IB Edexcel Mathematics: Simple Harmonic Motion Exam Focus | IB Edexcel 数学：简谐运动考点精讲

📚 IB Edexcel Mathematics: Simple Harmonic Motion Exam Focus | IB Edexcel 数学：简谐运动考点精讲

Simple Harmonic Motion (SHM) is a core topic that bridges pure mathematics and mechanics in both the IB and Edexcel A Level syllabuses. It revolves around a second-order differential equation and its trigonometric solutions, making it a favourite for questions that test modelling, calculus, and algebraic manipulation. This article breaks down every essential technique and common pitfall so you can approach SHM problems with confidence.

简谐运动（SHM）是 IB 和 Edexcel A Level 课程中连接纯数学与力学的核心主题。它围绕一个二阶微分方程及其三角函数解展开，因此常常成为考查建模、微积分和代数运算能力的理想题型。本文逐一梳理关键技巧和常见易错点，帮助你有把握地应对简谐运动问题。

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

Simple Harmonic Motion describes the oscillatory motion of a particle whose acceleration is directly proportional to its displacement from a fixed point and is always directed towards that point. Mathematically, this is expressed as a = −k x, where k is a positive constant. In the standard form used in examinations, we set k = ω², giving the defining equation a = −ω²x.

简谐运动描述的是一个质点的振荡运动，其加速度与它到某一固定点的位移成正比，且方向始终指向该固定点。数学上表示为 a = −k x，其中 k 是正数。考试中通常设 k = ω²，从而得到定义方程 a = −ω²x。

SHM arises whenever a restoring force obeys Hooke’s Law without damping. For a horizontal spring-mass system or a simple pendulum with small amplitude, the motion is approximately simple harmonic. However, in the mathematics papers, the physical context is often given, while the actual work centres on solving the differential equation.

只要回复力遵循胡克定律且没有阻尼，就会产生简谐运动。对于水平弹簧振子或小振幅的单摆，其运动近似为简谐运动。不过在数学试卷中，通常会提供物理背景，而实际解题核心在于求解微分方程。

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

The hallmark of SHM is the relationship between acceleration a and displacement x. Since acceleration is the second derivative of displacement with respect to time, the governing differential equation can be written as

d²x/dt² = −ω²x

简谐运动的标志是加速度 a 与位移 x 之间的关系。加速度是位移对时间的二阶导数，因此控制微分方程可写为

d²x/dt² = −ω²x

Here ω is the angular frequency, a constant that determines how rapidly the oscillation occurs. It is crucial to note that the negative sign ensures the acceleration is always opposite in sign to the displacement, pulling the particle back towards the equilibrium position at x = 0.

其中 ω 是角频率，一个决定振荡快慢的常数。务必注意，负号保证了加速度始终与位移符号相反，从而把质点拉回平衡位置 x = 0。

In exam questions, you may be given this equation in the form ẍ = −ω²x or asked to show that a particular system leads to it. Always write a = v dv/dx when working with velocity as a function of displacement.

在考试题目中，你可能会看到方程以 ẍ = −ω²x 的形式给出，或者被要求证明某个系统能导出此方程。当把速度表示为位移的函数时，请始终使用 a = v dv/dx。

3. The Second-Order Differential Equation | 二阶微分方程

The equation d²x/dt² + ω²x = 0 is a homogeneous second-order linear differential equation with constant coefficients. Its auxiliary equation is m² + ω² = 0, giving complex roots m = ± iω. Therefore, the general solution takes the form

x = A cos(ωt) + B sin(ωt)

方程 d²x/dt² + ω²x = 0 是一个常系数齐次二阶线性微分方程。其辅助方程为 m² + ω² = 0，得到一对共轭虚根 m = ± iω。因此通解可以写成

x = A cos(ωt) + B sin(ωt)

Here A and B are arbitrary constants determined by initial conditions. This solution shows that SHM is a superposition of sine and cosine waves with the same angular frequency ω.

这里的 A 和 B 是由初始条件确定的任意常数。这个解表明简谐运动是具有相同角频率 ω 的正弦波与余弦波的叠加。

This form is especially useful when initial displacement and initial velocity are given. For instance, if at t = 0, x = x₀ and v = v₀, then A = x₀ and B = v₀/ω.

当已知初始位移和初始速度时，这种形式尤其好用。例如，若 t = 0 时 x = x₀ 且 v = v₀，则有 A = x₀，B = v₀/ω。

4. Alternative Forms and Amplitude-Phase Expression | 振幅–相位形式与其他表达

The general solution can also be expressed as x = R sin(ωt + φ) or x = R cos(ωt − φ), where R is the amplitude (maximum displacement) and φ is the phase constant (or phase angle). These forms are linked by trigonometric identities:

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

通解也可表示为 x = R sin(ωt + φ) 或 x = R cos(ωt − φ)，其中 R 是振幅（最大位移），φ 是初相（或相位角）。这些形式可通过三角恒等式联系起来：

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

Thus, given x = A cos ωt + B sin ωt, we have R = √(A² + B²) and tan φ = A/B (or similar relations depending on the chosen form). Examiners often ask you to convert between these forms or to find the maximum speed and acceleration from them.

因此，对于 x = A cos ωt + B sin ωt，有 R = √(A² + B²) 以及 tan φ = A/B（或类似关系，取决于选用哪种形式）。考官常常要求你在这些形式之间转换，或据此求出最大速度和加速度。

Being fluent with both the standard sum-of-sines form and the R sin(ωt + φ) form is essential. When a question asks “find the first time the particle reaches maximum displacement”, the phase form quickly gives t = (π/2 − φ)/ω for sin form.

熟练掌握正弦余弦叠加形式与 R sin(ωt + φ) 形式至关重要。若题目问“质点何时第一次到达最大位移”，使用相量形式可以快速得到 sin 形式下的 t = (π/2 − φ)/ω。

5. Period, Frequency, and Angular Frequency | 周期、频率与角频率

From the argument ωt, we see that the motion repeats every time ωt increases by 2π. Hence the period T is

T = 2π/ω

从自变量 ωt 可以看出，每当 ωt 增加 2π 时，运动重复一次。因此周期 T 为

T = 2π/ω

The frequency f is the number of complete oscillations per unit time: f = 1/T = ω/(2π). The angular frequency ω (in rad s⁻¹) is related to the frequency by ω = 2πf. In physics-based contexts, these relationships are second nature, but in pure mathematics questions, you might be asked to derive T from the differential equation without explicit reference to the formula.

频率 f 是单位时间内完成的全振动次数：f = 1/T = ω/(2π)。角频率 ω（单位为 rad s⁻¹）满足 ω = 2πf。在物理情境下，这些关系是自然而然的，但在纯数学问题中，你可能会被要求从微分方程推导出 T，而不是直接套用公式。

Always check if the time variable in the solution is measured in seconds; if not, the period formula still holds as long as ω has units of (time)⁻¹. A common trick is to give a displacement equation like x = 4 cos(3t) and ask for the period: T = 2π/3 seconds.

务必检查解中的时间变量是否以秒为单位；如果不是，只要 ω 的量纲是 (时间)⁻¹，周期公式依然成立。一个常见的陷阱是给出位移方程如 x = 4 cos(3t) 并要求周期：T = 2π/3 秒。

6. Velocity in SHM: v = ±ω√(A² − x²) | 简谐运动中的速度：v = ±ω√(A² − x²)

A powerful relationship connects velocity v and displacement x, independent of time. Starting from a = v dv/dx = −ω²x, integrate both sides with respect to x:

∫ v dv = ∫ −ω²x dx ⇒ ½ v² = −½ ω² x² + C

一个强有力的关系式将速度 v 和位移 x 联系起来，且不含时间 t。从 a = v dv/dx = −ω²x 出发，两边对 x 积分：

∫ v dv = ∫ −ω²x dx ⇒ ½ v² = −½ ω² x² + C

If we define the amplitude A as the maximum displacement where v = 0, we obtain C = ½ ω² A². Thus,

v² = ω² (A² − x²) ⇒ v = ± ω √(A² − x²)

若将振幅 A 定义为 v = 0 时的最大位移，可得 C = ½ ω² A²。于是

v² = ω² (A² − x²) ⇒ v = ± ω √(A² − x²)

This formula tells us that the speed is greatest when x = 0 (passing through the centre) and zero when x = ± A (at the extremes). The sign depends on the direction of motion. This relation is extremely useful for finding the speed at a given position without dealing with time.

这个公式告诉我们，x = 0 时（经过中心）速率最大，x = ± A 时（达到端点）速率为零。正负号取决于运动方向。这一关系在已知位置求速率时非常有用，而无需处理时间变量。

7. Maximum Speed and Maximum Acceleration | 最大速度与最大加速度

From the velocity-displacement equation, the maximum speed occurs when x = 0:

v_max = ω A

从速度–位移方程可知，当 x = 0 时出现最大速度：

v_max = ω A

The acceleration is a = −ω²x, so its magnitude is maximal when |x| = A:

a_max = ω² A

加速度为 a = −ω²x，因此当 |x| = A 时其大小最大：

a_max = ω² A

These two expressions are frequently tested. For example, an exam question might give the maximum speed and maximum acceleration of a particle and ask you to determine ω and A. By taking the ratio a_max / v_max = ω, you can quickly find ω, then substitute back to find A.

这两个表达式经常被考查。例如，考题可能给出一个质点的最大速度和最大加速度，要求你求出 ω 和 A。通过比值 a_max / v_max = ω 可快速找出 ω，再代回求 A。

Be careful: in some contexts, maximum acceleration is expressed as (2πf)² A or (2π/T)² A. Knowing these alternative forms can save time in multi-step problems.

注意：在某些情境中，最大加速度会表示为 (2πf)² A 或 (2π/T)² A。熟悉这些不同形式能在多步解题中节省时间。

8. Phase Difference and Initial Conditions | 初相与初始条件

When comparing two SHM equations, the phase difference indicates the lead or lag between them. For example, x₁ = A sin ωt and x₂ = A cos ωt are ½π out of phase because cos ωt = sin(ωt + π/2). Understanding phase is essential when two particles oscillate with the same ω but different starting positions.

当比较两个简谐运动方程时，相位差表示它们之间的超前或滞后。例如，x₁ = A sin ωt 与 x₂ = A cos ωt 的相位差为 ½π，因为 cos ωt = sin(ωt + π/2)。当两个质点以相同 ω 但不同初始位置振动时，理解相位至关重要。

Initial conditions are typically given as x(0) and v(0). Use them to determine the constants A and B, or R and φ. Remember that if the particle is released from rest at x = A, then φ = 0 for the cosine form x = A cos ωt. If it passes through the centre with positive velocity at t = 0, the sine form x = (v₀/ω) sin ωt is more suitable.

初始条件通常以 x(0) 和 v(0) 的形式给出。利用它们确定常数 A、B 或 R、φ。记住，如果质点从 x = A 处静止释放，则余弦形式 x = A cos ωt 的初相 φ = 0。如果质点在 t = 0 时经过中心且速度为正，则正弦形式 x = (v₀/ω) sin ωt 更为合适。

In IB and Edexcel exams, a typical question reads: “A particle performs SHM with amplitude 0.5 m and period 4 s. At t = 0 it is at x = 0.25 m moving towards the centre. Find the equation of motion.” Always sketch a reference circle or think about the direction to assign the correct sign for velocity.

在 IB 和 Edexcel 考试中，一道典型的题目是：“一个质点以振幅 0.5 m、周期 4 s 作简谐运动。t = 0 时它位于 x = 0.25 m 且向中心运动。求运动方程。”务必画出参考圆或考虑运动方向，以正确确定速度的符号。

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

SHM can be viewed as the projection of uniform circular motion onto a diameter. If a point moves around a circle of radius A with constant angular velocity ω, its projection on a fixed diameter executes SHM. The displacement is given by x = A cos(ωt + φ) or A sin(ωt + φ) depending on the chosen axis.

简谐运动可视为匀速圆周运动在一条直径上的投影。如果一个点以恒定角速度 ω 在半径为 A 的圆上运动，它在某条固定直径上的投影即做简谐运动。位移可表示为 x = A cos(ωt + φ) 或 A sin(ωt + φ)，取决于所选坐标轴。

This geometric interpretation makes it easy to remember that the maximum speed is the tangential speed of the reference circle: v_max = ω A. It also explains why the period is T = 2π/ω – the time for one complete revolution.

这一几何解释有助于记忆：最大速度就是参考圆的切向速度 v_max = ω A。它还解释了为什么周期 T = 2π/ω——那是转动一周所需的时间。

Examiners sometimes use this link to ask questions like “Find the time taken for the particle to move from x = A/2 to x = −A/2”. Using the reference circle, you can calculate the angle swept and determine the time as angle / ω.

考官有时会利用这一联系出题，例如“求质点从 x = A/2 运动到 x = −A/2 所用的时间”。利用参考圆，可以计算出转过的角度，再通过角度 / ω 求得时间。

10. Energy in SHM (Mathematics Context) | 简谐运动中的能量（数学情境）

While energy considerations belong more naturally to physics, they appear in some mathematics syllabuses when modelling oscillations. The total mechanical energy E of a particle in SHM is the sum of kinetic and potential energy and can be shown to be constant:

E = ½ m v² + ½ m ω² x² = ½ m ω² A²

尽管能量讨论更自然地属于物理范畴，但在一些数学大纲中，当对振荡进行建模时也会涉及。简谐运动中一个质点的总机械能 E 是动能与势能之和，可以证明它是常数：

E = ½ m v² + ½ m ω² x² = ½ m ω² A²

An exam question might ask: “Using the expression for v², show that the total energy is constant.” You simply substitute v² = ω² (A² − x²) into the kinetic energy term and combine with ½ m ω² x² to obtain ½ m ω² A². This is standard algebraic manipulation.

考题可能会问：“利用 v² 的表达式，证明总能量是常数。”你只需将 v² = ω² (A² − x²) 代入动能项，并与 ½ m ω² x² 合并，即可得到 ½ m ω² A²。这属于常规的代数运算。

Occasionally, you might be given potential and kinetic energy at a specific displacement and asked to find the amplitude. Setting E = constant and using v² = ω² (A² − x²) is the key.

偶尔题目会给出某一位移下的势能与动能，要求求出振幅。关键在于令 E 为常数并使用 v² = ω² (A² − x²)。

11. Typical Exam Problem Techniques | 典型考题技巧

Technique 1: Verifying SHM. To show a motion is SHM, derive an equation of the form a = −ω²x. This often involves resolving forces and using Newton’s second law, or differentiating a given expression for displacement twice. Once in standard form, identify ω².

技巧 1：验证简谐运动。要证明一个运动是简谐运动，需推导出形如 a = −ω²x 的方程。这通常涉及力的分解并使用牛顿第二定律，或者将给定的位移表达式对时间求导两次。一旦达成标准形式，即可识别出 ω²。

Technique 2: Finding time intervals. When the question asks for the time between two positions, using the sine or cosine form directly can be messy. A cleaner method is to use the reference circle or to solve ωt₁ = arcsin(x₁/A) etc., carefully considering the quadrant.

技巧 2：计算时间间隔。当题目要求计算两个位置之间的时间时，直接使用正弦或余弦形式可能很繁琐。更简洁的方法是运用参考圆，或求解 ωt₁ = arcsin(x₁/A) 等，并仔细考虑象限。

Technique 3: Using v² = ω²(A² − x²). This relation is your go-to for linking velocity and position without time. Remember to take the correct sign based on direction. Some problems ask for the velocity when passing through a certain point for the nth time – here you need to combine with periodicity.

技巧 3：使用 v² = ω²(A² − x²)。当需要避开时间联系速度与位置时，这就是你的首选关系式。记得根据方向选取正确的符号。有些问题会问质点第 n 次通过某点时的速度——这时需要结合周期性来解答。

12. Common Mistakes and How to Avoid Them | 常见错误与避坑指南

Mistake 1: Forgetting the negative sign in a = −ω²x. Always check direction: acceleration is restoring. 中文：错误 1：忽略 a = −ω²x 中的负号。务必检查方向：加速度是回复性的。
Mistake 2: Mixing up amplitude A with coefficient of sin or cos. After writing x = A sin ωt + B cos ωt, the amplitude is R = √(A²+B²), not A or B individually. 中文：错误 2：混淆振幅 A 与 sin 或 cos 的系数。写出 x = A sin ωt + B cos ωt 后，振幅是 R = √(A²+B²)，而不是单独的 A 或 B。
Mistake 3: Assuming v = 0 at t = 0 without checking initial conditions. If the particle starts at the centre with speed, v is not zero. 中文：错误 3：未经检查初始条件就假设 t = 0 时 v = 0。如果质点从中心以一定速度开始，速度就不为零。
Mistake 4: Using degrees instead of radians in calculus when ωt is involved. Derivatives of sin and cos are only correct when angles are in radians. 中文：错误 4：在涉及 ωt 的微积分中使用角度制而非弧度制。只有在弧度量纲下，sin 和 cos 的导数才是正确的。

To avoid these, always write down the given data, convert everything to consistent units, and draw a quick sketch of the motion. Tracing the reference circle can resolve sign confusion instantly.

为避免这些错误，请务必写下已知数据，将所有量统一到一致的单位，并快速画出运动示意图。追踪参考圆可以立即消除符号混淆。

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IB Edexcel Mathematics: Simple Harmonic Motion Exam Focus | IB Edexcel 数学：简谐运动 考点精讲