Simple Harmonic Motion: CCEA A-Level Maths Key Points | 简谐运动考点精讲

📚 Simple Harmonic Motion: CCEA A-Level Maths Key Points | 简谐运动考点精讲

Simple Harmonic Motion (SHM) is a fundamental topic in CCEA A-Level Mathematics, appearing in mechanics and differential equations. It models oscillatory systems where the restoring force is proportional to displacement. Mastering SHM requires understanding its defining equation, solutions, energy transformations, and applications like springs and pendulums. This revision guide breaks down the key concepts tested in the CCEA specification, with clear explanations and practical tips.

简谐运动是CCEA A-Level数学中力学和微分方程部分的核心内容。它描述了回复力与位移成正比的振动系统。要掌握SHM，必须理解其定义方程、通解、能量转化以及在弹簧振子和单摆等模型中的应用。本文拆解CCEA考试涉及的考点，提供清晰要旨和解题技巧。

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

An object undergoes simple harmonic motion when its acceleration a is directly proportional to its displacement x from a fixed equilibrium point and is always directed towards that point. Mathematically, a ∝ −x. The negative sign indicates that acceleration and displacement are in opposite directions. This type of motion is periodic and results from a linear restoring force, such as the force exerted by an ideal spring.

当物体的加速度a与其离开固定平衡位置的位移x成正比，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 ∝ −x，负号说明加速度与位移方向相反。这种运动是周期性的，源于线性的回复力，例如理想弹簧产生的力。

2. Defining Equation: a = –ω²x | 定义方程：a = –ω²x

The constant of proportionality is written as ω², where ω is the angular frequency (rad s⁻¹). This gives the central defining equation of SHM:

a = –ω²x

The quantity ω is not just a constant; it connects to the physical period and frequency of the oscillation. Every SHM system, regardless of its physical origin, follows this relation. In the CCEA exam, you may be asked to identify whether a given motion satisfies this condition.

比例常数记为ω²，其中ω是角频率（单位rad s⁻¹）。由此得到简谐运动的核心定义方程：

a = –ω²x

ω不仅是一个常数，它还与振动的周期和频率有关。任何简谐运动系统，不论物理来源如何，都遵循这个关系。在CCEA考试中，考生可能需要判断给定的运动是否满足此条件。

3. 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 = d²x/dt², we obtain a second-order linear differential equation:

d²x/dt² = –ω²x

This equation appears in the CCEA Mechanics and Differential Equations components. Any function whose second derivative is a negative multiple of itself satisfies SHM. Solving this differential equation yields the general description of the motion.

因为加速度是位移对时间的二阶导数a = d²x/dt²，我们得到一个二阶线性微分方程：

d²x/dt² = –ω²x

该方程出现在CCEA力学和微分方程模块中。任何二阶导数与自身成负倍数关系的函数都满足简谐运动。求解此微分方程即可得到运动的通解。

4. General Solutions of the Differential Equation | 微分方程的通解

The general solution can be expressed in two common forms. The first uses sine and cosine terms: x = A sin(ωt) + B cos(ωt), where A and B are constants determined by initial conditions. An equivalent, often more convenient form is the amplitude-phase representation:

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

R is the amplitude (maximum displacement), ω is the angular frequency, and φ is the phase constant (initial phase). The choice of sine or cosine depends on the starting position: if the particle starts at equilibrium with a positive velocity, the sine form with φ = 0 is a natural fit. CCEA questions often require you to find R and φ from given initial displacement and velocity.

通解通常有两种表达形式。第一种使用正弦和余弦项：x = A sin(ωt) + B cos(ωt)，其中A和B由初始条件确定。另一种等效且更常用的形式是振幅–相位表示法：

x = R sin(ωt + φ) 或 x = R cos(ωt + φ)

R为振幅（最大位移），ω为角频率，φ为初相。选择正弦还是余弦取决于初始位置：若质点从平衡位置向正方向开始运动，采用φ = 0的正弦形式很自然。CCEA考题常要求根据给定的初始位移和速度求出R和φ。

5. Period and Frequency | 周期和频率

From the angular frequency ω, the period T and frequency f of the oscillation are derived:

T = 2π/ω, f = 1/T = ω/(2π), ω = 2πf

A defining feature of SHM is that the period is independent of the amplitude—a property known as isochronism. This means that for a given system (e.g. a specific spring-mass setup or pendulum length), the oscillation period remains constant even if the amplitude changes. In CCEA problems, you will use these relations to find ω when the period is known, or vice versa.

由角频率ω可得振动的周期T和频率f：

T = 2π/ω, f = 1/T = ω/(2π), ω = 2πf

简谐运动的一个标志性特征是周期与振幅无关，这个性质称为等时性。这意味着对于特定系统（例如某个弹簧振子或特定摆长），即使振幅改变，周期仍保持不变。在CCEA考题中，你需要运用这些关系在已知周期时求ω，或反之。

6. Velocity in SHM | 简谐运动中的速度

Differentiating the displacement x = R sin(ωt + φ) with respect to time gives the velocity:

v = dx/dt = Rω cos(ωt + φ)

The maximum speed is v_max = Rω and occurs as the particle passes through the equilibrium position (x = 0). Using the identity sin²θ + cos²θ = 1, we can eliminate time and express velocity in terms of displacement:

v = &pm; ω √(R² − x²)

This form is extremely useful when time is not given. The ± sign reflects the two possible directions of motion at a given displacement. In CCEA exams, you will often use this expression to compute speed at a specific point or to deduce the displacement when the speed is known.

将位移 x = R sin(ωt + φ) 对时间求导得到速度：

v = dx/dt = Rω cos(ωt + φ)

最大速率 v_max = Rω，出现在质点经过平衡位置（x = 0）时。利用三角恒等式 sin²θ + cos²θ = 1，可以消去时间，用位移表示速度：

v = &pm; ω √(R² − x²)

在未给出时间的情况下，该形式特别实用。正负号反映了在指定位移处运动的两个可能方向。在CCEA试题中，常用此式计算某位置的速度，或由已知速度反求位移。

7. Acceleration in SHM | 简谐运动中的加速度

Further differentiation of velocity gives the acceleration:

a = dv/dt = −Rω² sin(ωt + φ) = −ω²x

The acceleration is always directed towards the equilibrium point (hence the negative sign) and its magnitude is maximum at the extreme positions x = ±R, where a_max = ω²R. At the equilibrium position (x = 0), the acceleration is zero. Graphs of acceleration against displacement yield a straight line through the origin with a negative gradient of −ω², a key test for SHM.

继续对速度求导得到加速度：

a = dv/dt = −Rω² sin(ωt + φ) = −ω²x

加速度总是指向平衡位置（因而带负号），其大小在端点 x = ±R 处达到最大，a_max = ω²R。在平衡位置（x = 0）加速度为零。加速度对位移的图像是一条穿过原点、斜率为 −ω² 的直线，这是检验简谐运动的重要手段。

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

For an ideal spring-mass system, the kinetic energy (K) and potential energy (U) are:

K = ½ m v², ; U = ½ k x² = ½ m ω² x²

since the spring constant k = mω². The total mechanical energy E is conserved and can be expressed in terms of amplitude:

E = K + U = ½ m ω² R² = constant

Energy continuously transforms between kinetic and potential forms: at x = 0, all energy is kinetic; at x = ±R, all energy is potential. Many CCEA problems ask you to use energy conservation to find velocity at a given displacement without solving differential equations.

对于理想弹簧振子系统，动能(K)和势能(U)分别为：

K = ½ m v², ; U = ½ k x² = ½ m ω² x²

因为劲度系数 k = mω²。总机械能E守恒，可用振幅表示为：

E = K + U = ½ m ω² R² = 常量

能量在动能和势能之间不断转化：在 x = 0 处，全部能量为动能；在 x = ±R 处，全部能量为势能。许多CCEA问题要求考生利用能量守恒求某位移处的速度，而无需直接求解微分方程。

9. Horizontal Spring-Mass System | 水平弹簧振子系统

Consider a mass m attached to a light spring with spring constant k, resting on a smooth horizontal surface. The restoring force is F = −kx, giving m d²x/dt² = −kx. Comparing with the standard SHM equation yields:

ω = √(k/m), ; T = 2π √(m/k)

The equilibrium position is the natural length of the spring. This system is a standard SH

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Simple Harmonic Motion: CCEA A-Level Maths Key Points | 简谐运动考点精讲