📚 Simple Harmonic Motion for CIE IGCSE Additional Mathematics | 简谐运动考点精讲
Simple harmonic motion (SHM) is a fundamental topic in CIE IGCSE Additional Mathematics (0606). It describes a special type of periodic motion where the acceleration of a particle is directly proportional to its displacement from a fixed point and always directed towards that point. This article provides a thorough overview of the key concepts, formulas, graphical interpretations, and common problem types that appear in the examination. Mastering SHM will not only strengthen your understanding of applied calculus and trigonometry but also lay a solid foundation for further study in physics and advanced mathematics.
简谐运动 (SHM) 是 CIE IGCSE 附加数学 (0606) 中的一个重要课题。它描述了一类特殊的周期性运动：质点的加速度与其偏离固定点的位移成正比，并且始终指向该固定点。本文全面梳理了考试中常见的核心概念、公式、图像解读以及典型题型。熟练掌握简谐运动不仅能加深你对微积分和三角学应用的理解，还为将来学习物理与高等数学打下扎实基础。
1. Definition of SHM | 简谐运动的定义
In SHM, a particle oscillates about a central equilibrium position. The defining characteristic is that the acceleration a is proportional to the displacement x from the equilibrium and acts in the opposite direction. Mathematically, this is written as:
在简谐运动中，质点围绕中心平衡位置振动。其定义性特征是加速度 a 与相对于平衡位置的位移 x 成正比，且方向相反。数学上可表示为：
a = –ω²x
Here ω (omega) is a positive constant called the angular frequency. The negative sign indicates that acceleration is always directed towards the equilibrium point.
这里 ω (欧米伽) 是一个正常数，称为角频率。负号表示加速度始终指向平衡位置。
2. Key Equations and Terms | 关键方程与术语
When a particle moves in SHM, its displacement x from the equilibrium position at time t can be expressed using either sine or cosine functions. The two standard forms are:
当质点作简谐运动时，其在时刻 t 相对于平衡位置的位移 x 可用正弦或余弦函数表示。两种标准形式为：
x = A sin(ωt + φ) or x = A cos(ωt + φ)
Where:
• A is the amplitude (maximum displacement)
• ω is the angular frequency (related to period and frequency)
• φ (phi) is the phase constant or initial phase angle, determining the starting position of the particle at t = 0.
其中：
• A 为振幅（最大位移）
• ω 为角频率（与周期和频率相关）
• φ (phi) 为初相或相位常数，决定了 t = 0 时质点的起始位置。
The velocity v and acceleration a are obtained by differentiating displacement with respect to time.
速度 v 与加速度 a 可通过位移对时间求导得出。
3. Differential Equation of SHM | 简谐运动的微分方程
The acceleration is the second derivative of displacement, so the defining equation can be written as a second-order differential equation:
加速度是位移的二阶导数，因此定义方程可写成二阶微分方程：
d²x/dt² = –ω²x
This equation states that the second derivative of x is a negative constant multiple of x itself. In CIE IGCSE Additional Mathematics, you are expected to recognise this form and know that its solutions are sinusoidal functions.
该方程表明 x 的二阶导数等于其自身的负常数倍。在 CIE IGCSE 附加数学中，你应能识别这一形式，并知道它的解是正弦或余弦函数。
4. Solution to the Differential Equation | 微分方程的解
The general solution of d²x/dt² = –ω²x can be written as x = A sin ωt + B cos ωt, where A and B are arbitrary constants determined by initial conditions. Alternatively, it can be expressed in the amplitude-phase form x = R sin(ωt + α) or x = R cos(ωt + α).
方程 d²x/dt² = –ω²x 的通解可写为 x = A sin ωt + B cos ωt，其中 A、B 为由初始条件决定的任意常数。或者，也可以表示为振幅-相位形式 x = R sin(ωt + α) 或 x = R cos(ωt + α)。
For a particle released from rest at maximum displacement (x = A when t = 0), the appropriate form is x = A cos ωt. For a particle passing through equilibrium with maximum velocity at t = 0, we often use x = A sin ωt.
对于在最大位移处静止释放的质点（t = 0 时 x = A），常用形式为 x = A cos ωt。对于 t = 0 时经过平衡位置且速度最大的质点，我们常用 x = A sin ωt。
5. Displacement, Velocity and Acceleration | 位移、速度与加速度
Consider the standard form x = A sin ωt. Differentiating with respect to t gives:
考虑标准形式 x = A sin ωt。对 t 求导得：
v = dx/dt = Aω cos ωt
a = dv/dt = –Aω² sin ωt = –ω²x
Notice that acceleration is always opposite in sign to displacement, confirming the restoring nature of SHM.
注意加速度的符号总是与位移相反，这验证了简谐运动的恢复特性。
An important relationship connecting velocity and displacement without time is:
一个不包含时间 t、关联速度与位移的重要关系式为：
v² = ω²(A² – x²)
This equation is extremely useful in solving problems where the time variable is not directly involved.
该方程在求解不直接涉及时间变量的问题时极其有用。
6. Maximum Values | 最大值
From the velocity and acceleration expressions, the maximum speed occurs when the particle passes through the equilibrium position (x = 0):
由速度和加速度表达式可知，最大速率出现在质点通过平衡位置（x = 0）时：
v_max = ωA
The maximum acceleration occurs at the extreme positions (x = ±A):
最大加速度出现在极限位置（x = ±A）处：
a_max = ω²A
These formulas are essential for quickly determining the bounds of motion in exam questions.
这些公式对于在考试中快速确定运动范围至关重要。
7. Period and Frequency | 周期与频率
The period T is the time taken for one complete oscillation. It is related to the angular frequency by:
周期 T 为完成一次完整振动所需的时间。它与角频率的关系为：
T = 2π/ω
The frequency f (number of oscillations per unit time) is the reciprocal of the period:
频率 f（单位时间内的振动次数）是周期的倒数：
f = 1/T = ω/2π
In CIE questions, you often need to find ω from a given period, or use ω to write down the full equation of motion.
在 CIE 考题中，你常需根据给定周期求 ω，或利用 ω 写出完整的运动方程。
8. Energy in SHM | 简谐运动的能量
Although not always examined in depth at IGCSE Additional Mathematics level, energy considerations can provide quick insights. The total mechanical energy in an undamped SHM system is constant:
尽管 IGCSE 附加数学并不总深入考查能量，但能量分析可提供快速洞见。在无阻尼简谐运动系统中，总机械能守恒：
Total Energy = (1/2)mω²A²
At any displacement x, the kinetic energy is ½ m v² = ½ m ω² (A² – x²) and the potential energy is ½ m ω² x². These formulas are useful for contextual problems linking SHM with work and energy.
在任意位移 x 处，动能为 ½ m v² = ½ m ω² (A² – x²)，势能为 ½ m ω² x²。这些公式在将简谐运动与功和能量结合的应用题中很有用。
9. Graphical Representations | 图像表示
You must be comfortable interpreting displacement–time, velocity–time, and acceleration–time graphs for SHM. If x = A sin ωt, then:
你必须能够熟练解读简谐运动的位移-时间、速度-时间以及加速度-时间图像。若 x = A sin ωt，则：
- Displacement graph is a sine wave starting at zero.
- Velocity graph is a cosine wave, shifted by π/2 (quarter period) ahead of displacement.
- Acceleration graph is an inverted sine wave, exactly out of phase with displacement by π.
- 位移图像是一条起始于零的正弦波。
- 速度图像是一条余弦波，较位移超前 π/2（四分之一周期）。
- 加速度图像是一条倒置的正弦波，与位移恰好反相 π。
Questions often ask you to sketch these graphs on the same axes or to deduce one motion parameter from a given graph.
考题经常要求你在同一坐标系内画出这些图像，或根据给定的图像推断某一运动参量。
10. Summary of Formulas | 公式总结
| Quantity / 量 | Formula / 公式 |
|---|---|
| Defining equation | a = –ω²x |
| Differential form | d²x/dt² = –ω²x |
| Displacement | x = A sin ωt or A cos ωt |
| Velocity | v = ω√(A² – x²) |
| v_max | ωA |
| a_max | ω²A |
| Period T | T = 2π/ω |
Memorising these formulas and understanding how to derive them quickly will save time in the exam.
熟记这些公式并理解其快速推导方法，能在考试中为你节省时间。
11. Example Problem Walkthrough | 典型例题讲解
A particle moves in SHM with amplitude 0.2 m and period 4 s. Find the maximum speed and the speed when the displacement is 0.1 m.
一个质点作简谐运动，振幅为 0.2 m，周期为 4 s。求最大速率以及位移为 0.1 m 时的速率。
First, ω = 2π/T = 2π/4 = π/2 rad/s.
v_max = ωA = (π/2) × 0.2 = 0.1π ≈ 0.314 m/s.
Using v = ω√(A² – x²), at x = 0.1 m: v = (π/2)√(0.2² – 0.1²) = (π/2)√(0.04 – 0.01) = (π/2)√0.03 = (π/2) × 0.1732 ≈ 0.272 m/s.
首先，ω = 2π/T = 2π/4 = π/2 rad/s。
v_max = ωA = (π/2) × 0.2 = 0.1π ≈ 0.314 m/s。
利用 v = ω√(A² – x²)，当 x = 0.1 m 时：v = (π/2)√(0.04 – 0.01) = (π/2)√0.03 ≈ 0.272 m/s。
12. Common Mistakes and Exam Tips | 常见错误与应试技巧
Many students confuse the relationship between phase angle and initial conditions. Always check whether the particle starts at equilibrium or extreme position before writing the displacement equation. Also, remember that the angular frequency ω must be in radians per unit time; if a problem gives frequency in Hz or period in seconds, convert appropriately. Another pitfall is forgetting the negative sign in a = –ω²x when calculating acceleration from displacement.
许多学生容易混淆初相与初始条件的关系。务必在写出位移方程之前，先确认质点是从平衡位置还是极限位置开始运动。同时记住，角频率 ω 的单位是弧度每单位时间；若题目给出的频率单位是 Hz，或周期以秒计，请进行相应换算。另一个易错点是在由位移计算加速度时忘记 a = –ω²x 中的负号。
A smart exam tip: use the energy approach (v² = ω²(A² – x²)) whenever time is not asked for; it often simplifies the algebra significantly.
一个明智的应试技巧：当题目不要求求时间时，尽量使用能量关系式 v² = ω²(A² – x²)，这通常能显著简化代数运算。
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