IGCSE Maths: Simple Harmonic Motion Key Points Review | IGCSE 数学：简谐运动考点精讲

📚 IGCSE Maths: Simple Harmonic Motion Key Points Review | IGCSE 数学：简谐运动考点精讲

Simple harmonic motion (SHM) is a fundamental type of oscillatory motion found in many physical systems, from a mass on a spring to a pendulum. In the context of IGCSE Mathematics, particularly for students aiming for high grades in Additional Mathematics or preparing for A Level studies, understanding the mathematical description of SHM is essential. This article will break down the key concepts, equations, and typical exam questions related to simple harmonic motion, using clear notation and step‑by‑step reasoning.

简谐运动（SHM）是在许多物理系统中常见的一种基本振动形式，如弹簧振子或单摆。在 IGCSE 数学中，特别是对于追求高分的附加数学学生或为 A Level 学习做准备的同学，掌握简谐运动的数学描述至关重要。本文将使用清晰的符号和逐步推理，拆解简谐运动的核心概念、方程以及典型考题。

1. What is Simple Harmonic Motion? | 什么是简谐运动？

Simple harmonic motion is a type of periodic motion where the restoring force – and therefore the acceleration – is directly proportional to the displacement from an equilibrium position and is always directed towards that equilibrium. Mathematically, this can be written as a ∝ −x. The motion is sinusoidal in time, which means it can be described using sine or cosine functions.

简谐运动是一种周期性运动，其恢复力（及加速度）的大小与偏离平衡位置的位移成正比，方向始终指向平衡位置。数学上可表示为 a ∝ −x。该运动随时间呈正弦变化，因此可用正弦或余弦函数描述。

a ∝ −x

2. Defining Equation of SHM | 简谐运动的定义方程

The acceleration of a particle in simple harmonic motion satisfies a = −ω²x, where ω (omega) is a constant called the angular frequency. This is the fundamental differential equation of SHM: d²x/dt² = −ω²x. Solving this differential equation yields the displacement equation x = A sin(ωt + φ) or x = A cos(ωt + φ), where A is the amplitude and φ is the phase constant that depends on the initial conditions.

简谐运动中质点的加速度满足 a = −ω²x，其中 ω（欧米伽）是一个常数，称为角频率。这是简谐运动的基本微分方程：d²x/dt² = −ω²x。解此微分方程得到位移方程 x = A sin(ωt + φ) 或 x = A cos(ωt + φ)，式中 A 为振幅，φ 为依赖于初始条件的初相。

a = −ω²x

3. Amplitude, Period, and Frequency | 振幅、周期和频率

The amplitude A is the maximum displacement from the equilibrium position. The period T is the time taken for one complete oscillation. Since the sine or cosine function repeats every 2π radians, we have ωT = 2π, giving T = 2π/ω. Frequency f is the number of oscillations per unit time: f = 1/T = ω/(2π). Angular frequency ω is measured in radians per second (rad/s).

振幅 A 是离开平衡位置的最大位移。周期 T 是完成一次完整振动所需的时间。因为正弦或余弦函数每 2π 弧度重复一次，所以有 ωT = 2π，即 T = 2π/ω。频率 f 是单位时间内的振动次数：f = 1/T = ω/(2π)。角频率 ω 的单位是弧度每秒（rad/s）。

T = 2π/ω f = ω/(2π)

4. Displacement, Velocity, and Acceleration Functions | 位移、速度和加速度函数

Starting from a convenient form x = A sin(ωt), we can differentiate to find velocity and acceleration. Velocity is v = dx/dt = Aω cos(ωt). It can also be expressed as v = Aω sin(ωt + π/2), showing that velocity leads displacement by a phase of π/2. The maximum speed is v_max = Aω, occurring when the particle passes through the equilibrium position.

从一个方便的形式 x = A sin(ωt) 出发，通过求导可得速度和加速度。速度 v = dx/dt = Aω cos(ωt)。也可写作 v = Aω sin(ωt + π/2)，表明速度比位移超前 π/2 相位。最大速率为 v_max = Aω，出现在质点经过平衡位置时。

Acceleration is the derivative of velocity: a = dv/dt = −Aω² sin(ωt) = −ω²x. The maximum magnitude of acceleration is a_max = ω²A, which occurs at the extreme displacements. The relationship a = −ω²x is the hallmark of SHM and can be used to verify whether a given motion is simple harmonic.

加速度是速度的导数：a = dv/dt = −Aω² sin(ωt) = −ω²x。加速度的最大大小为 a_max = ω²A，出现在最大位移处。a = −ω²x 是简谐运动的标志性关系，可用于检验某个运动是否为简谐运动。

Quantity	Expression (x = A sin ωt)	Maximum value
Displacement x	A sin ωt	A
Velocity v	Aω cos ωt	Aω
Acceleration a	−Aω² sin ωt	ω²A

5. The Phase and Initial Phase | 相位与初相

The argument (ωt + φ) inside the sine or cosine function is called the phase of the motion. The constant φ is the initial phase (or phase constant), which determines the state at t = 0. For x = A sin(ωt + φ), when t = 0 we have x₀ = A sin φ. If the particle starts at the equilibrium position and moves in the positive direction, φ = 0. If it starts at the positive amplitude, φ = π/2.

正弦或余弦函数内部的辐角 (ωt + φ) 称为相位。常数 φ 是初相，决定了 t = 0 时的状态。对于 x = A sin(ωt + φ)，当 t = 0 时有 x₀ = A sin φ。若质点从平衡位置向正方向开始运动，则 φ = 0；若从正的最大位移处开始，则 φ = π/2。

Selecting the appropriate trigonometric function can simplify problems. Using x = A cos(ωt) automatically gives an initial maximum displacement. Examination questions often ask you to determine φ from given initial values of x and v.

选择合适的三角函数可以简化问题。使用 x = A cos(ωt) 时自动得到初始最大位移。考题经常会要求根据给定的初始 x 和 v 确定 φ。

6. Graphical Representation | 图形表示

When you plot displacement, velocity, and acceleration against time on the same axes, you obtain three sinusoidal waveforms with distinct phase relationships. With x = A sin ωt, the velocity v = Aω cos ωt = Aω sin(ωt + π/2) leads the displacement by π/2. The acceleration a = −Aω² sin ωt = ω²A sin(ωt + π) is completely out of phase with the displacement (phase difference of π). This means that when displacement is at a maximum, acceleration is at a maximum in the opposite direction.

如果在同一坐标轴上绘制位移、速度和加速度随时间变化的曲线，会得到三条具有明确相位关系的正弦波形。当 x = A sin ωt 时，速度 v = Aω cos ωt = Aω sin(ωt + π/2) 超前位移 π/2。加速度 a = −Aω² sin ωt = ω²A sin(ωt + π) 与位移完全反相（相差 π）。这意味着位移最大时，加速度在反方向达到最大。

Recognising these phase shifts is extremely useful for interpreting multiple‑choice questions where graphs of x, v and a are given, and you must identify which is which.

识别这些相位差对解读多选题中的 x、v 和 a 图形非常有帮助，在这类题目中你需要判断哪条曲线代表哪个物理量。

7. The Link to Uniform Circular Motion | 与匀速圆周运动的联系

Simple harmonic motion can be visualised as the projection of uniform circular motion onto a diameter. Imagine a particle moving counter‑clockwise around a circle of radius A with constant angular speed ω. The projection of the radius vector onto a horizontal or vertical axis performs SHM with amplitude A and angular frequency ω. This geometric interpretation makes it easy to understand why the velocity is Aω (the tangential speed of the reference circle) and why the maximum acceleration is ω²A (the centripetal acceleration of the circular motion).

简谐运动可视为匀速圆周运动在直径上的投影。设想一个质点以恒定角速度 ω 在半径为 A 的圆周上逆时针运动。半径矢量在水平或竖直轴上的投影即做振幅为 A、角频率为 ω 的简谐运动。这一几何解释有助于理解为何最大速率为 Aω（对应参考圆的线速度）以及为何最大加速度为 ω²A（对应圆周运动的向心加速度）。

Many students find that linking SHM to the reference circle clarifies the phase constants and makes it easier to derive the velocity‑displacement relation v² = ω²(A² − x²).

许多学生发现将简谐运动与参考圆联系起来能理清初相的概念，并更容易推导出速度‑位移关系 v² = ω²(A² − x²)。

8. The Velocity–Displacement Relation | 速度与位移的关系

A very important formula that connects velocity v and displacement x without involving time explicitly is v² = ω²(A² − x²). This can be derived from the energy of the system or by eliminating t from the parametric equations. It tells you that the speed is maximum when x = 0 (v_max = Aω) and zero when x = ±A. This relation is extremely handy for finding the speed at a given position.

一个非常重要的公式将速度 v 和位移 x 联系起来而不显含时间：v² = ω²(A² − x²)。这可通过系统能量消去 t 或由参数方程推导得到。该式表明当 x = 0 时速率最大（v_max = Aω），当 x = ±A 时速率为零。这一关系在求特定位置处的速率时极为实用。

v² = ω²(A² − x²)

9. Energy Considerations (Mathematical Perspective) | 能量观点（数学视角）

Although energy is often discussed in physics, the mathematical structure of SHM directly gives constant total mechanical energy in the absence of damping. The sum of kinetic energy (½mv²) and potential energy (½mω²x² for a spring‑type system) results in a constant: ½mω²A². From a mathematical standpoint, the equation v² = ω²(A² − x²) multiplied by ½m yields ½mv² + ½mω²x² = ½mω²A², which is an identity. This can appear in problems that require you to find the speed at a particular displacement without using calculus.

虽然能量通常在物理中讨论，但简谐运动的数学结构直接给出无阻尼情况下恒定的总机械能。动能（½mv²）与势能（对于弹簧系统为 ½mω²x²）之和等于常数 ½mω²A²。从数学角度看，将 v² = ω²(A² − x²) 两边乘以 ½m 即得到 ½mv² + ½mω²x² = ½mω²A²，这是一个恒等式。这在不需要微积分而求特定位移处速度的问题中可能出现。

10. Common Exam Question Types | 常见题型

IGCSE and Additional Mathematics exams frequently test SHM through the following question styles:

Given an equation such as x = 4 cos(3t), identify the amplitude, angular frequency, period, and frequency.
Determine velocity or acceleration at a specific time or displacement.
Use v² = ω²(A² − x²) to find speed at a point.
Find the phase constant from initial conditions.
Interpret displacement‑time, velocity‑time, or acceleration‑time graphs.
Verify that a given restoring force or acceleration satisfies a ∝ −x.

IGCSE 及附加数学考试常通过以下题型考查简谐运动：

给定方程如 x = 4 cos(3t)，求出振幅、角频率、周期和频率。
求特定时刻或位移处的速度或加速度。
利用 v² = ω²(A² − x²) 求任一点的速率。
根据初始条件求初相。
解读位移‑时间、速度‑时间或加速度‑时间图线。
验证给定的恢复力或加速度满足 a ∝ −x。

11. Worked Example | 例题解析

A particle moves with SHM according to the equation x = 5 sin(2πt + π/6) cm, where t is in seconds. Determine:
(a) the amplitude, angular frequency, and period;
(b) the maximum speed and the speed at t = 0;
(c) the acceleration when the displacement is 3 cm.

一质点按方程 x = 5 sin(2πt + π/6) cm 做简谐运动（t 单位为秒）。求：
(a) 振幅、角频率和周期；
(b) 最大速率及 t = 0 时的速率；
(c) 位移为 3 cm 时的加速度。

Solution / 解答

(a) The equation is in the form x = A sin(ωt + φ). Comparing, A = 5 cm, ω = 2π rad/s. Period T = 2π/ω = 2π/(2π) = 1 s.

(a) 方程形如 x = A sin(ωt + φ)，对比可得 A = 5 cm，ω = 2π rad/s。周期 T = 2π/ω = 2π/(2π) = 1 s。

(b) Maximum speed v_max = Aω = 5 × 2π = 10π cm/s ≈ 31.4 cm/s. At t = 0, v = dx/dt = Aω cos(ωt + φ) = 10π cos(π/6) = 10π × √3/2 = 5π√3 cm/s ≈ 27.2 cm/s.

(b) 最大速率 v_max = Aω = 5 × 2π = 10π cm/s ≈ 31.4 cm/s。t = 0 时，v = dx/dt = Aω cos(ωt + φ) = 10π cos(π/6) = 10π × √3/2 = 5π√3 cm/s ≈ 27.2 cm/s。

(c) Using a = –ω²x, we have a = –(2π)² × 3 = –4π² × 3 = –12π² cm/s² ≈ –118.4 cm/s² (the negative sign indicates direction towards equilibrium).

12. Tips for Success and Common Pitfalls | 应考技巧与常见陷阱

To score full marks on SHM questions, keep the following points in mind:

Always check whether the motion starts from equilibrium or an extreme point – this determines whether you should use sine or cosine without an added phase, or include the appropriate initial phase.
Remember that angular frequency ω must be in rad/s, and phase angles must be in radians when performing differentiation or evaluation.
Do not confuse maximum speed Aω with maximum acceleration ω²A. The former occurs at x = 0, the latter at x = ±A.
When a problem asks for the speed rather than velocity, give the magnitude only; the direction may be omitted or specified by context.
Practice converting between forms: x = A sin(ωt + φ) can always be written as a combination of sine and cosine terms, and vice versa.

要在简谐运动题目中拿到满分，请牢记以下几点：

一定检查运动是从平衡位置还是端点开始——这决定了你应使用不带额外初相的正弦或余弦，还是需要加入合适的初相。
请记住，角频率 ω 必须以 rad/s 为单位，进行求导或代入计算时相位角必须使用弧度。
切勿混淆最大速率 Aω 和最大加速度 ω²A。前者出现在 x = 0 处，后者出现在 x = ±A 处。
若题目要求求的是速率（speed）而非速度（velocity），只给出大小即可；方向可省略或根据语境说明。
多练习形式转换：x = A sin(ωt + φ) 总能写为正弦项和余弦项的组合，反之亦然。

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IGCSE Maths: Simple Harmonic Motion Key Points Review | IGCSE 数学：简谐运动 考点精讲