt)² 与动能成正比，½ ω² x² 与弹性势能成正比。两者之和在整个运动过程中保持不变，证实无能量耗散。
This first-order relationship can also be used to sketch phase portraits (velocity vs. displacement), which are ellipses for the undamped oscillator.
此一阶关系也可用于绘制相图（速度对位移），对无阻尼振子而言为椭圆。
10. Connection to Complex Exponentials | 与复指数的联系
Another representation uses Euler’s formula: x(t) = Re{ X eiωt } where X is a complex constant X = A eiφ. This form is concise and often used in higher-level mathematics and physics.
另一种表示利用欧拉公式：x(t) = Re{ X eiωt }，其中 X 为复常数 X = A eiφ。该形式简洁，常用于更高阶的数学和物理中。
The complex solution approach directly shows that r = ±iω and that the real part of Ceiωt covers all solutions. IB students may encounter this when exploring the connection between trigonometric and exponential functions.
复解法直接表明 r = ±iω，且 Ceiωt 的实部涵盖所有解。IB 学生在探索三角函数与指数函数的联系时可能会遇到该方法。
11. Summary and Exam Tips | 总结与考试技巧
Free oscillations of a linear oscillator are governed by d2x/dt2 + ω2x = 0, with general solution x = C₁ cos(ωt) + C₂ sin(ωt) = A cos(ωt + φ). The constants are determined by initial position and velocity.
线性振子的自由振荡由 d2x/dt2 + ω2x = 0 支配，通解为 x = C₁ cos(ωt) + C₂ sin(ωt) = A cos(ωt + φ)。常数由初始位置和速度确定。
In the IB exam, you should be able to: (1) write down the characteristic equation and its roots; (2) give the general solution; (3) use initial conditions to find particular solutions; (4) convert between the cosine–sine and amplitude–phase forms; and (5) interpret ω, T, f, A, φ in context.
在 IB 考试中，你应能：(1) 写出特征方程及其根；(2) 给出通解；(3) 利用初始条件求特解；(4) 在正余弦形式与振幅–相位形式之间转换；(5) 在具体情境中解释 ω, T, f, A, φ。
Memorise the key formulas A = √(x0² + (v0/ω)²) and T = 2π/ω, and remember to check quadrant for the phase angle. Algebraic manipulation and differentiation are essential skills tested alongside this topic.
牢记关键公式 A = √(x0² + (v0/ω)²) 和 T = 2π/ω，并记住检查相位角的象限。代数处理与微分是与本主题一同考查的基本技能。
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