IB Math: Simple Harmonic Motion Key Points | IB 数学：简谐运动考点精讲

📚 IB Math: Simple Harmonic Motion Key Points | IB 数学：简谐运动考点精讲

In the IB Mathematics Analysis & Approaches (and Applications & Interpretation) syllabus, Simple Harmonic Motion (SHM) is a core application of trigonometric functions and differential equations. It serves as a brilliant bridge between pure mathematics and the physical world, modelling everything from a mass on a spring to the oscillation of a pendulum. Understanding SHM requires fluency with the second-order differential equation d²x/dt² = −ω²x, its general solution, and the subsequent derivation of velocity, acceleration, and energy relationships. This revision guide breaks down the essential key points, common pitfalls, and exam strategies you need to master SHM with confidence.

在 IB 数学分析与方法（以及应用与解释）课程中，简谐运动（SHM）是三角函数与微分方程的核心应用。它巧妙地将纯数学与物理世界连接起来，可以用来描述弹簧振子、单摆振荡等现象。要真正掌握简谐运动，你必须熟悉二阶微分方程 d²x/dt² = −ω²x、它的通解，以及由此推导出的速度、加速度和能量关系。本篇文章将梳理所有关键考点、常见错误和应试技巧，帮助你扎实拿下这部分内容。

1. The Definition of SHM | 简谐运动的定义

In physics and mathematics, a particle is said to undergo Simple Harmonic Motion if its acceleration is always directed towards a fixed point (the equilibrium position) and is proportional to its displacement from that point. Mathematically, this condition is expressed as the relationship a ∝ −x. The negative sign indicates that the acceleration and displacement are always in opposite directions, creating the restoring force that drives the oscillation.

在物理和数学中，如果一个质点的加速度始终指向某个固定点（平衡位置），并且加速度的大小与其相对于该点的位移成正比，我们就说该质点在做简谐运动。这个条件可以用数学关系 a ∝ −x 来表示。负号表明加速度与位移方向始终相反，正是这个恢复力维持着往复振荡。

There is no requirement for the particle to slow down due to friction in ideal SHM; the motion continues indefinitely with constant amplitude. This idealised model is the foundation for understanding more complex oscillatory systems, and IB exam questions often start with this definition before asking you to derive further equations.

在理想简谐运动中，质点不会因摩擦而减速，振荡将永远以恒定的振幅持续进行。这个理想化模型是理解更复杂振动系统的基础，IB 试题经常从这一定义入手，要求你进一步推导其他方程。

2. The Key Equation: a = −ω²x | 核心公式：a = −ω²x

The proportionality in the definition is turned into an equality by introducing a positive constant ω², known as the angular frequency. The key equation of SHM is written as:

在定义中的比例关系通过引入一个正常数 ω²（称为角频率）转变为等式。简谐运动的核心公式写作：

a = −ω²x

Here, a represents acceleration, x is displacement from the equilibrium position, and ω is the angular frequency measured in rad s⁻¹. This equation holds for all forms of SHM and is the starting point for solving differential equations. It is vital to remember that ω is constant for a given system—it does not change with time or displacement.

这里 a 代表加速度，x 是相对于平衡位置的位移，ω 是角频率，单位为 rad s⁻¹。这个方程对所有形式的简谐运动都成立，也是求解微分方程的起点。务必记住，对于一个给定的振动系统，ω 是常数——它不随时间或位移变化。

Many students confuse angular frequency ω with velocity v, but they are distinct quantities. While velocity tells you how fast the position is changing, angular frequency characterises how rapidly the oscillation completes its cycles, linked to period T by ω = 2π/T.

很多同学会把角频率 ω 与速度 v 混淆，但它们是截然不同的物理量。速度描述位置变化的快慢，而角频率则表征振荡完成一个完整周期的快慢，与周期 T 的关系为 ω = 2π/T。

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, the defining equation can be rewritten as a second-order linear differential equation:

因为加速度是位移对时间的二阶导数，定义方程可以改写为一个二阶线性微分方程：

d²x/dt² = −ω²x

or equivalently,

或等价地，

d²x/dt² + ω²x = 0

This is one of the most important differential equations in applied mathematics. Its solution yields the sinusoidal functions that describe SHM. In IB Mathematics, you are expected to recognise this form and know that any equation reducible to d²x/dt² = −k²x (with k > 0) models SHM. The constant k is then identified as ω.

这是应用数学中最重要的微分方程之一。它的解就是描述简谐运动的正弦或余弦函数。在 IB 数学中，你需要能够识别这一形式，并且知道任何可以化为 d²x/dt² = −k²x（k > 0）的方程都表示简谐运动，这时常数 k 即为 ω。

Verifying that a given function satisfies the SHM differential equation is a common exam task. For example, if x = 3 cos(2t) + 4 sin(2t), you differentiate twice and show that d²x/dt² equals −4x, hence ω = 2. This is a routine but high-mark question, so practise the differentiation carefully, especially when the chain rule is involved.

验证某个给定函数是否满足简谐运动微分方程是考试中的常见题型。例如，若 x = 3 cos(2t) + 4 sin(2t)，你进行两次求导并证明 d²x/dt² 等于 −4x，由此得出 ω = 2。这类题目虽然套路化，但分值较高，务必仔细求导，尤其要注意链式法则的应用。

4. General Solution: x = A cos(ωt + φ) | 通解：x = A cos(ωt + φ)

The general solution to the differential equation d²x/dt² + ω²x = 0 can be expressed in two equivalent forms, both of which you must be fully comfortable with:

微分方程 d²x/dt² + ω²x = 0 的通解有两种等价形式，你必须对这两种形式都非常熟悉：

x = A cos(ωt + φ)

x = A sin(ωt + φ)

In either form, A represents the amplitude (maximum displacement from equilibrium) and φ is the phase constant (or initial phase), which determines the starting position at t = 0. The choice between cosine and sine depends on the initial conditions given in the problem. If the particle starts at maximum displacement (x = A when t = 0), then x = A cos(ωt) (with φ = 0) is the natural choice. If it starts at the equilibrium position moving in the positive direction, x = A sin(ωt) fits perfectly.

在这两种形式中，A 代表振幅（距离平衡位置的最大位移），φ 为相位常数（或初相），它决定了 t = 0 时的初始位置。选择余弦还是正弦形式取决于题目给定的初始条件。如果质点从最大位移处开始运动（t=0 时 x=A），那么自然选用 x = A cos(ωt)（φ=0）。如果质点从平衡位置开始向正方向运动，则 x = A sin(ωt) 恰好满足。

It is also possible to write the solution as x = C cos(ωt) + D sin(ωt), where C and D are constants determined by initial conditions. This linear combination form is especially useful when the initial displacement and initial velocity are both given. The amplitude A can then be found from A = √(C² + D²) and the phase φ from tan φ = D/C (depending on the chosen form). These manipulations appear frequently in IB problem-solving.

通解也可以写成 x = C cos(ωt) + D sin(ωt) 的形式，其中 C 和 D 是由初始条件决定的常数。当初始位移和初始速度都已知时，这种线性组合形式尤其方便。振幅 A 可以通过 A = √(C² + D²) 求得，相位 φ 通过 tan φ = D/C 确定（取决于所选形式）。这些变形在 IB 解题中经常出现。

5. Velocity and Acceleration Functions | 速度与加速度函数

Once the displacement function is known, velocity and acceleration are obtained by straightforward differentiation. For x = A cos(ωt + φ), we have:

一旦确定了位移函数，速度和加速度就可以通过直接求导得到。对于 x = A cos(ωt + φ)，有：

v = dx/dt = −Aω sin(ωt + φ)

a = d²x/dt² = −Aω² cos(ωt + φ) = −ω²x

The velocity varies sinusoidally with time, and its maximum magnitude is v_max = Aω. This maximum speed occurs as the particle passes through the equilibrium position (x = 0). Conversely, the acceleration has its maximum magnitude a_max = Aω² at the extreme points where displacement is largest (x = ±A). At the equilibrium point, acceleration is zero.

速度随时间按正弦规律变化，其最大值为 v_max = Aω。这个最大速度出现在质点经过平衡位置（x=0）的时刻。相反，加速度的最大值 a_max = Aω² 出现在位移最大处（x=±A）。在平衡位置，加速度为零。

An extremely useful relationship that avoids time entirely is the velocity-displacement equation:

一个可以完全消去时间变量的非常有用的关系式是速度-位移方程：

v = ± ω √(A² − x²)

This equation is derived using the trigonometric identity sin²θ + cos²θ = 1. It tells you the speed at any displacement x and is particularly helpful for energy calculations and for finding the speed when only position is known. The sign (±) indicates the direction of motion: positive when moving away from the equilibrium in the positive direction, negative when moving towards the negative extreme.

这个方程是利用三角恒等式 sin²θ + cos²θ = 1 推导出来的。它能直接给出任意位移 x 处的速度大小，在能量计算和仅知位置而不知时间的题目中非常实用。正负号表示运动方向：朝正方向远离平衡位置时取正，朝负方向极端运动时取负。

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

The quantities that characterise the timing of SHM are the period T, the frequency f, and the angular frequency ω. Their interrelations are fundamental and must be memorised:

描述简谐运动时间特征的物理量包括周期 T、频率 f 和角频率 ω。它们之间的相互关系非常基础，必须牢记：

ω = 2πf = 2π/T

The period T is the time taken for one complete oscillation, so the motion repeats itself after time T: x(t + T) = x(t). Because the cosine and sine functions have period 2π, we have ωT = 2π, leading directly to T = 2π/ω. Frequency f is the number of oscillations per unit time, measured in hertz (Hz), with f = 1/T.

周期 T 是完成一次完整振荡所需的时间，因此经过时间 T 后运动完全重复：x(t+T)=x(t)。由于余弦和正弦函数的周期为 2π，我们有 ωT = 2π，从而直接得到 T = 2π/ω。频率 f 是单位时间内的振荡次数，单位是赫兹（Hz），且 f = 1/T。

In IB problems, you might be given a second-order differential equation such as d²x/dt² + 4x = 0 and asked to find the period. Comparing with the standard form gives ω² = 4, so ω = 2 rad s⁻¹. Then T = 2π/ω = π s. This is a quick one-step deduction that can save time in multiple-choice sections.

在 IB 题目中，你可能会遇到诸如 d²x/dt² + 4x = 0 的二阶微分方程，要求求出周期。与标准形式对比可得 ω² = 4，因此 ω = 2 rad s⁻¹。随后 T = 2π/ω = π s。这种一步到位的推理在选择题中能节省大量时间。

Be aware that ω is called angular frequency because it is analogous to the angular speed in uniform circular motion, which provides a powerful geometric interpretation of SHM. However, the particle in SHM is not moving in a circle; it is the projection of circular motion onto a line.

注意，ω 之所以称为角频率，是因为它类似于匀速圆周运动中的角速度，这为简谐运动提供了一种强有力的几何解释。但要注意，做简谐运动的质点本身并不是在圆周上运动；它只是圆周运动在一条直线上的投影。

7. Phase and Initial Conditions | 相位与初始条件

The concept of phase is critical for distinguishing between two SHM systems that have the same amplitude and period but start from different positions. The quantity (ωt + φ) is called the phase at time t, and φ is the initial phase (phase at t = 0). A thorough understanding allows you to incorporate initial displacement and initial velocity accurately into the solution.

相位的概念对于区分两个振幅和周期相同但起始位置不同的简谐运动系统至关重要。量 (ωt + φ) 称为 t 时刻的相位，φ 则是初相（t=0 时的相位）。透彻理解相位能让你准确地将初始位移和初始速度代入通解。

When a problem states: ‘At t = 0, the particle is 2 cm from the equilibrium and moving towards it with speed 5 cm s⁻¹’, you should set up equations: x(0) = A cos φ = 2 and v(0) = −Aω sin φ = −5 (negative because moving towards equilibrium from positive displacement). Solving these simultaneously will yield A and φ. The arctangent function tan φ = −v(0)/(ω x(0)) is a handy shortcut, but you must carefully determine the correct quadrant for φ based on the signs of x(0) and v(0).

当题目给出：“t=0 时，质点距离平衡位置 2 cm 并正以 5 cm s⁻¹ 的速度朝向平衡位置运动”，你应该建立方程组：x(0) = A cos φ = 2 且 v(0) = −Aω sin φ = −5（负号是因为从正位移向平衡位置运动）。联立求解即可得到 A 和 φ。正切关系式 tan φ = −v(0)/(ω x(0)) 是个便捷的捷径，但你必须根据 x(0) 和 v(0) 的正负号仔细确定 φ 的正确象限。

Questions about phase difference between two oscillators also appear. For example, if one oscillator has displacement x₁ = A cos(ωt) and another has x₂ = A cos(ωt + π/2), the second is said to lead the first by π/2 radians (or 90°). This implies that x₂ reaches its peak earlier than x₁ by a quarter period. Sketching the two curves is often the best way to visualise this relationship.

关于两个振子之间相位差的题目也会出现。例如，若一个振子的位移为 x₁ = A cos(ωt)，另一个为 x₂ = A cos(ωt + π/2)，则称第二个振子的相位超前第一个 π/2 弧度（或 90°）。这意味着 x₂ 比 x₁ 早四分之一周期到达峰值。画出两条曲线通常是直观理解这种关系的最好方式。

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

Although energy considerations are primarily a physics topic, they are heavily embedded in the applied mathematics context of IB. The total mechanical energy in an undamped SHM system remains constant, being continuously converted between kinetic energy (KE) and potential energy (PE).

虽然能量分析主要属于物理范畴，但它已深深融入 IB 应用数学的背景中。在无阻尼简谐运动系统中，总机械能保持不变，动能（KE）与势能（PE）之间持续转化。

The kinetic energy as a function of displacement is given by:

动能作为位移的函数可表示为：

KE = ½ m v² = ½ m ω² (A² − x²)

The potential energy (elastic potential in a spring, for example) is:

势能（例如弹簧的弹性势能）为：

PE = ½ m ω² x²

Adding these gives the constant total energy:

将两者相加得到恒定的总能量：

E_total = ½ m ω² A²

From these expressions, you can see that at x = 0 all energy is kinetic; at x = ±A all energy is potential. Understanding these forms enables you to find velocity without calculus: simply set ½ m v² = ½ m ω² (A² − x²) and solve for v. This is mathematically identical to the velocity-displacement relation derived earlier.

由这些表达式可以看出，在 x=0 处所有能量为动能；在 x=±A 处所有能量为势能。理解这些形式后，你可以不依赖微积分求出速度：只需令 ½ m v² = ½ m ω² (A² − x²) 并求解 v。这与之前推导的速度-位移关系在数学上是完全等价的。

If a damping force is introduced (HL only), the total energy decreases exponentially over time. While detailed damping differential equations might be beyond the core, you should know that the amplitude decreases but the frequency may slightly shift (or remain approximately constant for light damping).

如果引入阻尼力（仅限高水平课程），总能量随时间呈指数衰减。尽管详细的阻尼微分方程可能超出核心范围，但你应当了解，振幅会减小，而频率可能会略微改变（对于弱阻尼情况，频率近似保持不变）。

9. Graphical Representations | 图形表示

IB exam questions frequently ask you to sketch or interpret graphs of displacement, velocity, and acceleration against time for an SHM system. All three are sinusoidal and share the same period, but they are shifted relative to one another.

IB 考试中常要求你画出或解读简谐运动系统的位移-时间、速度-时间和加速度-时间图像。这三者都是正弦曲线且具有相同的周期，但它们彼此之间存在相位偏移。

For x = A cos(ωt) (starting at maximum positive displacement):

Displacement x(t) is a cosine curve starting at A.

位移 x(t) 是一条从 A 开始的余弦曲线。
Velocity v(t) is a negative sine curve, starting at 0 and first going negative. It leads displacement by π/2.

速度 v(t) 是一条负正弦曲线，从 0 开始先向负方向。它比位移超前 π/2。
Acceleration a(t) is a negative cosine curve, exactly out of phase with displacement (phase difference of π).

加速度 a(t) 是一条负余弦曲线，与位移完全反相（相位差为 π）。

Being able to transform between x-t, v-t, and a-t graphs is a key skill. The zero-crossings of the velocity graph correspond to the turning points (maxima and minima) of the displacement graph. The zero-crossings of the acceleration graph correspond to the points where displacement is zero. Always label axes correctly and indicate the amplitude and period clearly on your sketches.

能够在 x-t、v-t 和 a-t 图之间进行转换是一项关键技能。速度图与时间轴的交点对应位移图的极值点（极大值和极小值）。加速度图与时间轴的交点对应位移为零的位置。画草图时务必正确标注坐标轴，并清楚标明振幅和周期。

10. Connection to Uniform Circular Motion | 与匀速圆周运动的联系

One of the most insightful ways to understand SHM is through its geometric link to uniform circular motion. Imagine a point P moving with constant angular speed ω around a circle of radius A. The projection of P onto a diameter of the circle executes simple harmonic motion.

理解简谐运动最直观的方法之一，就是通过它与匀速圆周运动的几何联系。设想一个点 P 以恒定角速度 ω 在半径为 A 的圆周上运动，那么 P 在直径上的投影就做简谐运动。

If the diameter is along the x-axis and the centre is the origin, the x-coordinate of P at time t is x = A cos(ωt + φ), exactly the SHM equation. The velocity of the projection is the component of the circular velocity Aω along the x-axis, which gives v = −Aω sin(ωt + φ). The acceleration is the component of the centripetal acceleration Aω² towards the centre, yielding a = −Aω² cos(ωt + φ).

若该直径沿 x 轴且圆心为原点，则 P 在时刻 t 的 x 坐标为 x = A cos(ωt + φ)，正是简谐运动方程。投影的速度是圆周运动线速度 Aω 在 x 轴上的分量，从而得出 v = −Aω sin(ωt + φ)。加速度则是向心加速度 Aω² 在 x 轴上的分量，即 a = −Aω² cos(ωt + φ)。

This circular reference model not only helps in deriving the SHM equations but also provides a mnemonic for remembering the maximum values: max speed is the circular speed Aω; max acceleration is the centripetal acceleration Aω². When you forget a formula, a quick mental sketch of the reference circle can bring it back.

这个参考圆模型不仅有助于推导简谐运动方程，也为记住极值提供了一个记忆技巧：最大速度等于圆周线速度 Aω；最大加速度等于向心加速度 Aω²。当你记不清公式时，脑海中快速画一个参考圆往往能帮你回忆起来。

11. Solving SHM Problems: A Step-by-Step Strategy | 简谐运动解题策略

Approaching an SHM problem systematically can prevent many common mistakes. Follow this general strategy:

Identify ω: Extract ω from the given differential equation or from the period/frequency. Write down ω² clearly.

确定 ω：从给定的微分方程或周期/频率中提取 ω。清晰写下 ω²。
Write general solution: choose the form that best matches initial conditions, e.g., x = A cos(ωt + φ) or x = C cos(ωt) + D sin(ωt).

写出通解：选择最匹配初始条件的形式，例如 x = A cos(ωt + φ) 或 x = C cos(ωt) + D sin(ωt)。
Apply initial conditions: plug in t = 0 for displacement and velocity. Solve simultaneously for A and φ (or C and D). Pay close attention to the sign of velocity.

代入初始条件：将 t=0 代入位移和速度的表达式。联立求解 A 和 φ（或 C 和 D）。特别注意速度的正负号。
Find required quantities: whether it is the time to reach a certain position, the maximum speed, or the total energy, use the relationships v_max = Aω, a_max = Aω², and v = ±ω√(A² − x²).

计算所求量：无论是求到达某一位置的时间、最大速度还是总能量，使用关系式 v_max = Aω, a_max = Aω² 以及 v = ±ω√(A² − x²)。
Check for reasonableness: amplitude must be positive; period T must be positive; the maximum speed cannot exceed Aω. If your calculated phase puts the particle in the wrong direction at t=0, revisit the quadrant choice.

合理性检查：振幅必须为正；周期 T 必须为正；最大速度不能超过 Aω。如果计算出的相位导致 t=0 时质点运动方向错误，请重新审视象限的选择。

This structured method will serve you well across both Paper 1 and Paper 2 questions. In calculator papers, you might need to solve trigonometric equations numerically, but the algebraic setup remains the same.

这种有条理的方法在试卷一和试卷二中都会让你受益匪浅。在允许使用计算器的试卷中，你可能需要数值求解三角方程，但代数框架的搭建步骤是完全相同的。

12. Common Pitfalls and Exam Tips | 常见错误与考试技巧

Even students confident with calculus often lose marks on SHM questions due to small but critical errors. Here are the most frequent pitfalls to avoid:

Confusing amplitude and displacement: A is a constant, while x varies with time. In energy formulas, always use A² – x², never A² – A².

混淆振幅与位移：A 是常数，而 x 随时间变化。在能量公式中，始终使用 A² – x²，绝不能用 A² – A²。
Forgetting units: angular frequency ω is in rad s⁻¹, not Hz. Period T is in seconds, frequency f in Hz. Mixing these up leads to dimensionally incorrect answers.

忘记单位：角频率 ω 的单位是 rad s⁻¹，不是 Hz。周期 T 的单位是秒，频率 f 的单位是 Hz。混淆单位会导致答案量纲错误。
Sign errors in velocity: if the particle is moving towards the equilibrium from a positive displacement, the velocity is negative. Always visualise the physical scenario.

速度的正负号错误：如果质点从正位移处向平衡位置运动，速度应为负值。始终在脑海中想象物理情景。
Incorrect differentiation: when differentiating cos(ωt + φ), the derivative is −ω sin(ωt + φ); don’t forget the factor ω. Similarly, the second derivative of cos is −ω² cos.

求导错误：对 cos(ωt + φ) 求导时，导数是 −ω sin(ωt + φ)，不要漏掉因子 ω。同理，cos 的二阶导数为 −ω² cos。
Misidentifying ω: in an equation like d²x/dt² = −9x, ω² = 9, so ω = 3, not 9. This elementary slip can cascade through the whole problem.

错误识别 ω：在像 d²x/dt² = −9x 的方程中，ω² = 9，所以 ω = 3，而不是 9。这种低级错误会连带导致整道题出错。

Practise with official IB past papers to become familiar with the phrasing and expected level of algebraic detail. Show all steps clearly, especially the verification that a given function satisfies the SHM differential equation—examiners award method marks even if a minor slip occurs.

使用 IB 官方历年真题进行练习，熟悉题目措辞和预期的代数推导细致程度。务必清晰展示所有步骤，尤其是在验证给定函数满足简谐运动微分方程时——即使出现小失误，考官仍会按解题方法给分。

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IB Math: Simple Harmonic Motion Key Points | IB 数学：简谐运动 考点精讲