📚 IB Mathematics: Free Oscillations of the Linear Harmonic Oscillator | IB数学：线性谐振子的自由振荡
The linear harmonic oscillator is one of the most important models in applied mathematics. In physics, it describes a mass attached to a spring that obeys Hooke's law, or a pendulum swinging with small amplitude. In this article, we focus on the mathematical treatment of free, undamped oscillations — meaning no external driving force and no friction. The problem reduces to solving a simple second-order differential equation, yet it offers deep insight into linear systems, eigenvalues, and periodic motion.
线性谐振子是应用数学中最重要的模型之一。在物理中，它描述服从胡克定律的弹簧振子，或小角度摆动的单摆。本文将重点讨论无阻尼自由振荡的数学处理——即没有外力驱动、也没有摩擦。该问题最终归结为求解一个简单的二阶微分方程，而它同时为线性系统、特征值和周期运动提供了深刻见解。
1. The Origin: Newton's Second Law and Hooke's Law | 起源：牛顿第二定律与胡克定律
displacement x from the equilibrium position and acts in the opposite direction: F = −kx.
对于劲度系数为 k 的弹簧上连接的质量 m，弹簧所产生的回复力与偏离平衡位置的位移 x 成正比，且方向相反：F = −kx。
Newton's second law states F = ma, where a = x'' he is the acceleration. Combining these two laws gives the equation of motion for the oscillator.
牛顿第二定律给出 F = ma，其中 a = x'' 为加速度。将两个定律结合，即得到振子的运动方程。
2. Setting Up the Differential Equation | 建立微分方程
Because F = ma and F = −kx, we have m x'' and rearranging yields the standard linear homogeneous second-order differential equation:
由于 F = ma 且 F = −kx，我们得到 m x'' = −kx。两边除以 m 并整理，得到标准的线性齐次二阶微分方程：
x'' + ω² x = 0, where ω² = k/m
Here ω is called the angular frequency. It is a positive constant determined only by the physical parameters m and k, not by the initial conditions.
这里的 ω 称为角频率。它是由物理参数 m 和 k 唯一确定的正常数，与初始条件无关。
3. Solving the Differential Equation Using Characteristic Roots | 用特征根求解微分方程
To solve x'' + ω² x = 0, assume a trial solution of the form x = ert. Substituting into the equation gives the characteristic equation r² + ω² = 0.
为求解 x'' + ω² x = 0，假设尝试解为 x = ert。代入方程后得到特征方程 r² + ω² = 0。
e purely imaginary. The general solution is therefore a linear combination of cos ωt and sin ωt:
特征根为 r = iω 和 r = −iω，均为纯虚数。因此通解为 cos ωt 与 sin ωt 的线性组合：
x(t) = C₁ cos ωt + C₂ sin ωt
where C₁ and C₂ are arbitrary constants determined by the initial position and velocity.
其中 C₁ 和 C₂ 是由初始位置和初始速度确定的任意常数。
4. General Solution in Amplitude–Phase Form | 振幅-相位形式的通解
tity cos(ωt + φ) = cos ωt cos φ − sin ωt sin φ, we can set C₁ = A cos φ and C₂ = −A sin φ.
同频率的余弦与正弦之和总可以写成带相位偏移的单一余弦。利用恒等式 cos(ωt + φ) = cos ωt cos φ − sin ωt sin φ，我们可令 C₁ = A cos φ，C₂ = −A sin φ。
Thus the general solution becomes:
因此通解可化为：
x(t) = A cos(ωt + φ)
Here A = √C₁² + C₂² an(−C₂/C₁) is the initial phase. Both are constants that depend on the initial conditions.
其中 A = √C₁² + C₂² 为振幅，φ = arctan(−C₂/C₁) 为初相位。二者均为取决于初始条件的常数。
5. Interpreting the Parameters: Period, Frequency, and Angular Frequency | 参数解读：周期、频率与角频率
time T required for one complete oscillation is found from ωT = 2π. This gives the period:
余弦函数每隔 2π 重复一次，因此完成一次完整振荡所需的时间 T 由 ωT = 2π 得到。于是周期为：
T = 2π/ω = 2π√m/k
of oscillations per second, f = 1/T = ω/(2π). The table below summarizes these quantities.
普通频率 f 是每秒振荡次数，f = 1/T = ω/(2π)。下表总结了这些量。
| Quantity | 物理量 | Symbol | 符号 | Formula | 公式 |
| Angular frequency | 角频率 | ω | √k/m |
| Period | 周期 | T | 2π/ω = 2π√m/k |
| Frequency | 频率 | f | ω/(2π) = 1/T |
6. Initial Conditions Determine Amplitude and Phase | 初始条件决定振幅与相位
Suppose at t = 0 we know the initial displacement x(0) = x₀ and the initial velocity x' he form x(t) = C₁ cos ωt + C₂ sin ωt, we immediately obtain C₁ = x₀ and C₂ = v₀/ω, since x'(t) = −ω C₁ sin ωt + ω C₂ cos ωt.
假设在 t = 0 时已知初始位移 x(0) = x₀ 和初始速度 x'(0) = v₀。对于形式 x(t) = C₁ cos ωt + C₂ sin ωt，我们直接得到 C₁ = x₀，C₂ = v₀/ω，因为 x'(t) = −ω C₁ sin ωt + ω C₂ cos ωt。
phase satisfies tan φ = −v₀/(ω x₀). These expressions are often needed in exam questions.
因此振幅为 A = √(x₀² + (v₀/ω)²)，相位满足 tan φ = −v₀/(ω x₀)。这些表达式在考试中经常用到。
7. Energy Conservation in Free Oscillation | 自由振荡中的能量守恒
os(ωt + φ), the potential energy stored in the spring is V = ½ k x² = ½ k A² cos²(ωt + φ).
系统的总机械能是动能与势能之和。利用 x = A cos(ωt + φ)，弹簧储存的势能为 V = ½ k x² = ½ k A² cos²(ωt + φ)。
The velocity is x' = −ω A sin(ωt + φ), so the kinetic energy is K = ½ m (x')² = ½ m ω² A² sin²(ωt + φ). Since ω² = k/m, this becomes K = ½ k A² sin²(ωt + φ).
速度为 x' = −ω A sin(ωt + φ)，所以动能为 K = ½ m (x')² = ½ m ω² A² sin²(ωt + φ)。由于 ω² = k/m，可化为 K = ½ k A² sin²(ωt + φ)。
Adding V and K gives the total energy:
将 V 与 K 相加得到总能量：
E = ½ k A² = ½ m ω² A²
This value is constant in time, illustrating that energy is conserved for undamped free oscillations.
该值不随时间变化，说明无阻尼自由振荡中能量守恒。
8. Velocity and Acceleration: Phase Relationships | 速度与加速度：相位关系
x = A cos(ωt + φ), then differentiating once gives velocity and again gives acceleration:
若位移为 x = A cos(ωt + φ)，则一次微分得到速度，再次微分得到加速度：
x'(t) = −ω A sin(ωt + φ), x''(t) = −ω² A cos(ωt + φ)
Notice that x'' iginal differential equation. The velocity leads the displacement by π/2 radians, while the acceleration is exactly out of phase (shifted by π) with the displacement.
注意 x''(t) = −ω² x(t)，这正是原始微分方程。速度领先位移 π/2 弧度，而加速度与位移的相位差为 π（方向相反）。
9. Complex Exponential Representation | 复指数表示
Euler's formula eiθ + i sin θ allows us to write the solution in a compact complex form. We set x(t) = Re( C eiωt ), where C = A eiφ is a complex constant containing the amplitude and phase.
欧拉公式 eiθ = cos θ + i sin θ 允许我们将解写成紧凑的复数形式。令 x(t) = Re( C eiωt )，其中 C = A eiφ 是包含振幅与相位的复常数。
Differentiating the complex exponential twice gives (iω)² C eiωt = −ω² C eiωt, so the differential equation is satisfied immediately. This method is extremely useful in IB mathematics, especially when combining oscillations or analysing systems with damping.
对复指数求两次导数得到 (iω)² C eiωt = −ω² C eiωt，因此微分方程立刻被满足。这种方法在 IB 数学中极其有用，尤其是在叠加振荡或分析含阻尼系统时。
10. Connection to Uniform Circular Motion | 与匀速圆周运动的联系
A simple harmonic oscillator can be viewed as the projection of uniform circular motion. A particle moving on a circle of radius A with constant angular speed ω has position (A cos(ωt + φ), A sin(ωt + φ)). Its x-coordinate is exactly the oscillator solution.
简谐振动可以看作匀速圆周运动的投影。粒子在半径为 A 的圆上以恒定角速度 ω 运动，其位置为 (A cos(ωt + φ), A sin(ωt + φ))。它的 x 坐标正好就是谐振子的解。
This geometric interpretation helps explain why the period is independent of amplitude. It also provides a visual way to recall the phase relationships between displacement, velocity, and acceleration.
这种几何解释帮助理解为什么周期与振幅无关，同时也为记忆位移、速度、加速度之间的相位关系提供了直观方法。
11. Worked Example | 典型例题
t of equilibrium and released from rest. Find the displacement x(t), the amplitude, the period, and the maximum speed.
考虑一个 0.5 kg 的物体连接在劲度系数为 2 N m⁻¹ 的弹簧上。在 t = 0 时，物体被拉至平衡位置右侧 0.3 m 处，并从静止释放。求位移 x(t)、振幅、周期和最大速度。
First, compute ω = √k/m = √2/0.5 = √4 ocity is zero, v₀ = 0, we have C₁ = x₀ = 0.3 and C₂ = 0. Thus the motion is a pure cosine:
首先计算 ω = √k/m = √2/0.5 = √4 = 2 rad s⁻¹。由于初速度为零，v₀ = 0，因此 C₁ = x₀ = 0.3，C₂ = 0。于是运动为纯余弦：
x(t) = 0.3 cos(2t)
the period is T = 2π/ω = π s ≈ 3.14 s. The maximum speed occurs when sin(2t) = ±1, giving |x'|max = ωA = 2 × 0.3 = 0.6 m s⁻¹.
振幅为 0.3 m；周期为 T = 2π/ω = π s ≈ 3.14 s。最大速度出现在 sin(2t) = ±1 时，|x'|max = ωA = 2 × 0.3 = 0.6 m s⁻¹。
12. Summary and Exam Tips | 总结与考试提示
The free motion of a linear harmonic oscillator is governed by x'' sin ωt or as x = A cos(ωt + φ). Always use radians, not degrees, when performing calculus.
线性谐振子的自由运动由 x'' + ω² x = 0 控制。其通解可写为 x = C₁ cos ωt + C₂ sin ωt 或 x = A cos(ωt + φ)。进行微积分运算时务必使用弧度制，而非角度制。
Identify ω before everything else: ω = √k/m for a mass-spring system or ω = √g/L for a small-angle pendulum.
首先确定 ω：弹簧质量系统 ω = √k/m，小角度单摆 ω = √g/L。
itial conditions to find the two constants; do not assume C₁ = A unless the phase is zero.
用初始条件求两个常数；除非相位为零，否则不要假设 C₁ = A。
Check units: ω has units rad s⁻¹, T in seconds, f in Hz (s⁻¹).
检查单位：ω 单位为 rad s⁻¹，T 单位为秒，f 单位为 Hz（s⁻¹）。
ms, write E = ½ k A² and use conservation to relate speed and displacement at any instant.
对于能量问题，写出 E = ½ k A²，并利用守恒关系联立任意时刻的速度与位移。
Mastering these concepts will allow you to solve any free-oscillation problem in the IB mathematics syllabus, whether it appears in the Analysis and Approaches or Applications and Interpretation course.
掌握这些概念后，你就能解决 IB 数学课程中任何自由振荡问题，无论该问题出现在分析与方法（Analysis and Approaches）还是应用与解释（Applications and Interpretation）课程中。
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