📚 A-Level AQA Mathematics: Simple Harmonic Motion Key Points | A-Level AQA 数学：简谐运动 考点精讲
Simple Harmonic Motion (SHM) forms a core part of the AQA A-Level Mathematics mechanics syllabus. Students are expected to model systems that oscillate about an equilibrium position with an acceleration proportional to, and directed towards, that equilibrium. This article walks through every major concept, equation, and practical application, helping you build a solid understanding for the exam. From the defining differential equation to energy considerations and damping, we cover the full range of topics in clear, bilingual form.
简谐运动（SHM）是 AQA A-Level 数学力学模块的重点内容。学生需要掌握围绕平衡位置振动、且加速度始终指向平衡位置并与位移成正比的运动模型。本文逐一梳理每一个核心概念、公式和实际应用，帮助你在考试中扎实拿分。从定义性的微分方程到能量分析、再到阻尼振动，我们以中英双语清晰讲解全部考点。
1. Definition and Basic Equation | 定义与基本方程
In SHM, the resultant force and acceleration are always directed towards a fixed equilibrium point. Mathematically, the condition for SHM is that the acceleration a is proportional to the displacement x from equilibrium, and opposite in sign:
在简谐运动中，合外力和加速度始终指向一个固定的平衡点。数学上，SHM 的条件是加速度 a 与离开平衡位置的位移 x 成正比，且方向相反：
a ∝ −x or a = −ω² x
Here, ω is the angular frequency, a constant that depends on the physical properties of the system (e.g. spring constant and mass). The negative sign shows that acceleration acts in the opposite direction to displacement, always pulling the particle back towards equilibrium.
其中 ω 是角频率，一个由系统物理特性（如弹簧劲度系数和质量）决定的常数。负号表明加速度的方向总是与位移相反，即始终将质点拉回平衡位置。
The motion can be described by the second-order differential equation:
该运动可用如下二阶微分方程描述：
d²x/dt² = −ω² x
This equation emerges from combining Newton' with the force law F = −kx (for a spring) or the tangential component in a pendulum.
该方程源自牛顿第二定律与力的规律（如弹簧的 F = −kx 或单摆的切向分量）的结合。
2. Displacement Equation | 位移方程
The general solution for the displacement x as a function of time t is sinusoidal. Two common forms are:
位移 x 随时间 t 变化的一般解为正弦形式。常用的两种表达式为：
x = A cos(ωt) or x = A sin(ωt)
he particle starts at equilibrium (x=0) when t=0. A more general phase-shifted version is:
采用余弦形式时，质点从最大位移 A 处（t=0 时）开始运动；采用正弦形式时，质点从平衡位置（t=0 时 x=0）出发。更一般的含初相位的形式为：
x = A sin(ωt + φ) or x = A cos(ωt + φ)
Here A is the amplitude (maximum displacement), ω the angular frequency, and φ the phase constant determined by initial conditions. The quantity (ωt + φ) is called the phase.
其中 A 为振幅（最大位移），ω 为角频率，φ 为由初始条件决定的初相。(ωt + φ) 称为相位。
3. Velocity in SHM | 简谐运动中的速度
is obtained by differentiating displacement with respect to time. For x = A cos(ωt):
速度 v 可由位移对时间求导得到。以 x = A cos(ωt) 为例：
v = dx/dt = −Aω sin(ωt)
A very useful velocity–displacement relationship, independent of time, is:
一个不依赖时间的、极其有用的速度—位移关系式为：
v = ± ω √(A² − x²)
quilibrium position (x=0) and zero at the extremes (x = ±A). The maximum speed is vmax = ωA.
该公式表明：质点经过平衡位置（x=0）时速率最大，在端点（x = ±A）时速度为零。最大速率 vmax = ωA。
Students must be able to use the velocity equation to find speed at any given displacement, or to determine the displacement when the speed is given. The ± sign accounts for the direction of motion.
学生需掌握运用速度方程求任意位移下的速率，或已知速率时求位移。正负号表示运动方向。
4. Acceleration in SHM | 简谐运动中的加速度
Acceleration a is the second derivative of displacement, or the derivative of velocity:
加速度 a 是位移的二阶导数，或速度的一阶导数：
dv/dt = d²x/dt² = −ω² A cos(ωt) = −ω² x
Thus, the maximum magnitude of acceleration is amax aximum displacement where the restoring force is strongest. At equilibrium, acceleration is zero.
因此，最大加速度的大小为 amax = ω²A，出现在位移最大处，此时回复力最强。在平衡位置加速度为零。
tions, if you are told that a = −4x, then ω² = 4 and ω = 2 rad·s⁻¹.
关系式 a = −ω²x 是简谐运动的标志性定义。在考题中，若给出 a = −4x，则 ω² = 4，ω = 2 rad·s⁻¹。
5. Period and Frequency | 周期与频率
The period T is the time for one complete oscillation. In SHM, angular frequency ω, period T, and ordinary frequency f are related by:
周期 T 是完成一次全振动所需的时间。在简谐运动中，角频率 ω、周期 T 和普通频率 f 的关系为：
T = 2π/ω and f = 1/T = ω/(2π)
Because the motion repeats every 2π radians of phase, these relations are independent of amplitude — an important characteristic of simple harmonic oscillators.
由于运动每经过 2π 弧度的相位就重复一次，这些关系与振幅无关 —— 这是简谐振子的一个重要特征。
). Notice that the period does not depend on amplitude (for small angles in the pendulum).
对于弹簧振子，T = 2π√(m/k)；对于单摆，T = 2π√(l/g)。注意周期与振幅无关（单摆在摆角很小时成立）。
6. Mass-Spring System | 弹簧振子系统
A mass m attached to a light spring of stiffness k, oscillating horizontally on a smooth surface or vertically, executes SHM provided the spring obeys Hooke's law.
质量为 m 的物体连接在劲度系数为 k 的轻弹簧上，在光滑水平面上或竖直方向上振动，只要弹簧服从胡克定律，该物体便做简谐运动。
The restoring force is F = −kx. Using Newton' gives −kx = ma, so a = −(k/m)x. Hence ω² = k/m, and the period is:
回复力为 F = −kx。应用牛顿第二定律得 −kx = ma，因此 a = −(k/m)x。于是 ω² = k/m，周期为：
T = 2π √(m/k)
The motion is still SHM because the additional constant weight simply shifts the equilibrium position without affecting ω.
对于竖直弹簧振子，平衡伸长量为 e = mg/k。运动仍为简谐运动，因为恒定的重力只使平衡位置发生平移，不影响 ω。
Typical exam questions ask for T given m and k, or to find k from a measured period. Remember to use SI units: mass in kg, k in N·m⁻¹.
常见考题为已知 m 和 k 求 T，或由测量周期求 k。注意使用国际单位制：质量用 kg，k 用 N·m⁻¹。
7. Simple Pendulum | 单摆
A simple pendulum consists of a point mass (bob) suspended from a light, inextensible string of length l. For small angular displacements (θ < approx 10°), the motion approximates SHM.
单摆由悬挂于长为 l 的轻质不可伸长的细线上的质点（摆球）构成。在角位移很小（θ < 约 10°）时，运动近似为简谐运动。
isplacement s = lθ, we get a = −(g/l)s, giving ω² = g/l and period:
回复力由重力的一个分量 mg sin θ 提供。利用小角近似 sin θ ≈ θ，切向加速度为 a = −gθ。又因为弧位移 s = lθ，可得 a = −(g/l)s，故 ω² = g/l，周期为：
T = 2π √(l/g)
Note that the period depends only on the length of the pendulum and the local gravitational field strength, not on the mass of the bob or the amplitude (assuming small angles). This makes the pendulum a useful tool for measuring g.
注意，周期仅取决于摆长和当地的重力加速度，与摆球质量及振幅（小角度假设下）无关。这使得单摆成为测量 g 的有效工具。
For larger amplitudes, the motion is still periodic but no longer simple harmonic; the period increases slightly with amplitude.
对于较大振幅，运动虽然仍有周期性，但不再属于简谐运动；周期随振幅略有增加。
8. Energy in SHM | 简谐运动中的能量
In an undamped simple harmonic oscillator, the total mechanical energy remains constant, continuously transforming between kinetic and potential forms.
在无阻尼简谐振子中，总机械能保持不变，动能与势能之间不断相互转化。
Kinetic energy (KE) at displacement x:
在位移 x 处的动能：
KE = ½; m v² = ½; m ω² (A²; − x²)
Potential energy (PE) stored in the oscillator (e.g. elastic potential for a spring, measured from equilibrium) is:
振子中储存的势能（例如弹簧的弹性势能，从平衡位置量起）为：
PE = ½; k x² = ½; m ω² x²;
Thus the total energy E is:
因此总能量 E 为：
E = KE + PE = ½; m ω² A²;
is zero and KE is maximum; at the extremes (x = ±A), KE is zero and PE is maximum.
总能量与振幅的平方成正比。在平衡位置（x=0），势能为零，动能最大；在端点（x = ±A），动能为零，势能最大。
Students should be able to sketch energy–displacement or energy–time graphs, and use energy conservation to find speed at a given position without dealing with time.
学生应能画出能量—位移或能量—时间关系图，并运用能量守恒在不涉及时间的情况下求指定位置的速度。
9. Graphical Representations | 图像表示
The exam routinely requires interpretation and sketching of displacement–time, velocity–time, and acceleration–time graphs for SHM.
考试时常要求解释和绘制简谐运动的位移—时间、速度—时间、加速度—时间图像。
- Displacement vs time: a sine or cosine wave; amplitude A, period T.
- 速度 vs 时间: a sine or cosine wave shifted by π/2 relative to displacement; amplitude vmax = ωA.
- 加速度 vs 时间: inverted version of the displacement graph; amplitude amax = ω²A, phase difference of π from displacement.
英文要点：
- Displacement: sinusoidal with amplitude A.
- Velocity: sinusoidal, leading displacement by π/2 (or lagging by 3π/2), with maximum magnitude ωA.
- Acceleration: sinusoidal, opposite in sign to displacement (phase difference of π), with maximum magnitude ω²A.
中文对照：
- 位移：正弦波，振幅 A。
- 速度：正弦波，比位移超前 π/2（或滞后 3π/2），最大值为 ωA。
- 加速度：正弦波，与位移反相（相位差 π），最大值为 ω²A。
n error is mixing up sine and cosine forms; always identify the starting conditions (t=0).
准确掌握相位关系至关重要。常见错误是混淆正弦和余弦形式；务必先确定初始条件（t=0 时刻）。
10. Damping in SHM | 简谐运动中的阻尼
In real systems, dissipative forces such as friction or air resistance cause the amplitude of oscillation to decrease over time. This is called damping. AQA distinguishes three cases:
在真实系统中，摩擦力或空气阻力等耗散力导致振幅随时间减小，这称为阻尼。AQA 大纲区分三种情况：
Light damping: the system oscillates with gradually decreasing amplitude; oscillations continue for many cycles. The frequency is slightly lower than the natural frequency.
轻阻尼：系统持续振动，振幅逐渐减小；振动可维持多个周期，频率比固有频率略低。
Critical damping: the system returns to equilibrium in the shortest possible time without oscillating. This is desirable in car suspension and door closers.
临界阻尼：系统在最短时间内回到平衡位置而不发生振动。汽车悬架和闭门器的设计希望达到此状态。
Heavy damping: the system returns to equilibrium without oscillation, but more slowly than in the critical case.
过阻尼：系统不发生振动地返回平衡位置，但比临界阻尼更慢。
Damping affects energy: the total energy decays exponentially for light damping. In exam questions, you may be asked to sketch displacement–time graphs for each case.
阻尼影响能量：轻阻尼时，总能量呈指数衰减。考题可能要求画出每种情形下的位移—时间草图。
11. Forced Vibrations and Resonance | 受迫振动与共振
When an external periodic force drives an oscillator at a frequency fdriving, the system vibrates at that driving frequency. The amplitude of the forced oscillations depends on how close fdriving is to the natural frequency f0.
当外部周期性驱动力以频率 fdriving 作用于振子时，系统按该驱动频率振动。受迫振动的振幅取决于 fdriving 与固有频率 f0 的接近程度。
Resonance occurs when the driving frequency equals the natural frequency, resulting in a maximum transfer of energy and peak amplitude. The resonance curve (amplitude vs driving frequency) shows a sharp peak for lightly damped systems and a broader peak for heavily damped ones.
当驱动频率等于固有频率时发生共振，此时能量转移最大，振幅达到峰值。共振曲线（振幅—驱动频率）在轻阻尼时呈现尖锐峰值，重阻尼时峰值较宽缓。
Key points for AQA:
- The phase difference between driver and oscillator is π/2 at resonance.
- Increased damping reduces the maximum amplitude and shifts the resonant frequency slightly lower.
- Examples include a child on a swing, the Tacoma Narrows Bridge collapse, and microwave heating of water molecules.
AQA 重点：
- 共振时驱动源与振子间的相位差为 π/2。
- 增大阻尼会降低最大振幅并使共振频率略向低频偏移。
- 实例包括荡秋千、塔科马海峡大桥坍塌、微波加热水分子等。
12. Summary of Key Formulas and Exam Tips | 核心公式总结与应试技巧
Memorising these essential relationships will give you quick access to most marks in AQA SHM questions:
熟记以下关键关系式，能帮你迅速拿下 AQA 简谐运动考题的大部分分数：
| Equation / 公式 | Notes / 备注 |
| a = −ω² x | Defining equation of SHM / SHM 定义式 |
| x = A cos(ωt + φ) | Displacement solution / 位移通解 |
| v = ± ω √(A² − x²) | Velocity at a given x / 给定 x 的速度 |
| vmax = ωA | Maximum speed at equilibrium / 平衡处最大速率 |
| amax = ω²A | Maximum acceleration at extremes / 端点最大加速度 |
| T = 2π/ω = 2π√(m/k) = 2π√(l/g) | Period; choose form for spring or pendulum / 周期；弹簧或单摆选用不同形式 |
| E = ½; m ω² A²; | Total mechanical energy / 总机械能 |
(A² − x²), remember to assign the correct sign based on the direction of motion. Show all steps clearly, particularly when differentiating trig functions to switch between displacement, velocity, and acceleration.
解题时要先从已知信息中识别 ω。若给出周期，则用 ω = 2π/T。使用 v = ω√(A²; − x²) 时，须根据运动方向确定正负号。清晰展示所有步骤，尤其是在对三角函数求导以切换位移、速度和加速度时。
Finally, practise linking graphical features to equations: the gradient of a displacement–time graph gives velocity; the gradient of velocity–time gives acceleration. This visual understanding often saves time in multiple-choice sections.
最后，多加练习图像特征与方程的关联：位移—时间图的斜率表示速度；速度—时间图的斜率表示加速度。这种直观理解常能在选择题部分节省时间。
Published by TutorHao | AQA Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图