IB Maths: Simple Harmonic Motion | IB数学：简谐振荡

📚 IB Maths: Simple Harmonic Motion | IB数学：简谐振荡

Simple harmonic motion (SHM) is one of the few topics that sits right at the meeting point of IB Mathematics and IB Physics. In mathematics it is a differential equation topic: you are asked to show that a given expression satisfies d²x/dt² = −ω²x, to solve that equation with given initial conditions, and to extract amplitude, period, maximum speed and maximum acceleration from the solution. In Physics the same equations appear as the motion of a mass on a spring or a small-angle pendulum.

简谐振荡（SHM）是少数同时横跨 IB 数学与 IB 物理的考点之一。在数学里它是一个微分方程专题：你需要证明某个表达式满足 d²x/dt² = −ω²x，在给定初始条件下求解该方程，并从解中提取振幅、周期、最大速率与最大加速度。在物理里，同一组方程则表现为弹簧振子或小角度单摆的运动。

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1. What Is Simple Harmonic Motion? | 什么是简谐振荡

Simple harmonic motion is the motion of a particle whose acceleration is always directed towards a fixed point and is proportional to its displacement from that point. The fixed point is called the equilibrium position. Because the acceleration always points back towards equilibrium, the particle is constantly being pulled back whenever it moves away, so it oscillates to and fro indefinitely (in the absence of damping).

简谐振荡是指质点的加速度始终指向某一固定点，且大小与它离开该点的位移成正比。这个固定点称为平衡位置。由于加速度始终指回平衡位置，质点一旦偏离就会被拉回，因此在没有阻尼的情况下会永远往复振荡。

• Direction of acceleration: always opposite to the displacement, towards equilibrium. | 加速度方向：始终与位移方向相反，指向平衡位置。
• Magnitude of acceleration: proportional to the distance from equilibrium. | 加速度大小：与离开平衡位置的距离成正比。
• Not every periodic motion is SHM. Circular motion, for example, is periodic but not simple harmonic in the x-direction unless you look at a single component. | 并非所有周期运动都是简谐振荡。例如匀速圆周运动本身是周期的，但只有在取某一坐标分量时才是简谐的。
• Typical IB examples: mass on a spring, small oscillations of a pendulum, a particle attached between two stretched elastic strings, a floating cylinder bobbing vertically. | IB 常见例子：弹簧振子、单摆的小幅摆动、两端被拉伸的弹性绳所系的质点、竖直浮沉的小圆柱。

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2. The Defining Differential Equation | 定义性微分方程

Writing the definition as an equation, the acceleration is a = −ω²x where ω² is a positive constant. Since a = d²x/dt², this gives the defining second-order differential equation of simple harmonic motion. If you can rearrange a given equation of motion into this form and read off a positive value of ω², you have proved the motion is simple harmonic.

把定义写成方程，加速度为 a = −ω²x，其中 ω² 是正常数。由于 a = d²x/dt²，就得到简谐振荡的定义性二阶微分方程。只要能把手头的运动方程整理成这个形式并读出一个正的 ω²，就证明了该运动是简谐的。

a = d²x/dt² = −ω²x

d²x/dt² + ω²x = 0

The constant ω is called the angular frequency and is measured in radians per second (rad s⁻¹). It is not the same as the frequency f. For a mass m on a spring of stiffness k, Newton’s second law gives ma = −kx, so ω² = k/m and therefore ω = √(k/m).

常数 ω 称为角频率，单位为弧度每秒（rad s⁻¹），它并不等于频率 f。对于劲度系数为 k 的弹簧上质量为 m 的物体，牛顿第二定律给出 ma = −kx，因此 ω² = k/m，即 ω = √(k/m)。

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3. Solving the Equation: General Solutions | 求解方程：通解

The auxiliary (characteristic) equation of d²x/dt² + ω²x = 0 is λ² + ω² = 0, giving λ = ±iω. Because the roots are purely imaginary and distinct, the general solution is a linear combination of sine and cosine, which can be written in a single cosine with a phase shift.

方程 d²x/dt² + ω²x = 0 的特征方程是 λ² + ω² = 0，解得 λ = ±iω。由于两根为纯虚数且互异，通解是正弦与余弦的线性组合，也可以写成带相位差的单一余弦形式。

x = A cos(ωt + ε)  or  x = A sin(ωt + φ)  or  x = P cos ωt + Q sin ωt

• All three forms are acceptable in the exam; choose the one that makes the given initial conditions easiest. | 三种形式在考试中都被接受，选择使初始条件最易处理的那种。
• A is the amplitude (always taken positive); ε and φ are phase angles in radians. | A 是振幅（通常取正值）；ε 与 φ 是相位角，单位弧度。
• There are two arbitrary constants (A and ε, or P and Q), which are fixed by two initial conditions, usually x(0) and v(0). | 通解中有两个任意常数（A 与 ε，或 P 与 Q），由两个初始条件确定，通常是 x(0) 与 v(0)。
• To verify a candidate solution, differentiate it twice and substitute back; you should recover a = −ω²x. | 验证一个候选解时，把它求导两次再代回，应得到 a = −ω²x。

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4. Amplitude, Period and Frequency | 振幅、周期与频率

The amplitude A is the maximum displacement from the equilibrium position, so the particle moves between x = −A and x = +A. The period T is the time for one complete oscillation, and the frequency f is the number of oscillations per second, measured in hertz (Hz). The angular frequency ω, the period and the frequency are linked by a set of simple relations.

振幅 A 是离开平衡位置的最大位移，因此质点运动范围是 −A ≤ x ≤ A。周期 T 是完成一次全振荡所需时间，频率 f 是每秒振荡次数，单位为赫兹（Hz）。角频率 ω、周期与频率之间由一组简单关系联系。

T = 2π ÷ ω    f = 1 ÷ T = ω ÷ (2π)    ω = 2πf = 2π ÷ T

A key property of SHM is that the period is independent of the amplitude. A pendulum swinging through 2° has the same period as one swinging through 5° (provided the angle stays small). This property is called isochronism and is a favourite exam discussion point.

简谐振荡的一个重要性质是：周期与振幅无关。摆动 2° 的单摆与摆动 5° 的单摆周期相同（只要角度足够小）。这一性质称为等时性，是考试中常见的讨论点。

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5. Phase Angle and Initial Conditions | 相位角与初始条件

Taking the form x = A cos(ωt + ε), differentiate once to get v = −Aω sin(ωt + ε). Setting t = 0 gives x₀ = A cos ε and v₀ = −Aω sin ε. Dividing the second by the first eliminates A and gives the phase angle directly.

取形式 x = A cos(ωt + ε)，求导一次得 v = −Aω sin(ωt + ε)。令 t = 0 得 x₀ = A cos ε 与 v₀ = −Aω sin ε。第二式除以第一式即可消去 A，直接得到相位角。

tan ε = −v₀ ÷ (ωx₀)

• Started at maximum displacement with zero velocity: x = A, v = 0, so ε = 0. | 从最大位移处静止释放：x = A，v = 0，故 ε = 0。
• Started at equilibrium moving in the positive direction: x = 0, v = Aω, so use x = A sin ωt, i.e. ε = −π/2 in the cosine form. | 从平衡位置向正方向运动：x = 0，v = Aω，用 x = A sin ωt，即在余弦形式中 ε = −π/2。
• Always check that your A and ε give the correct sign of v₀; the arctangent function on your GDC returns only one of two possible angles. | 一定要检验所选的 A 与 ε 是否给出正确的 v₀ 符号；计算器的反正切只返回两个可能角中的一个。

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6. Velocity and Acceleration | 速度与加速度

Differentiating x = A cos(ωt + ε) once gives the velocity and twice gives the acceleration. Notice that the acceleration is simply −ω² multiplied by the displacement, which is exactly the defining equation again – a useful self-check.

对 x = A cos(ωt + ε) 求导一次得速度，求导两次得加速度。注意加速度正好等于 −ω² 乘以位移，这正是定义方程本身，可作为自检。

v = −Aω sin(ωt + ε)    a = −Aω² cos(ωt + ε) = −ω²x

vₘₐₓ = Aω  at  x = 0     aₘₐₓ = Aω²  at  x = ±A

Position | 位置	Speed | 速率	Acceleration | 加速度
x = 0 (equilibrium | 平衡位置)	maximum, Aω | 最大，Aω	zero | 为零
x = ±A (extreme | 极端位置)	zero | 为零	maximum, Aω² | 最大，Aω²
x = ±A/2	(√3 ÷ 2)Aω	Aω² ÷ 2

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7. The Displacement-Velocity Relationship | 位移—速度关系

Because cos²(ωt + ε) + sin²(ωt + ε) = 1, you can eliminate time from the expressions for x and v. Dividing each by the appropriate amplitude and adding the squares gives a very useful identity that links speed to position without any trigonometry.

由于 cos²(ωt + ε) + sin²(ωt + ε) = 1，可以从 x 与 v 的表达式中消去时间。分别除以相应的振幅再平方相加，就得到一个非常实用的恒等式，无需三角函数即可把速率与位置联系起来。

(x ÷ A)² + (v ÷ Aω)² = 1

v² = ω²(A² − x²)    v = ±ω√(A² − x²)

• Use this when a question gives a position and asks for a speed, or vice versa, without mentioning time. | 当题目给出位置求速率（或反之）而不涉及时间时，用这个关系最快。
• The ± sign is essential: the particle passes through the same position twice per period, once in each direction. | ± 号不可省略：质点每周期两次经过同一位置，方向相反。
• Note that x² ≤ A², so |x| can never exceed the amplitude – a quick check on your answer. | 注意 x² ≤ A²，故 |x| 不可能超过振幅，这可用于快速检查答案。

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8. Energy in Simple Harmonic Motion | 简谐振荡中的能量

For a mass m oscillating with angular frequency ω, the kinetic energy depends on the speed and the potential energy depends on the square of the displacement, measured from equilibrium. The total mechanical energy is constant and is proportional to the square of the amplitude.

对于以角频率 ω 振荡的质量 m，动能取决于速率，势能取决于位移的平方（自平衡位置量起）。总机械能守恒，且与振幅的平方成正比。

Eₖ = ½mv² = ½mω²(A² − x²)

Eₚ = ½mω²x²     Eₜₒₜₐₗ = ½mω²A²

• At x = 0 all the energy is kinetic; at x = ±A all the energy is potential. | 在 x = 0 处能量全为动能；在 x = ±A 处能量全为势能。
• The average kinetic energy and the average potential energy are each ¼mω²A², so each is half the total. | 平均动能与平均势能均为 ¼mω²A²，各占总能量的一半。
• If the amplitude is doubled while ω stays fixed, the total energy is multiplied by 4. | 若 ω 不变而振幅加倍，总能量变为原来的 4 倍。
• Energy questions are often pure calculus in IB: differentiate or integrate the given expression to find a maximum. | IB 中的能量题常是纯微积分题：对给定表达式求导或积分以找最大值。

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9. Physical Models: Springs and Pendulums | 物理模型：弹簧振子与单摆

Two standard models generate the SHM equation directly and are worth memorising because they let you convert physical data into ω immediately. The mass-spring system comes from Hooke’s law, and the simple pendulum comes from the small-angle approximation sin θ ≈ θ (with θ in radians).

有两个标准模型能直接导出简谐振荡方程，值得记住，因为它们可以立刻把物理数据转换为 ω。弹簧振子来自胡克定律，单摆来自小角度近似 sin θ ≈ θ（θ 用弧度）。

Mass on a spring | 弹簧振子：ω = √(k ÷ m)    T = 2π√(m ÷ k)

Simple pendulum | 单摆：ω = √(g ÷ L)    T = 2π√(L ÷ g)

• For the pendulum, T depends only on length and the gravitational field strength, not on the mass of the bob. | 对单摆而言，T 只与摆长和重力加速度有关，与摆球质量无关。
• The small-angle approximation is usually quoted as valid for θ below about 10° (0.17 rad). | 小角度近似通常认为在 θ 小于约 10°（0.17 rad）时成立。
• A vertical spring stretches by an extra amount e = mg/k at equilibrium; oscillation then occurs about that new equilibrium point. | 竖直弹簧在平衡时额外伸长 e = mg/k，随后围绕这个新平衡位置振荡。

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10. SHM in the IB Syllabus and Exams | 简谐振荡在 IB 大纲与考试中

In IB Mathematics: Analysis and Approaches at HL, simple harmonic motion appears within the calculus topic, where the emphasis is on setting up and solving the differential equation d²x/dt² = −ω²x and interpreting the solution. In Applications and Interpretation, the sinusoidal model x = A sin(ωt + φ) is met earlier as a trigonometric modelling tool, and SHM questions then focus on amplitude, period, phase shift and maximum values.

在 IB 数学《分析与方法》（AA）HL 中，简谐振荡出现在微积分部分，重点是建立并求解微分方程 d²x/dt² = −ω²x 并解释其解。在《应用与解释》（AI）中，正弦模型 x = A sin(ωt + φ) 更早作为三角建模工具出现，简谐振荡题目则聚焦于振幅、周期、相位平移与最大值。

• “Show that the motion is simple harmonic” – differentiate twice, substitute, and state ω² clearly. | “证明该运动是简谐的”——求导两次、代入、并明确写出 ω²。
• “Find the amplitude and the period” – read A and ω from the given expression, then use T = 2π ÷ ω. | “求振幅与周期”——从表达式读出 A 与 ω，再用 T = 2π ÷ ω。
• “Given that the particle starts from rest at …” – use initial conditions to find A and ε. | “已知质点自……处静止开始”——用初始条件求 A 与 ε。
• “Find the maximum speed / the speed when x = …” – use Aω or v² = ω²(A² − x²). | “求最大速率／当 x = …… 时的速率”——用 Aω 或 v² = ω²(A² − x²)。
• “Find the first time at which …” – solve a trigonometric equation in radians and give the smallest positive t. | “求第一次出现……的时刻”——解弧度制三角方程，取最小的正 t。

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11. Common Mistakes and Exam Tips | 常见错误与应试技巧

Most lost marks in SHM questions come from a small set of recurring errors rather than from conceptual difficulty. Checking these before you move on is worth more than any extra practice question.

简谐振荡题的失分大多来自少数几类反复出现的错误，而不是概念上的困难。动笔前一分钟逐一排查这些点，比多做一道题更划算。

• Radian mode. Almost every IB SHM question is in radians; a calculator left in degree mode will produce nonsense for ωt. | 弧度模式。IB 的简谐振荡题几乎都用弧度，计算器若停在角度模式，ωt 的计算会全错。
• Confusing ω with f. ω = 2πf, so a particle with f = 2 Hz has ω = 4π rad s⁻¹, not 2. | 混淆 ω 与 f。ω = 2πf，例如 f = 2 Hz 对应 ω = 4π rad s⁻¹，而不是 2。
• Dropping the ± in v = ±ω√(A² − x²). If the question asks for a velocity with direction, you must decide the sign from the context. | 漏掉 v = ±ω√(A² − x²) 中的 ±。若题目要求带方向的速（速）度，必须根据情境判断符号。
• Forgetting the phase angle when the particle does not start at an extreme or at equilibrium. | 质点不是从极端位置或平衡位置开始时，忘记相位角。
• Assuming any oscillation is SHM. Always verify a ∝ −x with a positive constant of proportionality. | 假定任何振荡都是简谐的。必须验证 a ∝ −x 且比例常数为正。
• Rounding too early. Keep full precision on the GDC and round only the final answer, usually to three significant figures. | 过早四舍五入。计算器中间过程保留全精度，只对最终答案取三（或题目要求的）位有效数字。

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12. Full Worked Example | 完整例题

A particle P moves in a straight line so that its displacement x metres from a fixed point O at time t seconds is given by x = 0.8 cos(4πt + π/3). Find the amplitude, the period, the initial displacement, the maximum speed, the maximum magnitude of the acceleration, and the speed when x = 0.4.

质点 P 沿直线运动，在 t 秒时离开固定点 O 的位移为 x = 0.8 cos(4πt + π/3) 米。求振幅、周期、初始位移、最大速率、加速度的最大大小，以及当 x = 0.4 时的速率。

Step 1 – compare with x = A cos(ωt + ε): A = 0.8 m and ω = 4π rad s⁻¹. Step 2 – the period is T = 2π ÷ ω = 2π ÷ 4π = 0.5 s, so f = 2 Hz. Step 3 – the initial displacement is x(0) = 0.8 cos(π/3) = 0.8 × ½ = 0.4 m, and the particle starts on the positive side of O.

第一步——与 x = A cos(ωt + ε) 对照：A = 0.8 m，ω = 4π rad s⁻¹。第二步——周期 T = 2π ÷ ω = 2π ÷ 4π = 0.5 s，故 f = 2 Hz。第三步——初始位移 x(0) = 0.8 cos(π/3) = 0.8 × ½ = 0.4 m，质点从 O 的正侧开始运动。

Step 4 – the maximum speed is Aω = 0.8 × 4π = 3.2π ≈ 10.1 m s⁻¹. Step 5 – the maximum acceleration magnitude is Aω² = 0.8 × (4π)² = 0.8 × 16π² = 12

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