Simple Harmonic Motion in IGCSE AQA Maths | IGCSE AQA 数学：简谐运动 考点精讲

📚 Simple Harmonic Motion in IGCSE AQA Maths | IGCSE AQA 数学：简谐运动 考点精讲

lied contexts. In the IGCSE AQA Mathematics specification, SHM is not just a physics topic — understanding how displacement, velocity, and acceleration relate to one another through the key equation a = – ial concepts, formulas, graphical interpretations, and typical exam questions to help you achieve top marks.

简谐运动（SHM）是一种基础的振荡运动，在纯数学和应用情境中都经常出现。在 IGCSE AQA 数学大纲中，SHM 不仅仅是一个物理主题——它结合了微分方程、三角函数和微积分来为现实世界的周期行为建模。掌握这个主题意味着要理解位移、速度和加速度如何通过关键方程 a = –ω²x 相互关联，并能够解决涉及振幅、周期、频率和相位的问题。本文全面拆解了基本概念、公式、图形解读以及典型考题，助力你获得高分。

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 hence the acceleration) is directly proportional to the displacement from a fixed equilibrium position and is always directed towards that equilibrium. In mathematical terms, this condition is expressed by a second-order differential equation. Many systems, such as a mass on a spring or a simple pendulum swinging with small amplitude, can be approximated as SHM. The motion is sinusoidal in time and is characterised by constant amplitude and constant period.

简谐运动是一种周期运动，其中的回复力（以及加速度）与离开固定平衡位置的位移成正比，且总是指向该平衡位置。从数学角度，这一条件由一个二阶微分方程表达。许多系统，如弹簧上的质量块或小幅度摆动的单摆，都可以近似为简谐运动。该运动在时间上呈正弦曲线，并以恒定振幅和恒定周期为特征。

2. The Defining Differential Equation | SHM 的定义微分方程

The central equation for SHM is d²x/dt² = – acceleration always opposes the displacement. This equation is derived from Newton’s second law and Hooke’s law for a spring-mass system, but in mathematics it is treated as the starting point for solving SHM problems. Students must recognise that any motion satisfying this differential equation is SHM.

简谐运动的核心方程是 d²x/dt² = –ω²x，其中 x 是离开平衡位置的位移，t 是时间，ω 是角频率（一个正常数）。负号表示加速度总是与位移反向。该方程源自弹簧-质量系统的牛顿第二定律和胡克定律，但在数学中，它被当作求解 SHM 问题的起点。学生必须认识到，任何满足该微分方程的运动都是简谐运动。

d²x/dt² = –ω²x

3. General Solution for Displacement | 位移的通解

ent forms: x = A cos(ωt) + B sin(ωt), or more commonly x = A cos(ωt + φ) or x = A sin(ωt + φ), where A is the amplitude (maximum displacement) and φ is the phase constant determined by initial conditions. If the motion starts from maximum displacement, we usually choose x = A cos(ωt); if it starts from equilibrium with positive velocity, we use x = A sin(ωt). Being comfortable with both forms is essential for handling different initial conditions in exam questions.

求解该微分方程可得到一个包含正弦或余弦的位移函数。通解可以写成两种等价形式：x = A cos(ωt) + B sin(ωt)，或者更常见的 x = A cos(ωt + φ) 或 x = A sin(ωt + φ)，其中 A 是振幅（最大位移），φ 是由初始条件决定的相位常数。如果运动从最大位移处开始，我们通常选择 x = A cos(ωt)；如果从平衡位置以正向速度开始，则用 x = A sin(ωt)。熟练掌握这两种形式对于应对考题中不同的初始条件至关重要。

x = A cos(ωt + φ) or x = A sin(ωt + φ)

4. Velocity and Acceleration Functions | 速度和加速度函数

und by successive differentiation with respect to time. If x = A cos(ωt), then v = dx/dt = –Aω sin(ωt), and a = dv/dt = –Aω² cos(ωt) = –ω²x. The maximum speed occurs when passing through equilibrium (x = 0) and is v_max = Aω. The maximum acceleration occurs at the extreme displacements (x = ±A) and is a_max onships allow you to switch between displacement, velocity, and acceleration forms easily.

一旦已知位移函数，速度 v 和加速度 a 便可通过对时间逐次求导得到。若 x = A cos(ωt)，则 v = dx/dt = –Aω sin(ωt)，而 a = dv/dt = –Aω² cos(ωt) = –ω²x。最大速度出现在经过平衡位置（x = 0）时，为 v_max = Aω。最大加速度出现在极端位移处（x = ±A），为 a_max = Aω²。这些关系让你能轻松地在位移、速度和加速度形式之间切换。

v = –Aω sin(ωt), a = –Aω² cos(ωt) = –ω²x

v_max = Aω, a_max = Aω²

5. The Velocity-Displacement Relation | 速度与位移的关系

A very useful equation that eliminates time t is v² = ω²(A² – depends only on the displacement x and the constants A and ω. When x = 0, v² = ω²A², giving maximum speed. When x = ±A, v = 0. This formula is particularly handy for finding the speed at a given displacement without needing to compute the time first. Make sure you can derive it and apply it confidently.

一个十分有用的消去时间 t 的方程是 v² = ω²(A² – x²)。它可利用链式法则和积分从微分方程推导，或从能量角度得出。它表明速率仅取决于位移 x 以及常量 A 和 ω。当 x = 0 时，v² = ω²A²，得到最大速率；当 x = ±A 时，v = 0。该公式特别适用于求给定位移下的速率，而无需先计算时间。务必能自行推导并自信地应用它。

v² = ω²(A² – x²)

6. Period, Frequency, and Angular Frequency | 周期、频率和角频率

. The frequency f is the number of oscillations per unit time: f = 1/T = ω/(2π). The angular frequency ω is measured in rad/s and links directly to the differential equation constant. In many exam problems, you are given ω or you must find it from a given period. Remember that ω is not the same as frequency f; they differ by a factor of 2π.

周期 T 是完成一次完整振动所需的时间。由于正弦和余弦函数每 2π 弧度重复一次，于是有 ωT = 2π，因此 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 rad/s，并直接与微分方程中的常数关联。在许多考题中，你会直接得到 ω，或者需要从给定的周期求出 ω。请记住 ω 与频率 f 不同，它们相差 2π 倍。

T = 2π/ω, f = 1/T = ω/(2π)

7. Amplitude and Phase Angle | 振幅与相位角

aph horizontally. For example, x = A sin(ωt + φ) has a phase difference of φ compared to x = A sin(ωt). If φ is positive, the graph is shifted to the left. The value of φ is often determined by initial conditions such as initial displacement x₀ and initial velocity v₀. In exams, you may be asked to find A and φ given x₀ and v₀, using relationships like x₀ = A sin φ (or A cos φ) and v₀ = Aω cos φ (or –Aω sin φ).

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振幅 A 是离开平衡位置的最大距离，它总是一个正数。相位角 φ（或相位常数）使正弦图像发生水平平移。例如，x = A sin(ωt + φ) 相对于 x = A sin(ωt) 有 φ 的相位差。若 φ 为正，图像向左平移。φ 的值通常由初始条件决定，如初始位移 x₀ 和初始速度 v₀。在考试中，你可能会被要求根据 x₀ 和 v₀ 求出 A 和 φ，所用关系如 x₀ = A sin φ（或 A cos φ）以及 v₀ = Aω cos φ（或 –Aω sin φ）。

8. Graphical Interpretation of SHM | 简谐运动的图形解读

and a is a negative cosine function (phase difference of π from x). The velocity graph leads the displacement graph by π/2, and the acceleration graph leads the velocity graph by another π/2, or is in anti-phase with displacement. Understanding these phase relationships helps with sketching and interpreting graphs quickly. The amplitude on the v-t graph is Aω, and on the a-t graph is Aω². Displacement from a-t graph can be obtained by noting that a = –ω²x, so x = –a/ω².

位移 x、速度 v 和加速度 a 随时间 t 变化的图像都是正弦曲线。通常，若 x 是余弦函数，则 v 是负的正弦函数（相位差 π/2），而 a 是负的余弦函数（与 x 的相位差 π）。速度图像比位移图像超前 π/2，加速度图像又比速度图像超前 π/2，或者说与位移反相。理解这些相位关系有助于快速作图与读图。v-t 图像上的振幅是 Aω，a-t 图像上的振幅是 Aω²。通过 a-t 图像求位移时，可利用 a = –ω²x，从而 x = –a/ω²。

9. Working with Initial Conditions | 初值条件的处理

Exam questions often ask: 'Find the displacement at time t given initial displacement and velocity.' 0)/ω. This method avoids solving for φ directly and is very straightforward. Alternatively, if you prefer the amplitude-phase form, you can compute A = √(x₀² + (v₀/ω)²) and φ such that cos φ = x₀/A, sin φ = – n the chosen form). Both methods are acceptable; choose the one that feels more efficient.

考题经常问：“已知初始位移和初始速度，求某一时刻的位移”。处理方法为使用通解 x = A cos(ωt) + B sin(ωt)。那么 x(0) = A，而 v(0) = t=0 时的 dx/dt = Bω。因此 A 就是初始位移，而 B = v(0)/ω。这种方法避免了直接求解 φ，非常直接。另外，如果你偏爱振幅-相位形式，可以计算 A = √(x₀² + (v₀/ω)²) 以及 φ 满足 cos φ = x₀/A，sin φ = –v₀/(Aω)（取决于所选形式）。两种方法都可接受，选择你觉得更省时的一种。

10. Common Exam Pitfalls and Tips | 常见考试陷阱与建议

f in Hz. Another pitfall is forgetting the negative sign in the acceleration equation a = –ω²x; losing the sign will cost marks in explanation questions. When using v² = ω²(A² – ts when determining the phase angle from sine and cosine values. Finally, always verify that your calculator is in radian mode when evaluating trigonometric functions, as SHM is always modelled in radians unless stated otherwise.

一个常见错误是混淆角频率 ω 和频率 f。始终检查单位：ω 是 rad/s，f 是 Hz。另一个陷阱是忘记加速度方程 a = –ω²x 中的负号；丢掉负号会在解释题中失分。使用 v² = ω²(A² – x²) 时，记住 v 是速率大小，因此若题目要求速度矢量，还需单独考虑方向。同时，根据正弦和余弦值确定相位角时要注意象限。最后，在计算三角函数值之前务必确认计算器处于弧度模式，因为除非特别说明，SHM 始终以弧度建模。

11. Worked Example | 示例精讲

Example: A particle moves with SHM of period 4 seconds. It passes through the equilibrium point with speed 2 m/s. Find (a) the amplitude, (b) the maximum acceleration, and (c) the speed when the particle is 0.5 m from equilibrium.
Solution: T = 4 s ⇒ ω = 2π/T = π/2 rad/s. At equilibrium, speed is maximum, so v_max = Aω = 2 m/s ⇒ A = 2 / (π/2) = 4/π m. Maximum acceleration a_max = Aω² = (4/π) × (π/2)² = (4/π) × (π²/4) = π m/s². For part (c), use v² = ω²(A² – x²) = (π/2)² × ((4/π)² – 0.5²) = (π²/4) × (16/π² – 0.25) = (π²/4) × (5.092... – 0.25) — but exact calculation: 16/π² – 1/4. Multiply: (π²/4)×(16/π²) = 4, and (π²/4)×(–1/4) = –π²/16. So v² = 4 – π²/16. Then v = √4 – π²/16 ≈ √4 – 0.61685 ≈ √3.38315 ≈ 1.84 m/s. Always leave exact answers where possible.

示例：一质点以周期 4 秒做简谐运动。它以 2 m/s 的速度通过平衡点。求 (a) 振幅，(b) 最大加速度，以及 (c) 质点离平衡位置 0.5 m 时的速率。
解：T = 4 s ⇒ ω = 2π/T = π/2 rad/s。在平衡位置，速率最大，故 v_max = Aω = 2 m/s ⇒ A = 2 / (π/2) = 4/π m。最大加速度 a_max = Aω² = (4/π) × (π/2)² = (4/π) × (π²/4) = π m/s²。对于 (c)，使用 v² = ω²(A² – x²) = (π/2)² × ((4/π)² – 0.5²) = (π²/4) × (16/π² – 0.25) = (π²/4) × (16/π²) – (π²/4) × 0.25 = 4 – π²/16。于是 v = √4 – π²/16 ≈ √4 – 0.61685 ≈ √3.38315 ≈ 1.84 m/s。在可能的情况下尽量保留精确值。

12. Summary of Key Formulas | 重要公式总结

Concept	Formula	Notes
Defining equation	d²x/dt² = –ω²x	Acceleration proportional to displacement, directed to equilibrium.
Displacement	x = A cos(ωt + φ) or x = A sin(ωt + φ)	A amplitude, φ phase constant.
Velocity	v = –Aω sin(ωt) (if x = A cos ωt)	Max speed Aω at equilibrium.
Acceleration	a = –Aω² cos(ωt) = –ω²x	Max acceleration Aω² at extremes.
Speed-displacement	v² = ω²(A² – x²)	v = 0 at x = ±A.
Period	T = 2π/ω	ω in rad/s.
Frequency	f = 1/T = ω/(2π)	Unit: Hz.
Maximum speed	vmax = Aω	At x = 0.
Maximum acceleration	amax = Aω²	At x = ±A.

Memorise these core relationships and practise deriving them fluently. Mastery of SHM opens the door to tackling more advanced mechanics and differential equation topics with confidence.

记住这些核心关系，并熟练地推导它们。掌握简谐运动将为你自信地应对更高阶的力学和微分方程主题打开大门。

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