Simple Harmonic Motion: IB & CCEA A-Level Maths Key Points | 简谐运动：IB与CCEA数学考点精讲

📚 Simple Harmonic Motion: IB & CCEA A-Level Maths Key Points | 简谐运动：IB与CCEA数学考点精讲

Simple Harmonic Motion (SHM) is a fundamental topic in both IB Mathematics (Analysis & Approaches / Applications & Interpretation) and CCEA A-Level Mathematics. It bridges pure calculus with real-world oscillatory systems, making it a key area for modelling questions and differential equations. Understanding SHM from first principles – through second-order differential equations, trigonometric solutions, and energy considerations – is essential for top marks in both syllabi.

简谐运动（SHM）是 IB 数学（分析与方法/应用与解释）以及 CCEA A-Level 数学的核心课题。它将纯微积分与现实振动系统联系起来，因此成为建模题和微分方程应用题的重点考查内容。从基本原理出发理解 SHM——通过二阶微分方程、三角函数解以及能量分析——对在两个考试局中取得高分至关重要。

1. Defining Simple Harmonic Motion | 简谐运动的定义

SHM occurs when the acceleration of a particle is directly proportional to its displacement from a fixed equilibrium point and is always directed towards that equilibrium. Mathematically, this is expressed as a = −k x, where k is a positive constant. In standard notation, we write a = −ω² x, where ω is the angular frequency. The negative sign indicates that the acceleration opposes the displacement, which is the restoring condition that drives oscillatory behaviour.

当质点的加速度与其相对于某个固定平衡位置的位移成正比，并且方向始终指向该平衡点时，物体做简谐运动。数学上表示为 a = −k x，其中 k 为正的常数。在标准记号中，我们记为 a = −ω² x，其中 ω 为角频率。负号表明加速度与位移方向相反，正是这种恢复作用驱动了振动。

This definition forms the starting point for deriving the equations of motion. Both IB and CCEA exam questions often begin by asking you to recognise or verify that a given physical situation satisfies a = −ω² x, then proceed to find displacement, velocity, or time expressions.

这一定义是推导运动方程的起点。IB 和 CCEA 的试题常常先要求考生识别或验证某一物理情境是否满足 a = −ω² x，继而求出位移、速度或时间表达式。

Key condition: resultant force (or acceleration) ∝ −displacement
关键条件：合力（或加速度）正比于负的位移

2. The Differential Equation of SHM | 简谐运动的微分方程

Since acceleration is the second derivative of displacement with respect to time, a = −ω² x becomes the second-order linear differential equation d²x/dt² = −ω² x, or equivalently d²x/dt² + ω² x = 0. This is the standard form examined in both syllabi. IB Analysis & Approaches requires students to solve this ODE analytically, while CCEA also expects the solution starting from the auxiliary equation m² + ω² = 0, giving m = ± iω, and therefore a general solution of the form x = A cos ωt + B sin ωt.

加速度是位移对时间的二阶导数，因此 a = −ω² x 化为二阶线性微分方程 d²x/dt² = −ω² x，或等价地 d²x/dt² + ω² x = 0。这是两个考试局均考查的标准形式。IB 分析与方法要求学生解析求解该常微分方程，而 CCEA 也要求从辅助方程 m² + ω² = 0 出发，得到 m = ± iω，从而得出通解形式 x = A cos ωt + B sin ωt。

Many problems provide initial conditions such as t = 0, x = x₀, v = 0, allowing the arbitrary constants to be determined. Being fluent in applying these conditions is a core skill for SHM problems in both curricula.

许多问题会提供初始条件，例如 t = 0, x = x₀, v = 0，从而确定任意常数。熟练运用这些条件是掌握两个课程中 SHM 问题的核心技能。

d²x/dt² + ω² x = 0

3. General Solution and Alternative Forms | 通解与等价形式

The general solution x = A cos ωt + B sin ωt can be expressed in the amplitude-phase form x = A sin(ωt + φ) or x = A cos(ωt + φ), depending on convention. In IB and CCEA exams, you may be asked to rewrite an expression like x = 3 cos ωt + 4 sin ωt into the form R cos(ωt − α) using compound angle identities. Here R = √(A² + B²) gives the amplitude, and α is the phase shift determined by tan α = B/A.

通解 x = A cos ωt + B sin ωt 可以写成振幅-相位形式 x = A sin(ωt + φ) 或 x = A cos(ωt + φ)，取决于习惯。在 IB 和 CCEA 考试中，考生可能需将如 x = 3 cos ωt + 4 sin ωt 的表达式通过复合角公式改写为 R cos(ωt − α) 的形式。此时 R = √(A² + B²) 为振幅，α 为由 tan α = B/A 确定的相位差。

Both syllabi test the ability to work across different trigonometric representations. Being able to interpret initial phase and relate the maximum displacement to the amplitude R is vital for modelling pendulum or spring systems.

两个教学大纲都考查在不同三角表示形式之间转换的能力。能够解释初相并且将最大位移与振幅 R 关联起来，对于单摆或弹簧系统的建模至关重要。

Amplitude: maximum displacement from equilibrium
Period: T = 2π/ω
Frequency: f = 1/T = ω/(2π)
振幅：相对于平衡位置的最大位移
周期：T = 2π/ω
频率：f = 1/T = ω/(2π)

4. Velocity and Acceleration Functions | 速度与加速度函数

Once the displacement x is known, the velocity v = dx/dt and acceleration a = d²x/dt² follow by differentiation. For x = A sin(ωt), we get v = Aω cos(ωt) and a = −Aω² sin(ωt) = −ω² x. An extremely useful relationship is v² = ω²(A² − x²), which comes from using trigonometric identities or from energy considerations. This equation allows you to find speed at any displacement without needing time t.

一旦已知位移 x，通过求导即可得到速度 v = dx/dt 与加速度 a = d²x/dt²。对于 x = A sin(ωt)，有 v = Aω cos(ωt)，a = −Aω² sin(ωt) = −ω² x。一个极为有用的关系式是 v² = ω²(A² − x²)，可借助三角恒等式或通过能量分析得出。该方程使我们在已知位移时无需用到时间 t 即可求出速率。

This v² equation is frequently set up in exam questions to find maximum speed v_max = ωA (at x = 0) and to prove that the motion is indeed SHM. CCEA papers often ask: ‘Show that v² = ω²(a² − x²)’ and then use it to find period or amplitude.

该 v² 方程频频出现在试题中，用以求出最大速率 v_max = ωA（在 x = 0 处），并证明运动确为 SHM。CCEA 试卷常出现：“证明 v² = ω²(a² − x²)” 并以此求周期或振幅。

v = dx/dt = ω√(A² − x²) | v_max = ωA at x = 0

5. Graphical Interpretation of Displacement, Velocity, and Acceleration | 位移、速度和加速度的图像解释

Understanding SHM graphs is essential. The displacement-time graph is a sine or cosine wave with maximum ±A. The velocity-time graph is also sinusoidal but leads the displacement by π/2 (a quarter of a cycle). The acceleration-time graph is a sine wave anti-phase with displacement (phase difference of π). When displacement is maximum, velocity is zero and acceleration is maximum in the opposite direction. When displacement is zero, speed is maximum and acceleration is zero.

理解 SHM 的图像至关重要。位移–时间图像为正弦或余弦波，最大值为 ±A。速度–时间图像也是正弦波，但超前位移四分之一周期（相位领先 π/2）。加速度–时间图像是与位移反相（相位差 π）的正弦波。位移最大时速度为零，加速度相反方向最大；位移为零时速率最大，加速度为零。

Exam questions frequently ask you to sketch these three graphs on the same axes or to identify the phase relationships. IB in particular likes linking SHM with wave behaviour, while CCEA tests interpretation of given velocity–displacement graphs.

考试常要求在相同坐标轴上绘制这三种图像或辨认相位关系。IB 尤喜将 SHM 与波动行为联系起来，CCEA 则考查对给定的速度–位移图像进行解读。

Displacement x: x = A sin ωt
Velocity v: v = ωA cos ωt = ωA sin(ωt + π/2)
Acceleration a: a = −ω²A sin ωt = ω²A sin(ωt + π)
位移 x：x = A sin ωt
速度 v：v = ωA cos ωt = ωA sin(ωt + π/2)
加速度 a：a = −ω²A sin ωt = ω²A sin(ωt + π)

6. Period, Frequency, and Angular Frequency | 周期、频率与角频率

The period T is the time for one complete oscillation: T = 2π/ω. Frequency f = 1/T = ω/(2π) is the number of oscillations per second. Angular frequency ω has units rad s⁻¹ and relates directly to the physical properties of the system. For a mass-spring system, ω = √(k/m), giving T = 2π√(m/k). For a simple pendulum, ω = √(g/L), giving T = 2π√(L/g) (for small angles).

周期 T 是一次完整振动所需的时间：T = 2π/ω。频率 f = 1/T = ω/(2π) 是每秒振动次数。角频率 ω 的单位为 rad s⁻¹，并直接与系统的物理性质相关。对于质量–弹簧系统，ω = √(k/m)，从而 T = 2π√(m/k)。对于单摆，ω = √(g/L)，从而 T = 2π√(L/g)（小角度近似）。

These formulas are derived from the defining differential equation by substituting the net force. In CCEA mechanics papers, you are often required to derive T for a horizontal spring or a simple pendulum from first principles, which demands linking Newton’s second law with the SHM condition a = −ω² x.

上述公式通过将合力的表达式代入 SHM 条件 a = −ω² x 推导而出。在 CCEA 力学试卷中，常要求从基本原理出发推导水平弹簧振子或单摆的周期 T，这需要将牛顿第二定律与 SHM 条件联系起来。

7. The Horizontal Mass-Spring System | 水平质量–弹簧系统

Consider a mass m attached to a spring of stiffness k on a smooth horizontal surface. The restoring force is F = −k x. By Newton’s second law, m d²x/dt² = −k x ⇒ d²x/dt² + (k/m) x = 0. This matches d²x/dt² + ω² x = 0 with ω² = k/m. Hence the motion is SHM with period T = 2π/ω = 2π√(m/k).

考虑一个连接在劲度系数为 k 的弹簧上的质量 m，置于光滑水平面上。恢复力为 F = −k x。由牛顿第二定律，m d²x/dt² = −k x ⇒ d²x/dt² + (k/m) x = 0。与 d²x/dt² + ω² x = 0 对照得 ω² = k/m，因此物体做简谐运动，周期 T = 2π/ω = 2π√(m/k)。

Extension to vertical oscillations simply shifts the equilibrium position by mg/k, but the SHM part remains identical. Both IB and CCEA examine vertical springs, requiring you to measure displacement from the equilibrium position, not the natural length.

竖直方向振动的推广无非将平衡位置下移 mg/k，但 SHM 部分完全一致。IB 和 CCEA 均考查竖直弹簧振子，要求从平衡位置（而非原长）测量位移。

8. The Simple Pendulum | 单摆

For a point mass m suspended by a light inextensible string of length L, the tangential restoring force for small angular displacement θ is −mg sin θ ≈ −mg θ. The tangential acceleration is L d²θ/dt², so L d²θ/dt² = −g θ ⇒ d²θ/dt² + (g/L) θ = 0. This is SHM in the angular variable with ω² = g/L, giving T = 2π/ω = 2π√(L/g).

对于长为 L 的轻质不可伸长细线悬挂的质点 m，当角位移 θ 较小时，切向恢复力为 −mg sin θ ≈ −mg θ。切向加速度为 L d²θ/dt²，故 L d²θ/dt² = −g θ ⇒ d²θ/dt² + (g/L) θ = 0。这是以角度为变量的简谐运动，ω² = g/L，从而 T = 2π/ω = 2π√(L/g)。

The small-angle approximation sin θ ≈ θ (in radians) is essential. IB mark schemes insist on stating this approximation. CCEA often embeds pendulum questions within differential equation problems, asking for the period and for the solution of θ as a function of time given initial conditions.

小角度近似 sin θ ≈ θ（以弧度计）是关键。IB 评分标准要求明确写明这一近似。CCEA 常将单摆问题嵌入微分方程大题中，要求给出周期以及在给定初始条件下求 θ 随时间变化的解。

9. Energy in Simple Harmonic Motion | 简谐运动中的能量

The total mechanical energy of an undamped SHM system is constant and can be expressed as E = ½ m ω² A². This arises from the sum of kinetic energy K = ½ m v² and potential energy U = ½ k x² for a spring system, or analogous gravitational potential for a pendulum. At maximum displacement, energy is entirely potential; at equilibrium, entirely kinetic.

无阻尼简谐运动系统的总机械能守恒，并可表示为 E = ½ m ω² A²。它由动能 K = ½ m v² 和弹簧系统的势能 U = ½ k x²（或单摆中对应重力势能）相加而得。在最大位移处，能量全部为势能；在平衡位置，能量全部为动能。

Using the energy equation, you can derive v² = ω²(A² − x²) without calculus. IB applications & interpretation and CCEA both include energy-based SHM questions, often linking to graphs of K and U against displacement.

利用能量方程可以在不借助微积分的情况下推导出 v² = ω²(A² − x²)。IB 应用与解释和 CCEA 均包含基于能量的 SHM 问题，常与动能和势能随位移变化的图像结合。

K = ½ m ω² (A² − x²) | U = ½ m ω² x² | E_total = ½ m ω² A²

10. Damped Simple Harmonic Motion | 阻尼简谐运动

In real systems, resistive forces cause the amplitude to decrease gradually. Light damping leads to underdamped oscillations, where the system still oscillates but with an exponentially decaying amplitude. The equation becomes d²x/dt² + 2β dx/dt + ω₀² x = 0, where β is the damping coefficient. The solution takes the form x = A e^(−β t) cos(ω₁ t + φ), with ω₁ = √(ω₀² − β²).

在真实系统中，阻力会导致振幅逐渐减小。弱阻尼引起欠阻尼振动，即系统仍会振荡，但振幅按指数衰减。方程变为 d²x/dt² + 2β dx/dt + ω₀² x = 0，其中 β 为阻尼系数。其解的形式为 x = A e^(−β t) cos(ω₁ t + φ)，ω₁ = √(ω₀² − β²)。

This topic appears mainly in IB Analysis & Approaches HL as an extension of second-order differential equations. CCEA may ask qualitative descriptions of light, critical, and heavy damping, but not usually the full analytical solution. Knowing how to classify damping by the discriminant of the auxiliary equation is tested in IB.

此主题主要在 IB 分析与方法 HL 中以二阶微分方程拓展的形式出现。CCEA 可能要求定性描述弱阻尼、临界阻尼与过阻尼，但通常不要求完整的解析解。IB 则会考查如何通过辅助方程的判别式对阻尼类型进行分类。

11. Forced Oscillations and Resonance | 受迫振动与共振

When an external periodic force is applied, the system oscillates at the driving frequency. The amplitude becomes large when the driving frequency approaches the natural frequency ω₀, a phenomenon called resonance. The equation is d²x/dt² + 2β dx/dt + ω₀² x = F₀ cos(ω t). The steady-state solution has amplitude that peaks near ω = ω₀. Sharpness of resonance depends on the damping: lighter damping gives a sharper peak.

当施加外部周期性驱动力时，系统以驱动频率振动。当驱动频率接近固有频率 ω₀ 时，振幅变得很大，此现象称为共振。方程为 d²x/dt² + 2β dx/dt + ω₀² x = F₀ cos(ω t)。其稳态解的振幅在 ω = ω₀ 附近达到峰值。共振的尖锐程度取决于阻尼：阻尼越小，峰值越尖锐。

IB HL and some CCEA mechanics modules explore resonance, often through contextual problems like buildings swaying, bridge oscillations, or mechanical vibrations. You need to be able to explain why the amplitude grows and the role of energy input matching the natural frequency.

IB HL 以及 CCEA 力学部分模块会探讨共振，常通过诸如建筑物摇晃、桥梁振动或机械振动的应用题进行考查。你需要能够解释振幅为什么增大，以及能量输入与固有频率相匹配的作用。

12. Exam Strategies and Common Mistakes | 考试策略与常见错误

First, always define the equilibrium position clearly. SHM must be measured from equilibrium, not from the spring’s natural length. For velocity and acceleration questions, double-check the phase relationships: v = 0 at extreme points, a = max; at the centre, v = max, a = 0. When solving differential equations, write the general solution before applying initial conditions to avoid sign errors.

首先，务必明确定义平衡位置。SHM 必须从平衡位置开始度量，而非弹簧的原长。处理速度与加速度问题时，请反复确认相位关系：极端位置处 v = 0，a 最大；中心位置处 v 最大，a = 0。在求解微分方程时，先写出通解再应用初始条件，以避免符号错误。

In IB, the use of the auxiliary equation and complex roots must be shown explicitly. For CCEA, derivations of T from first principles (using F = ma and matching with −ω² x) are frequently worth many marks. Always state the small-angle approximation for pendulum problems. Finally, remember to switch your calculator to radian mode – degrees will give wrong values for ω and trigonometric derivatives.

在 IB 中，须明确展示辅助方程与复根的使用。对于 CCEA，从第一性原理推导 T（利用 F = ma 并与 −ω² x 对比）常占据大量分值。单摆问题务必说明小角度近似。最后，记得将计算器切换为弧度模式——角度制会给出错误的 ω 和三角求导结果。

Measure x from equilibrium, not natural length
Check phase: v leads x by π/2, a is anti-phase
Use v² = ω²(A² − x²) to avoid time t
Pendulum: sin θ ≈ θ (radians) for small angles
从平衡位置开始度量 x，而非原长
核对相位：v 领先 x π/2，a 与 x 反相
善用 v² = ω²(A² − x²) 避开时间 t
单摆：小角度时 sin θ ≈ θ（弧度）

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Simple Harmonic Motion: IB & CCEA A-Level Maths Key Points | 简谐运动：IB与CCEA数学考点精讲