IB Mathematics: Free Oscillations of the Linear Harmonic Oscillator | IB数学：线性谐振子的自由振荡

📚 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

For a mass m attached to a spring with spring constant k, the restoring force exerted by the spring is proportional to the 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” 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” = −kx. Dividing by m 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 = e^(rt). 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 = e^(rt)。代入方程后得到特征方程 r² + ω² = 0。

The roots are r = iω and r = −iω, which are 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

The sum of a cosine and sine with the same frequency can always be written as a single cosine with a phase shift. Using the identity 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) is the amplitude, and φ = arctan(−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

The cosine function repeats every 2π, so the 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)

The ordinary frequency f is the number 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'(0) = v₀. For the 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。

Therefore the amplitude is A = √(x₀² + (v₀/ω)²), and the 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

The total mechanical energy of the system is the sum of kinetic energy and potential energy. Using x = A cos(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

If displacement is 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”(t) = −ω² x(t), which is exactly the original 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 e^(iθ) = cos θ + i sin θ allows us to write the solution in a compact complex form. We set x(t) = Re( C e^(iωt) ), where C = A e^(iφ) is a complex constant containing the amplitude and phase.

欧拉公式 e^(iθ) = 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 e^(iωt) )，其中 C = A e^(iφ) 是包含振幅与相位的复常数。

Differentiating the complex exponential twice gives (iω)² C e^(iωt) = −ω² C e^(iωt), 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 e^(iωt) = −ω² C e^(iωt)，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

Consider a mass of 0.5 kg attached to a spring of spring constant 2 N m⁻¹. At t = 0, the mass is pulled 0.3 m to the right 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 = 2 rad s⁻¹. Since the initial velocity 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 amplitude is 0.3 m; 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” + ω² x = 0. Its general solution can be written as x = C₁ cos ωt + C₂ 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)。

• Use initial 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）。

• For energy problems, 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。

---

Published by TutorHao | IB Mathematical Methods Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）