📚 Simple Harmonic Motion – IGCSE Edexcel Maths | IGCSE Edexcel 数学：简谐运动考点精讲
Simple Harmonic Motion (SHM) is a key topic in Edexcel IGCSE Further Pure Mathematics. It combines trigonometric modelling, differentiation, and algebraic manipulation to describe oscillatory systems. This revision guide walks you through every concept, formula, graph, and exam-style problem you will meet, with clear bilingual explanations to deepen your understanding.
简谐运动（SHM）是爱德思 IGCSE 进阶纯数学的重要课题，融合了三角函数建模、微分法和代数运算来描述振动系统。这篇精讲将带你逐一攻克每个概念、公式、图像和常见考题，通过清晰的中英双语解释助你彻底掌握。
1. What is Simple Harmonic Motion? | 什么是简谐运动？
Simple Harmonic Motion occurs when a particle moves back and forth about a fixed equilibrium point, and its acceleration is directly proportional to its displacement from that point, but acts in the opposite direction. Mathematically, this is expressed as a ∝ −x, which leads to the defining equation a = −ω²x, where ω is a constant called the angular frequency.
简谐运动是指物体围绕一个固定的平衡点来回运动，其加速度与相对于平衡点的位移成正比，但方向始终指向平衡点。数学上表示为 a ∝ −x，进而得到定义方程 a = −ω²x，其中的常数 ω 称为角频率。
a = −ω²x
The negative sign is crucial: it ensures the acceleration always restores the particle towards the centre. In SHM, the motion is periodic and can be modelled using sine or cosine functions.
负号至关重要：它确保加速度始终将物体拉回平衡位置。简谐运动是周期性的，可用正弦或余弦函数建模。
2. The Standard Equation of SHM | 简谐运动的标准方程
The displacement x of a particle moving in SHM is most commonly written as x = a sin(ωt + ε) or x = a cos(ωt + ε). Here a is the amplitude (maximum displacement), ω is the angular frequency, t is time, and ε (epsilon) is the phase angle. The choice between sine and cosine depends on the starting position of the particle.
做简谐运动的物体的位移 x 通常写作 x = a sin(ωt + ε) 或 x = a cos(ωt + ε)。其中 a 是振幅（最大位移），ω 是角频率，t 是时间，ε（epsilon）是初相位。选用正弦还是余弦取决于物体的初始位置。
x = a sin(ωt + ε)
x = a cos(ωt + ε)
If the particle starts at the equilibrium position at t = 0, we use x = a sin(ωt) with ε = 0. If it starts at the maximum displacement a, we use x = a cos(ωt) with ε = 0. When ε ≠ 0, the motion has a phase shift.
若物体在 t = 0 时从平衡位置出发，使用 x = a sin(ωt) 且 ε = 0。若从最大位移 a 处出发，则使用 x = a cos(ωt) 且 ε = 0。当 ε ≠ 0 时，运动具有相位移。
3. Amplitude and Period | 振幅与周期
Amplitude a is the maximum distance from the equilibrium point. The period T is the time taken to complete one full oscillation. In SHM, T is constant and depends only on the angular frequency: T = 2π/ω. The particle repeats its motion exactly after every period.
振幅 a 是物体离平衡点的最大距离。周期 T 是完成一次完整振动所需的时间。在简谐运动中，T 为定值，仅取决于角频率：T = 2π/ω。每经过一个周期，物体完全重复之前的运动。
T = 2π / ω
For example, if ω = 3 rad s⁻¹, then T = 2π/3 ≈ 2.09 seconds. This relationship is frequently tested; always ensure your calculator is in radian mode.
例如，若 ω = 3 rad s⁻¹，则 T = 2π/3 ≈ 2.09 秒。这一关系常考，务必确保计算器处于弧度制。
4. Frequency and Angular Frequency | 频率与角频率
Frequency f is the number of complete oscillations per unit time: f = 1/T. Thus f = ω/(2π) and ω = 2πf. Angular frequency ω has units of radians per second and links linear quantities with circular motion. In SHM, ω is not an angular velocity of a rotating wheel, but it plays the same mathematical role.
频率 f 是单位时间内完整振动的次数：f = 1/T。因此 f = ω/(2π) 且 ω = 2πf。角频率 ω 的单位是弧度每秒，它将线性量与圆周运动联系起来。在简谐运动中，ω 并非旋转的角速度，但数学地位相同。
ω = 2πf
Being comfortable converting between T, f, and ω will save you time in multi‑step problems. Remember that ω is the most frequently used parameter in the equations of SHM.
熟练掌握 T、f 和 ω 之间的转换能在多步问题中节省时间。记住，在各 SHM 方程中，ω 是最常用的参数。
5. Phase Angle and Initial Conditions | 初相位与初始条件
The phase angle ε (sometimes called the initial phase) determines where in its cycle the particle is at t = 0. You can find ε by substituting the known initial displacement and sometimes initial velocity into the displacement equation and solving for ε.
初相位 ε（有时称为初相）决定了 t = 0 时物体处于振动周期的哪个位置。通过将已知的初始位移（有时还有初速度）代入位移方程，即可解出 ε。
At t = 0, x = a sin ε or x = a cos ε
For instance, if a particle starts at x = 2 when a = 4 and we use x = a sin(ωt + ε), then 2 = 4 sin ε ⇒ sin ε = 0.5 ⇒ ε = π/6 or 5π/6, etc. Additional information such as the direction of initial velocity is needed to choose the right option.
例如，若物体振幅 a = 4，在 t = 0 时 x = 2，且采用 x = a sin(ωt + ε)，则 2 = 4 sin ε ⇒ sin ε = 0.5 ⇒ ε = π/6 或 5π/6。此时需要初速度方向等额外信息来选定唯一值。
6. Displacement, Velocity and Acceleration | 位移、速度与加速度
Differentiating the displacement equation gives the velocity v. For x = a sin(ωt + ε), we obtain v = dx/dt = aω cos(ωt + ε). For x = a cos(ωt + ε), v = −aω sin(ωt + ε). The acceleration a is obtained by differentiating velocity: a = dv/dt = −aω² sin(ωt + ε) = −ω²x, confirming the SHM definition.
对位移方程求导可得速度 v。对于 x = a sin(ωt + ε)，得到 v = dx/dt = aω cos(ωt + ε)。对于 x = a cos(ωt + ε)，v = −aω sin(ωt + ε)。再对速度求导得加速度：a = dv/dt = −aω² sin(ωt + ε) = −ω²x，印证了简谐运动的定义。
v = aω cos(ωt + ε) (when using sine for x)
a = −ω²x
An especially useful form linking speed and displacement without time is v² = ω²(a² − x²). This equation is derived by eliminating t using trigonometric identities, and it is a favourite in exams.
一个不显含时间、联系速度与位移的极其有用的公式是 v² = ω²(a² − x²)。该方程通过三角恒等式消去 t 推导而来，是考试中的高频考点。
v² = ω²(a² − x²)
7. Maximum Values | 最大速度与最大加速度
From the velocity equation, the maximum speed occurs when the cosine factor is ±1, i.e. when x = 0 (the equilibrium position). Hence v_max = aω. The maximum magnitude of acceleration occurs when the sine factor is ±1, i.e. at the extreme points x = ±a. Thus a_max = ω²a.
由速度方程可知，最大速率出现在余弦因子为 ±1 时，即当 x = 0（平衡位置），故 v_max = aω。加速度的最大绝对值出现在正弦因子为 ±1 时，即两端点 x = ±a 处，故 a_max = ω²a。
v_max = aω a_max = ω²a
These values are frequently used to determine a or ω when boundary conditions are given. Notice that v_max depends linearly on ω and a, while a_max depends on ω².
在已知边界条件时，这两个最大值常用来求 a 或 ω。注意 v_max 与 ω、a 线性相关，而 a_max 与 ω² 相关。
8. Using the v² = ω²(a² − x²) Formula | v² = ω²(a² – x²) 公式的应用
When time is not given, v² = ω²(a² − x²) is the best tool. For example, if you know the amplitude a, angular frequency ω, and the displacement x at a certain instant, you can directly find the speed v without finding t first.
当未给出时间时，v² = ω²(a² − x²) 是最佳工具。例如，若已知振幅 a、角频率 ω 和某一时刻的位移 x，无需先求出 t，即可直接得到速率 v。
This relationship also shows that the speed is zero at maximum displacement (x = ±a) and is maximum at the centre (x = 0). It reinforces the energy‑conservation idea in physical SHM: kinetic energy + potential energy = constant.
该关系式还表明，在最大位移处 (x = ±a) 速率为零，在平衡位置 (x = 0) 速率最大。这也印证了物理简谐运动中能量守恒的观点：动能 + 势能 = 恒量。
v = ω √(a² − x²) (taking the positive root for speed)
9. Differentiation Approach – Worked Example | 微分法实例
Consider a particle moving with x = 3 sin(2t + π/6), where x is in metres and t in seconds. To find velocity and acceleration, differentiate: v = dx/dt = 3 × 2 cos(2t + π/6) = 6 cos(2t + π/6). Then a = dv/dt = −6 × 2 sin(2t + π/6) = −12 sin(2t + π/6). Notice that a = −2² x = −4x, as expected.
设一物体的位移方程为 x = 3 sin(2t + π/6)，其中 x 以米计，t 以秒计。求速度和加速度：v = dx/dt = 3×2 cos(2t + π/6) = 6 cos(2t + π/6)。然后 a = dv/dt = −6×2 sin(2t + π/6) = −12 sin(2t + π/6)。注意 a = −2² x = −4x，与理论一致。
At t = 0: x = 3 sin(π/6) = 1.5 m, v = 6 cos(π/6) = 6 × (√3/2) = 3√3 ≈ 5.2 m s⁻¹. At the instant when x = 0, solve 3 sin(2t + π/6) = 0 ⇒ 2t + π/6 = nπ; the first positive t gives t = 5π/12 s. Then v = 6 cos(nπ) = ±6 m s⁻¹, so speed is 6 m s⁻¹, matching v_max = aω = 3 × 2 = 6.
在 t = 0 时：x = 3 sin(π/6) = 1.5 m, v = 6 cos(π/6) = 6×(√3/2) = 3√3 ≈ 5.2 m s⁻¹。当 x = 0 时，解 3 sin(2t+π/6)=0 ⇒ 2t+π/6 = nπ；第一个正时间 t = 5π/12 s，此时 v = 6 cos(nπ) = ±6 m s⁻¹，速率为 6 m s⁻¹，与 v_max = aω = 3×2 = 6 吻合。
10. Graphs of SHM Quantities | 简谐运动量的图像
The displacement‑time graph is a sine or cosine wave with amplitude a and period T. The velocity‑time graph is also sinusoidal but leads the displacement graph by π/2 (a quarter of a cycle). The acceleration‑time graph is a cosine wave that is π out of phase with displacement (i.e. acceleration is always opposite to displacement).
位移‑时间图像是振幅为 a、周期为 T 的正弦或余弦波。速度‑时间图像也是正弦型的，但超前位移 π/2（四分之一周期）。加速度‑时间图像则是与位移相差 π 的余弦波，即加速度始终与位移反向。
Sketching and interpreting these graphs is a common exam task. Key features to label are amplitude a, maximum speed aω, maximum acceleration ω²a, and the points where each quantity is zero. Also note that the gradient of the x‑t graph gives the velocity, and the gradient of the v‑t graph gives acceleration.
绘制并理解这些图像是常见考题。需要标注的关键特征包括振幅 a、最大速率 aω、最大加速 ω²a 以及各量取零值的点。还应注意，x‑t 图的斜率即为速度，v‑t 图的斜率即为加速度。
11. Exam Tips and Common Pitfalls | 考试技巧与常见错误
Always work in radians unless the question specifies degrees. Use v² = ω²(a
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