📚 AS Mathematics: Simple Harmonic Motion Exam Essentials | AS 数学：简谐运动考点精讲
Simple harmonic motion (SHM) is one of the most distinctive topics in AS Mathematics – it bridges mechanics, calculus and trigonometric modelling. In this revision guide, you will find every essential formula, graph and skill that examiners expect you to master, from the fundamental acceleration law to energy conservation in a spring-mass system.
简谐运动是 AS 数学中非常独特的一个专题——它将力学、微积分与三角函数建模完美地连接了起来。在这篇考点精讲中，你会看到从最基本的加速度规律到弹簧振子能量守恒的所有必备公式、图像与解题技巧，覆盖阅卷老师最关注的全部重点。
1. Definition of SHM | 简谐运动的定义
SHM occurs when a particle moves along a straight line such that its acceleration is always directed towards a fixed point (the centre of oscillation) and is proportional to its distance from that point. Mathematically, we write a ∝ –x. The negative sign means acceleration always opposes the displacement.
当质点沿直线运动，且加速度始终指向一个固定点（振动中心），大小与其到该固定点的距离成正比时，物体做简谐运动。数学上写作 a ∝ –x。负号表明加速度永远与位移方向相反。
In AS exam questions, candidates often lose marks by forgetting the direction. A complete definition must state both the proportionality and the fact that acceleration points toward the equilibrium position. Without the negative sign, the motion is not SHM.
在 AS 考试中，考生常因漏掉方向而丢分。完整的定义必须同时说明正比关系和加速度指向平衡位置这两个要点。如果缺少负号，运动就不属于简谐运动。
2. The Fundamental Equation a = –ω²x | 基本方程 a = –ω²x
By introducing a positive constant ω (angular frequency), the proportionality becomes a = –ω²x. This is the core differential equation of SHM. It tells us that the magnitude of acceleration grows with displacement, and the constant ω² determines how ‘stiff’ the oscillation is.
引入一个正常数 ω（角频率），比例关系就写成 a = –ω²x。这是简谐运动的核心微分方程。它告诉我们加速度的大小随位移增大而增大，而常数 ω² 决定了振动的“刚度”。
Almost every SHM problem begins with identifying ω. You will be given either a graph, a period, or a spring’s force constant, and from that you must extract ω². Remember: a is the second derivative of x with respect to time, so d²x/dt² = –ω²x.
几乎每一道简谐运动题目都要从确定 ω 开始。题目会给你图像、周期或弹簧的劲度系数，你必须从中求出 ω²。请记住：a 是 x 对时间的二阶导数，因此 d²x/dt² = –ω²x。
3. Displacement as a Function of Time | 位移随时间的变化
The general solution of d²x/dt² = –ω²x is x = A sin(ωt + φ) or x = A cos(ωt + φ). The two forms are mathematically equivalent; a phase shift φ can switch between sine and cosine. A represents the amplitude and φ is the phase constant.
微分方程 d²x/dt² = –ω²x 的通解为 x = A sin(ωt + φ) 或 x = A cos(ωt + φ)。这两种形式在数学上等价，通过调整初相 φ 可以实现正弦与余弦的转换。A 代表振幅，φ 为初相。
In an AS exam, you must choose the correct form based on initial conditions. If the particle starts at the equilibrium with positive velocity, use x = A sin ωt. If it starts at maximum displacement, use x = A cos ωt. Always check the starting position before writing the solution.
在 AS 考试中，你必须根据初始条件选择正确的形式。若质点从平衡位置以正方向速度出发，使用 x = A sin ωt。若从最大位移处出发，则使用 x = A cos ωt。写解之前务必先判断起始位置。
4. Velocity in SHM | 简谐运动中的速度
Differentiating displacement gives velocity. For x = A sin(ωt), v = dx/dt = ωA cos(ωt). The maximum speed is vₘₐₓ = ωA, which occurs at the centre. An alternative expression, independent of time, is v = ±ω √(A² – x²). The sign indicates direction.
将位移对时间求导即得速度。对于 x = A sin(ωt)，v = dx/dt = ωA cos(ωt)。最大速度 vₘₐₓ = ωA 出现在振动中心处。另一个与时间无关的速度公式为 v = ±ω √(A² – x²)，正负号表示速度方向。
The v²-x² relation is extremely useful for energy calculations and for finding speed at a specific point without solving for t. When x = 0, v = ±ωA; when x = A, v = 0. These extremes are frequently tested in structured questions.
v²-x² 关系式在能量计算以及不需求解时间 t 而直接求某点速度时非常有用。当 x = 0 时，v = ±ωA；当 x = A 时，v = 0。这些极值情况在结构化问题中频繁出现。
5. Amplitude, Period and Frequency | 振幅、周期与频率
The amplitude A is the maximum displacement from equilibrium. The period T is the time for one complete oscillation, given by T = 2π/ω. The frequency f is the number of oscillations per second: f = 1/T = ω/(2π). ω itself has units rad s⁻¹.
振幅 A 是物体离平衡位置的最大位移。周期 T 是完成一次完整振动所需的时间，公式为 T = 2π/ω。频率 f 是每秒振动的次数：f = 1/T = ω/(2π)。ω 本身的单位是 rad s⁻¹。
A common mistake is to confuse amplitude with the total length of oscillation (which is 2A). Also, remember that T depends only on ω; for a given ω, period is constant, which is the hallmark of isochronous motion.
一个常见错误是把振幅与振动总路程（2A）混淆。还要记住，周期 T 只依赖于 ω；给定 ω 后周期恒定，这正是简谐运动等时性的标志。
6. Spring-Mass Systems | 弹簧振子系统
For a mass m attached to a light spring of stiffness k, oscillating horizontally on a smooth surface, the restoring force is F = –kx. Using Newton’s second law, m d²x/dt² = –kx, hence ω² = k/m and T = 2π √(m/k).
对于连接在劲度系数为 k 的轻弹簧上、在光滑水平面上振动的质量块 m，恢复力为 F = –kx。利用牛顿第二定律，m d²x/dt² = –kx，因此 ω² = k/m，T = 2π √(m/k)。
In vertical spring-mass systems, the equilibrium position shifts due to gravity, but the SHM still occurs about the new equilibrium with the same ω. You must measure displacement from the equilibrium, not from the spring’s natural length. Examiners love to test this distinction.
在竖直弹簧振子中，重力会使平衡位置下移，但简谐运动仍以新平衡位置为中心，且 ω 不变。务必以平衡位置为位移零点，而不是弹簧原长。阅卷人特别偏爱考查这一区别。
7. The Simple Pendulum (Small-Angle Approximation) | 单摆（小角度近似）
A simple pendulum consists of a point mass on a light inextensible string. For small swings (θ < 10°), the motion is approximately SHM. The tangential restoring force gives ω² = g/l, so T = 2π √(l/g). Notice that period is independent of mass and amplitude.
单摆由质点悬挂在轻质不可伸长的细线上构成。当摆角很小（θ < 10°）时，运动可近似为简谐运动。切向恢复力导出 ω² = g/l，因此 T = 2π √(l/g)。注意周期与质量和振幅无关。
To measure g in a laboratory setting, students often plot T² against l, obtaining a straight line through the origin with gradient 4π²/g. You may be asked to find g from such a graph; always be precise about units and significant figures.
在实验测量重力加速度 g 时，学生通常绘制 T² 对 l 的图像，得到一条过原点的直线，斜率为 4π²/g。你可能会被要求从图像求 g，请注意单位和有效数字的准确性。
8. Energy in SHM | 简谐运动中的能量
The total mechanical energy in undamped SHM is constant: E_total = ½ m ω² A². Kinetic energy at displacement x is KE = ½ m ω² (A² – x²), and potential energy (elastic or gravitational) is PE = ½ m ω² x². The energy continuously swaps between kinetic and potential.
无阻尼简谐运动中的总机械能守恒：E_total = ½ m ω² A²。在位移 x 处的动能为 KE = ½ m ω² (A² – x²)，势能（弹性或重力势能）为 PE = ½ m ω² x²。能量在动能和势能之间不断转化。
At the centre (x=0), PE=0 and KE is maximum. At extremes (x=±A), KE=0 and PE is maximum. Energy bar charts and pie charts appear in multiple-choice questions to test your understanding of this swap.
在中心位置（x=0），PE=0，动能最大；在极端位置（x=±A），KE=0，势能最大。能量条形图和饼状图常出现在选择题中，考查你就这种转化关系的理解。
9. Graphical Representations | 图形表示
| Quantity | Shape vs time | Shape vs displacement |
|---|---|---|
| Displacement x | Sine or cosine curve | Straight line through origin for x-t? Not applicable; however, x vs x is a line. Better to note: a vs x is a straight line through origin with negative slope. |
Actually, let’s create a clear table:
| Graph | Appearance | Key feature |
|---|---|---|
| x vs t | Sinusoidal wave | Amplitude A, period T |
| v vs t | Sinusoidal, ½ π out of phase with x | v leads x by quarter period |
| a vs t | Sinusoidal, π out of phase with x | a is anti-phase with x |
| a vs x | Straight line through origin, negative gradient | Gradient = –ω² |
| v² vs x² | Linear: v² = ω²(A² – x²) | Intercepts give A² and ω²A² |
Being able to sketch and interpret these graphs quickly is essential. Always label axes, key values (A, T, ωA, ω²A) and note the phase relationships. In the exam, you might be asked to find ω from the gradient of an a–x graph.
能够迅速绘制并解读这些图像至关重要。一定要标注坐标轴、关键值（A、T、ωA、ω²A）并指明相位关系。考试中可能会要求你由 a–x 图像的斜率求出 ω。
10. Differential Equations and SHM Proofs | 微分方程与简谐运动证明
Many AS mechanics questions ask you to show that a system moves with SHM. You must derive an equation of the form d²x/dt² + ω²x = 0, or a = –ω²x. Start by resolving forces, write Newton’s second law, and manipulate until you obtain the required form. Always state ‘hence the motion is simple harmonic’.
许多 AS 力学题目要求你证明一个系统做简谐运动。你必须推导出 d²x/dt² + ω²x = 0 或 a = –ω²x 形式。步骤是：分解受力，写出牛顿第二定律，然后整理成所需形式。最后务必注明“所以该运动是简谐运动”。
Once you have ω², you can write down the period and angular frequency directly. In questions involving a particle slightly displaced in a fluid or a coupled spring system, the acceleration expression might look complex, but with algebraic skill you can reduce it to the SHM standard form.
一旦得到 ω²，就可以直接写出周期和角频率。在涉及流体中轻微偏移的质点或耦合弹簧系统的问题中，加速度表达式可能看起来复杂，但只要掌握代数技巧，就能将其化为简谐运动的标准形式。
11. Damping and Forced Oscillations – Brief AS Context | 阻尼与受迫振动 – AS 简要内容
While full damped and forced oscillations belong to A2, AS syllabuses often introduce light damping and the effect on amplitude. In light damping, the period remains almost unchanged but the amplitude decays exponentially. The key distinction is between natural frequency and driving frequency.
虽然完整的阻尼与受迫振动属于 A2 内容，但 AS 课程大纲常会介绍轻微阻尼及其对振幅的影响。轻微阻尼下，周期几乎不变，但振幅按指数衰减。关键区别在于固有频率与驱动频率的不同。
If a question describes the amplitude gradually decreasing over time, you can state that the motion is no longer ideal SHM because energy is lost to resistive forces. However, you can still treat individual oscillations as quasi-SHM if damping is light.
如果题目描述振幅随时间逐渐减小，你可以指出该运动不再是理想简谐运动，因为能量因阻力损失。不过，若阻尼轻微，仍可将每一次振动近似视为类简谐运动处理。
12. Exam Tips and Common Pitfalls | 考试技巧与常见失分点
1. Always write a = –ω²x before doing any calculation. 2. Check the initial conditions to decide sine or cosine form. 3. Convert degrees to radians when using ωt in trigonometric functions. 4. Use v = ±ω √(A² – x²) rather than differentiating every time. 5. For pendulum problems, verify that the angle is small; if not given, state the assumption.
1. 做任何计算之前，先写出 a = –ω²x。2. 根据初始条件判断使用正弦还是余弦形式。3. 三角函数中使用 ωt 时，角度务必用弧度制。4. 多用 v = ±ω √(A² – x²) 而不是每次都求导。5. 在单摆问题中，核实摆角是否很小；若题目未说明，则需声明假设。
Additionally, watch for sign errors in velocity – the direction matters when calculating time to reach a point. A neat sketch of the motion always helps. Finally, remember that the total energy is proportional to A², so doubling the amplitude quadruples the energy.
此外，注意速度的符号——在计算到达某点所需时间时，正负方向很重要。清晰地画出运动示意图总是有帮助的。最后，记住总能量正比于 A²，因此振幅加倍会使能量变为四倍。
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