📚 Simple Harmonic Motion | 简谐运动考点精讲
Simple Harmonic Motion (SHM) is a fundamental topic in the IGCSE OCR Further Mathematics syllabus. Understanding its differential equation and the resulting sinusoidal functions is essential for exam success. This revision guide covers all key concepts, formulas, and common exam pitfalls.
简谐运动（SHM）是IGCSE OCR附加数学课程中的重要主题。掌握其微分方程以及由此导出的正弦函数是考试成功的关键。本考点精讲涵盖所有核心概念、公式和常见考试错误。
1. What is Simple Harmonic Motion? | 什么是简谐运动？
ed point and is always directed towards that point. Mathematically, this is expressed as a ∝ –x, or a = –k x where k is a positive constant. The motion is oscillatory and sinusoidal in nature.
如果一个质点的加速度与其相对于某一固定点的位移成正比，且方向始终指向该固定点，则该质点在做简谐运动。数学上可表示为 a ∝ –x，或 a = –k x，其中 k 为正的常数。这种运动是周期性的，且位移随时间呈正弦变化。
2. The Differential Equation of SHM | 简谐运动的微分方程
e equation of motion for SHM is written as a second-order differential equation: d²x/dt² = – displacement. This defining equation links the acceleration directly to the displacement.
简谐运动的运动方程是一个二阶微分方程：d²x/dt² = –ω²x，其中 ω 为角频率。负号表明加速度与位移方向相反。这个定义方程直接将加速度与位移联系起来。
e between sine and cosine depends on initial conditions (e.g., at t=0, x=0 gives sine, x=A gives cosine).
该微分方程的通解为 x = A cos(ωt + φ) 或 x = A sin(ωt + φ)，其中 A 为振幅，φ 为初相。选择正弦还是余弦取决于初始条件（例如，t=0 时 x=0 选用正弦，x=A 选用余弦）。
3. Displacement, Amplitude and Phase | 位移、振幅和相位
The displacement x varies between –A and +A. The amplitude A is the maximum displacement from the equilibrium position. The term (ωt + φ) is the phase, and φ is the initial phase angle, determining the starting point of the oscillation.
位移 x 在 –A 到 +A 之间变化。振幅 A 是离开平衡位置的最大位移。(ωt + φ) 为相位，φ 是初相位，决定了振动的起始位置。
in the positive direction. With x = A cos(ωt), it starts at maximum positive displacement.
例如，采用 x = A sin(ωt)，在 t = 0 时质点位于平衡位置并向正方向运动；而采用 x = A cos(ωt)，质点从正的最大位移处开始运动。
4. Velocity and Acceleration Equations | 速度与加速度方程
, we obtain velocity: v = dx/dt. For x = A sin(ωt), v = Aω cos(ωt). For x = A cos(ωt), v = –Aω sin(ωt). The maximum speed is vmax = Aω, occurring at the equilibrium position (x=0).
通过微分位移函数可得到速度：v = dx/dt。对于 x = A sin(ωt)，v = Aω cos(ωt)；对于 x = A cos(ωt)，v = –Aω sin(ωt)。最大速率为 vmax = Aω，出现在平衡位置（x=0）。
A very useful formula linking velocity and displacement without time is: v = ± ω √A² – x². The acceleration is given by a = dv/dt = –ω²x. The maximum acceleration is amax = ω²A, at the extreme positions (x = ±A).
一个极为有用的不显含时间的速度和位移关系式是：v = ± ω √A² – x²。加速度为 a = dv/dt = –ω²x。最大加速度 amax = ω²A，出现在两端极限位置 (x = ±A)。
Below is a summary table of the key SHM formulas:
以下是简谐运动关键公式汇总表：
| Quantity | Formula |
|---|---|
| Displacement | x = A cos(ωt + φ) or A sin(ωt + φ) |
| Velocity | v = ± ω √A² – x² |
| Acceleration | a = –ω²x |
| Maximum speed | vmax = Aω |
| Maximum acceleration | amax = ω²A |
5. Period and Frequency | 周期与频率
T = 2π/ω, and the frequency is f = 1/T = ω/2π. These quantities depend only on the system's physical properties, not on the amplitude.
角频率 ω 与周期 T 和频率 f 的关系为 ω = 2π/T = 2πf。因此，简谐运动的周期为 T = 2π/ω，频率为 f = 1/T = ω/2π。这些量仅取决于系统的物理属性，与振幅无关。
6. The Spring–Mass System | 弹簧振子系统
attached to a spring of force constant k undergoes SHM (assuming no damping). From Newton's second law and Hooke's law, we get d²x/dt² = –(k/m)x, so ω² = k/m. Hence the period is T = 2π √m/k.
一个质量为 m 的物体连接在力常数为 k 的弹簧上，可做简谐运动（假设无阻尼）。由牛顿第二定律和胡克定律可得 d²x/dt² = –(k/m)x，因此 ω² = k/m。周期为 T = 2π √m/k。
In exam questions, you may be asked to find k from measured T and m, or to determine how T changes if the mass is doubled (T scales by √2).
在考试中，你可能会被要求根据测得的 T 和 m 求 k，或者判断当质量加倍时周期如何变化（T 按 √2 倍增）。
7. The Simple Pendulum | 单摆
small amplitude (θ < about 10°), the motion approximates SHM. The angular frequency is ω = √g/l, so the period is T = 2π √l/g. Note that T is independent of the mass and amplitude (for small swings).
对于一个摆长为 l、在小振幅（θ < 约 10°）下摆动的单摆，其运动可近似为简谐运动。角频率为 ω = √g/l，因此周期 T = 2π √l/g。注意 T 与质量和（小角度时的）振幅无关。
Make sure you can derive T from the restoring force expression: –(mg/l)x leads to ω² = g/l.
你需要能够从恢复力表达式 –(mg/l)x 推导出 ω² = g/l 并得到周期。
8. Energy in Simple Harmonic Motion | 简谐运动中的能量
nsists of kinetic and potential components. The kinetic energy is KE = ½ m v² = ½ m ω² (A² – x²). The potential energy is PE = ½ m ω² x². The total energy is E = ½ m ω² A².
简谐运动中的机械能总量守恒，由动能和势能组成。动能为 KE = ½ m v² = ½ m ω² (A² – x²)。势能为 PE = ½ m ω² x²。总能量为 E = ½ m ω² A²。
he extremes (x=±A), KE=0 and PE is maximum. Energy remains constant throughout the motion.
在平衡位置（x=0），动能最大而势能为零；在两端（x=±A），动能为零而势能最大。能量在整个运动过程中保持不变。
9. Graphical Representations | 图形表示
as x = A sin(ωt), then velocity v = Aω cos(ωt) leads x by π/2 (90°), and acceleration a = –ω²A sin(ωt) is π (180°) out of phase with displacement.
位移、速度和加速度随时间的变化图均为正弦曲线。若位移为 x = A sin(ωt)，则速度 v = Aω cos(ωt) 超前位移 π/2（90°），而加速度 a = –ω²A sin(ωt) 与位移反相，相位差 π（180°）。
Knowing how to sketch these graphs and label key values (amplitude, period, intercepts) is a common exam requirement. Make sure to indicate that acceleration is always opposite to displacement.
能够绘制这些图像并标定关键值（振幅、周期、截距）是常见的考试要求。
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