Simple Harmonic Motion for IGCSE CIE Mathematics | IGCSE CIE 数学：简谐运动考点精讲

📚 Simple Harmonic Motion for IGCSE CIE Mathematics | IGCSE CIE 数学：简谐运动考点精讲

Simple Harmonic Motion (SHM) is a core topic in the IGCSE CIE Additional Mathematics syllabus (0606). It describes a special type of periodic motion where the restoring force is directly proportional to the displacement from equilibrium and always acts towards that equilibrium. Understanding SHM not only helps you solve exam problems involving oscillations and waves, but also deepens your grasp of calculus applied to physical systems. In this article, we will break down every essential concept, formula and graph you need to master for the exam.

简谐运动（SHM）是 IGCSE CIE 附加数学（0606）大纲中的核心课题。它描述的是一种特殊的周期性运动：回复力与物体偏离平衡位置的位移成正比，且始终指向平衡位置。掌握简谐运动不仅能帮助你解决涉及振动和波的考试问题，还能加深你对微积分在物理系统中应用的理解。本文将逐一拆解你必须掌握的每一个基本概念、公式和图像。

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

An object moves with simple harmonic motion if its acceleration is directly proportional to its displacement from a fixed point (the equilibrium position) and is always directed towards that point. Mathematically, the defining condition is a ∝ −x, or a = −ω²x, where ω is a positive constant called the angular frequency.

如果一个物体的加速度与其偏离某一固定点（平衡位置）的位移成正比，且方向始终指向该固定点，则该物体在做简谐运动。数学上，定义条件为 a ∝ −x，或写成 a = −ω²x，其中 ω 是一个正常数，称为角频率。

The negative sign indicates that acceleration and displacement are in opposite directions. When the object is to the right of the equilibrium, the acceleration is to the left, pulling it back. This proportionality leads to sinusoidal oscillations of displacement, velocity and acceleration with time.

负号表示加速度与位移方向相反。当物体在平衡位置右侧时，加速度指向左侧，将它拉回。这种正比关系使得位移、速度和加速度都随时间呈正弦变化。

In IGCSE Additional Mathematics, we usually study SHM along a straight line, such as a mass on a spring or a simple pendulum (for small angles). Although real pendulums are not perfect SHM, the mathematical model provides an excellent approximation and forms the basis of many exam questions.

在 IGCSE 附加数学中，我们通常研究沿直线进行的简谐运动，例如弹簧上的物体或小角度摆动的单摆。尽管实际单摆并非完美的简谐运动，但这个数学模型提供了极好的近似，也是许多考题的基础。

2. Displacement as a Function of Time | 位移随时间的变化

The displacement x of a particle moving with SHM can be expressed using either sine or cosine functions, depending on the starting position. The two standard forms are:

做简谐运动的质点的位移 x 可以用正弦或余弦函数表示，具体取决于初始位置。两种标准形式是：

x = A sin(ωt) or x = A cos(ωt)

where A is the amplitude (maximum displacement), ω is the angular frequency, and t is time. If the object starts from the equilibrium position (x = 0) when t = 0, we use x = A sin(ωt). If it starts from the maximum displacement (x = A) when t = 0, we use x = A cos(ωt).

其中 A 是振幅（最大位移），ω 是角频率，t 是时间。如果物体在 t = 0 时从平衡位置（x = 0）出发，则使用 x = A sin(ωt)。如果物体在 t = 0 时从最大位移（x = A）出发，则使用 x = A cos(ωt)。

Sometimes a more general form x = A sin(ωt + φ) or x = A cos(ωt + φ) is used, where φ is the phase angle. In IGCSE exams, you are most likely to encounter the forms without φ, or you may be given initial conditions to determine the correct equation.

有时也会使用更一般的形式 x = A sin(ωt + φ) 或 x = A cos(ωt + φ)，其中 φ 是初相位。在 IGCSE 考试中，你更常遇到无 φ 的形式，或者题目会给出初始条件以便你确定正确的方程。

3. Velocity in SHM | 简谐运动中的速度

Velocity is the rate of change of displacement with time. Starting from x = A sin(ωt), we differentiate with respect to t:

速度是位移对时间的变化率。由 x = A sin(ωt) 出发，对 t 求导可得：

v = dx/dt = Aω cos(ωt)

If we use x = A cos(ωt), then v = −Aω sin(ωt). The velocity oscillates between +Aω and −Aω. Notice that velocity is maximum when passing through the equilibrium position (x = 0) and zero at the extreme points (x = ±A).

如果使用 x = A cos(ωt)，则 v = −Aω sin(ωt)。速度在 +Aω 与 −Aω 之间变化。注意，物体通过平衡位置（x = 0）时速度最大，在端点（x = ±A）处速度为零。

The maximum speed is given by v_max = Aω. This is a key formula that often appears in calculation questions. Make sure you know whether the question asks for speed or velocity, as speed is the magnitude and always positive.

最大速率为 v_max = Aω。这是一个经常出现在计算题中的关键公式。务必分清题目问的是速度还是速率，速率是大小，总为正值。

4. Acceleration in SHM | 简谐运动中的加速度

Acceleration is the rate of change of velocity. Differentiating v = Aω cos(ωt) gives:

加速度是速度的变化率。对 v = Aω cos(ωt) 求导得：

a = dv/dt = −Aω² sin(ωt) = −ω²x

This confirms the defining equation a = −ω²x. The acceleration is always directed towards the equilibrium position and its magnitude is proportional to x. Maximum acceleration occurs at x = ±A, and is given by a_max = ω²A. At the equilibrium position, acceleration is zero.

这就印证了定义方程 a = −ω²x。加速度始终指向平衡位置，其大小与 x 成正比。最大加速度出现在 x = ±A 处，大小为 a_max = ω²A。在平衡位置处，加速度为零。

The negative sign is crucial: it tells you that when x is positive, a is negative (directed leftwards). In an exam graph, the acceleration-time graph is a reflection of the displacement-time graph in the time axis, scaled by ω².

负号至关重要：它表明当 x 为正时，a 为负（指向左侧）。在考试图像题中，加速度-时间图像相当于位移-时间图像关于时间轴的反射，再乘以 ω²。

5. Relationship between Velocity and Displacement | 速度与位移的关系

A very useful equation links velocity v and displacement x without involving time t directly. Starting from v = Aω cos(ωt) and x = A sin(ωt), and using sin²θ + cos²θ = 1, we obtain:

有一个非常有用的方程直接将速度 v 与位移 x 联系起来，而不显含时间 t。由 v = Aω cos(ωt) 和 x = A sin(ωt)，并利用 sin²θ + cos²θ = 1，可推导出：

v² = ω²(A² − x²)

This relationship allows you to find the speed at any given displacement, or to find the displacement when the speed is known. It also highlights that v = 0 when x = ±A, and v = ±Aω when x = 0. Taking square roots, v = ±ω√(A² − x²), where the sign depends on the direction of motion.

利用这一关系，你可以在已知位移时求速率，或在已知速率时求位移。它还突显了：x = ±A 时 v = 0，x = 0 时 v = ±Aω。开方得 v = ±ω√(A² − x²)，正负号取决于运动方向。

This equation is often tested in multi-step problems where time is not given. You can also use it to check consistency between maximum speed and given values of A and ω.

在多步骤且未给出时间的题目中，经常考查这个方程。你也可以用它来检验最大速率与给定的 A、ω 值是否一致。

6. Period and Frequency | 周期与频率

The period T is the time taken for one complete oscillation. Since the sine and cosine functions repeat every 2π radians, we have ωT = 2π, giving:

周期 T 是完成一次全振动所需的时间。由于正弦和余弦函数每 2π 弧度重复一次，故有 ωT = 2π，从而：

T = 2π/ω

Frequency f is the number of oscillations per unit time and is the reciprocal of the period: f = 1/T = ω/(2π). Angular frequency ω is measured in rad s⁻¹, while f is in Hz (s⁻¹).

频率 f 是单位时间内的振动次数，是周期的倒数：f = 1/T = ω/(2π)。角频率 ω 的单位是 rad s⁻¹，而 f 的单位是 Hz（s⁻¹）。

In many IGCSE questions, you will be given either T or ω and asked to find the other. Remember that ω = 2π/T, and you can substitute this into acceleration and velocity formulas. For example, a_max = (2π/T)² A.

在 IGCSE 的许多考题中，会给出 T 或 ω 中的一个，要求计算另一个。请记住 ω = 2π/T，代入加速度和速度公式即可。例如 a_max = (2π/T)² A。

7. Energy in SHM | 简谐运动中的能量

For a particle of mass m undergoing SHM, the total mechanical energy is conserved. The kinetic energy KE and potential energy PE vary with position, but their sum remains constant.

一个质量为 m 的质点在做简谐运动时，总机械能是守恒的。动能 KE 和势能 PE 随位置变化，但它们的总和保持不变。

The kinetic energy at displacement x is:

在位移 x 处的动能为：

KE = ½ m v² = ½ m ω² (A² − x²)

The potential energy (stored in the spring or equivalent) is:

势能（储存在弹簧或类似系统中）为：

PE = ½ m ω² x²

Therefore, the total energy E is:

因此，总能量 E 为：

E = KE + PE = ½ m ω² A²

This shows that total energy is proportional to the square of the amplitude. At x = 0, PE = 0 and KE = E. At x = ±A, KE = 0 and PE = E. Energy questions in IGCSE usually involve finding v from a given x using energy conservation, which is equivalent to using v² = ω²(A² − x²).

这表明总能量与振幅的平方成正比。在 x = 0 处，PE = 0，KE = E；在 x = ±A 处，KE = 0，PE = E。IGCSE 中的能量题通常涉及利用能量守恒，由已知 x 求 v，这等价于运用 v² = ω²(A² − x²)。

8. Graphical Representations | 图像表示

Understanding and sketching the graphs of x, v and a against t is essential. For x = A sin(ωt), the corresponding graphs are:

理解并能够绘制 x-t、v-t 和 a-t 图像是必不可少的。对于 x = A sin(ωt)，相应的图像如下：

Displacement x: sine wave starting at 0, reaching A and −A.
Velocity v: cosine wave (starting at Aω), oscillating between ±Aω.
Acceleration a: inverted sine wave (−Aω² sin(ωt)), oscillating between ±Aω².

位移 x：正弦波形，从 0 出发，到达 A 和 −A。
速度 v：余弦波形（从 Aω 出发），在 ±Aω 之间振荡。
加速度 a：反相的正弦波形（−Aω² sin(ωt)），在 ±Aω² 之间振荡。

Notice the phase differences: v leads x by π/2 (quarter cycle), and a leads v by π/2 (or is in antiphase with x, i.e. phase difference of π). Exam questions may ask you to identify which graph corresponds to which quantity, or to sketch graphs for given conditions.

请注意相位差：v 领先 x π/2（四分之一周期），a 领先 v π/2（或与 x 反相，即相位差为 π）。考题可能要求你识别图像对应的物理量，或根据给定条件绘制图像。

A typical exam graph also includes the v²-x graph: from v² = ω²(A² − x²), this is a straight line with negative slope. Plotting v² on the y-axis and x² on the x-axis yields a straight line of slope −ω², which can be used to determine ω.

考试中常见的图像还有 v²-x 图：由 v² = ω²(A² − x²) 可知，这是一条斜率为负的直线。若以 v² 为 y 轴、x² 为 x 轴作图，得到斜率为 −ω² 的直线，可用于求 ω。

9. Worked Example | 典型例题

A particle moves with simple harmonic motion of amplitude 0.50 m and period 4.0 s. Find: (a) the maximum speed, (b) the speed when the particle is 0.20 m from the equilibrium position, and (c) the maximum acceleration.

一个质点做简谐运动，振幅为 0.50 m，周期为 4.0 s。求：(a) 最大速率，(b) 质点距平衡位置 0.20 m 时的速率，(c) 最大加速度。

First, calculate ω: T = 2π/ω ⇒ ω = 2π / T = 2π/4 = π/2 rad s⁻¹.

首先求 ω：T = 2π/ω ⇒ ω = 2π / T = 2π/4 = π/2 rad s⁻¹。

(a) Maximum speed v_max = Aω = 0.50 × (π/2) = 0.25π ≈ 0.785 m s⁻¹.

(b) Using v² = ω²(A² − x²): v² = (π/2)² (0.50² − 0.20²) = (π²/4)(0.25 − 0.04) = (π²/4)(0.21) = 0.0525π². So v = √(0.0525π²) = π√0.0525 ≈ 3.142 × 0.229 = 0.719 m s⁻¹ (taking positive root).

Always check if the question expects exact values in terms of π or decimals rounded to an appropriate number of significant figures. IGCSE CIE usually accepts either exact expressions or 2-3 significant figures.

解题时注意题目要求给出含 π 的精确值，还是保留适当有效数字的小数值。IGCSE CIE 通常两种均可，保留 2–3 位有效数字。

10. Summary of Key Formulas | 关键公式总结

Quantity	Formula
Defining equation	a = −ω²x
Displacement (equilibrium start)	x = A sin(ωt)
Displacement (extreme start)	x = A cos(ωt)
Velocity	v = Aω cos(ωt) [from x = A sin(ωt)]
Speed vs displacement	v = ω√(A² − x²)
Maximum speed	v_max = Aω
Maximum acceleration	a_max = ω²A
Period	T = 2π/ω
Frequency	f = 1/T = ω/(2π)
Total energy	E = ½ m ω² A²

Memorising these formulas is crucial, but equally important is understanding how they are derived and related. In the exam, check which variables you are given and select the appropriate relationship.

熟记这些公式至关重要，但同样重要的是理解它们的推导过程和相互联系。在考试中，要先看清题目给出的变量，再选择合适的公式。

Common pitfalls include forgetting the minus sign in a = −ω²x, mixing up maximum speed and maximum acceleration formulas, and misinterpreting the starting condition when choosing between sin and cos. Practise sketching graphs and labelling key values to avoid these errors.

常见错误包括：忘记 a = −ω²x 中的负号、混淆最大速率与最大加速度公式、以及在选择 sin 或 cos 时错误判断初始条件。多练习绘图并标出关键值，可以有效避免这些失误。

11. Interpreting SHM Questions | 解读简谐运动考题

IGCSE CIE exam questions on SHM often combine algebraic manipulation with graph interpretation. You might be asked to read amplitude and period from a displacement-time graph, then use them to find ω and write the equation of motion. Alternatively, a velocity-time graph may be given, from which you deduce amplitude using A = v_max/ω.

IGCSE CIE 的简谐运动考题常常将代数运算与图像解读结合在一起。你可能会被要求从位移-时间图像中读取振幅和周期，然后利用它们求出 ω 并写出运动方程。或者，题目可能给出速度-时间图像，让你通过 A = v_max/ω 反推振幅。

Word problems involving a spring or a pendulum often require you to convert physical quantities (e.g. spring constant k and mass m) into angular frequency using ω = √(k/m) for a mass-spring system, or ω = √(g/l) for a simple pendulum. While these specific formulas appear in physics, the CIE Additional Mathematics syllabus may give you such relationships in the stem; you are then expected to model the motion as SHM and apply the mathematical techniques above.

涉及弹簧或单摆的文字题，通常需要将物理量（如弹簧劲度系数 k 和质量 m）转化为角频率：对于弹簧振子用 ω = √(k/m)，对于单摆用 ω = √(g/l)。虽然这些具体公式出现在物理中，但 CIE 附加数学大纲可能会在题干中给出这些关系；接着你需要将运动建模为简谐运动，并运用上述数学方法求解。

Pay attention to units: always work in radians for ωt, keep lengths in metres and time in seconds unless otherwise specified. Double-check whether the answer requires the velocity (with sign) or speed (magnitude).

注意单位：ωt 始终用弧度，长度用米、时间用秒，除非题目另有说明。最后要核实答案是要求速度（带正负号）还是速率（大小）。

12. Final Tips for Success | 夺分技巧

Start by writing down the given data: A, T, ω, m, etc. Convert T to ω immediately if needed.
Identify which formula links the known quantities to the unknown.
If a graph is provided, mark the amplitude and period clearly on your sketch before answering.
When using v = ω√(A² − x²), take care with the square root and the ± sign; state the direction if velocity is required.
In questions involving energy, you can always check your result by seeing if KE + PE gives the same total as ½ m ω² A².

解题时先列出已知数据：A、T、ω、m 等。如有需要，立刻将 T 转化为 ω。
确定哪个公式能够将已知量与未知量联系起来。
如果题目提供图像，先在纸上标出振幅和周期，再作答。
使用 v = ω√(A² − x²) 时，注意开方和正负号；若要求速度，需指明方向。
涉及能量的问题，可以通过验证 KE + PE 是否等于 ½ m ω² A² 来检查结果。

With systematic practice, SHM questions can become some of the most predictable marks on your paper. Master the core equations, practise graph sketching, and always link back to the fundamental condition a = −ω²x.

通过系统练习，简谐运动题可以成为试卷上最可预测、最容易得分的部分。掌握核心方程，多练习绘图，并始终与基本条件 a = −ω²x 建立联系。

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Simple Harmonic Motion for IGCSE CIE Mathematics | IGCSE CIE 数学：简谐运动 考点精讲