📚 IB & AQA Maths: Simple Harmonic Motion Essentials | IB AQA 数学：简谐运动 考点精讲
Simple harmonic motion (SHM) is a fundamental type of oscillation found in many physical systems, and it appears in both IB Mathematics (Analysis & Approaches / Applications & Interpretation) and AQA A-level Mathematics (Mechanics). Understanding its mathematical description, differential equation, and energy aspects is essential for exam success. This article distils the key points you need to master, with examples and common pitfalls.
简谐运动（SHM）是许多物理系统中存在的一种基本振动形式，在 IB 数学（分析与方法/应用与解释）和 AQA A-level 数学（力学）中均有出现。理解其数学描述、微分方程和能量特点是考试成功的关键。本文精炼了需要掌握的核心考点，辅以示例和常见误区。
1. Definition and Conditions for SHM | 简谐运动的定义与条件
SHM occurs when the restoring force (or acceleration) on an object is directly proportional to its displacement from a fixed equilibrium point and always directed towards that point. The motion is periodic and the magnitude of the acceleration increases linearly with displacement.
当物体所受回复力（或加速度）与其对固定平衡点的位移成正比，并且始终指向该点时，就发生简谐运动。该运动是周期性的，加速度的大小随位移线性增大。
Mathematically, the defining condition is a ∝ −x, which can be written precisely as:
a = −ω²x
where a is acceleration, x is displacement from equilibrium, and ω (omega) is the angular frequency, a positive constant that determines how rapidly the system oscillates.
数学上，定义条件为 a ∝ −x，可精确写为：
a = −ω²x
其中 a 为加速度，x 为对平衡位置的位移，ω（角频率）是一个正常数，决定了系统振动的快慢。
The negative sign ensures that the acceleration is always opposite in direction to the displacement. The constant ω² is specific to the system and independent of the amplitude of motion. This linear restoring force condition is key to identifying SHM in exam questions.
负号确保加速度方向始终与位移相反。常数 ω² 取决于系统本身，与振幅无关。在考题中，识别出这种线性回复力条件是关键。
2. Differential Equation of SHM | 简谐运动的微分方程
Since acceleration is the second derivative of displacement with respect to time, the defining equation becomes a second-order linear differential equation:
d²x/dt² = −ω²x
Both IB and AQA specifications expect you to recognise this form and to be able to show that sinusoidal functions satisfy it.
因为加速度是位移对时间的二阶导数，定义式变为一个二阶线性微分方程：
d²x/dt² = −ω²x
IB 和 AQA 考纲都要求你识别这种形式，并能证明正弦型函数满足该方程。
The general solution is x = A sin(ωt) + B cos(ωt) or, equivalently, x = C cos(ωt + φ) or x = C sin(ωt + φ). The constants A, B or C, φ are determined by the initial displacement and velocity. In many textbook problems, timing starts when the particle passes through equilibrium or is at maximum displacement, simplifying the expressions to x = A sin(ωt) or x = A cos(ωt).
通解为 x = A sin(ωt) + B cos(ωt)，或等价地 x = C cos(ωt + φ) 或 x = C sin(ωt + φ)。常数 A、B 或 C、φ 由初始位移和速度决定。在许多教材问题中，计时从物体经过平衡位置或处于最大位移时开始，从而使表达式简化为 x = A sin(ωt) 或 x = A cos(ωt)。
In IB Analysis & Approaches, you may be asked to verify by differentiation that a given function satisfies the DE. In AQA Mechanics, you will more frequently use the standard formulas to compute displacement, velocity and acceleration at given times. Be prepared to substitute t=0, t=T/4 etc. to find constants.
在 IB 分析与方法中，可能要求通过求导验证给定函数满足微分方程。在 AQA 力学中，更常直接使用标准公式计算给定时点的位移、速度和加速度。要准备好代入 t=0, t=T/4 等求常数。
3. Displacement, Velocity and Acceleration Equations | 位移、速度和加速度方程
Choosing x = A cos(ωt) (starting at maximum positive displacement at t=0), differentiation yields the velocity and acceleration functions:
v = dx/dt = −Aω sin(ωt)
a = dv/dt = −Aω² cos(ωt) = −ω²x
选择 x = A cos(ωt)（在 t=0 时位于最大正位移），求导即可得到速度和加速度函数：
v = dx/dt = −Aω sin(ωt)
a = dv/dt = −Aω² cos(ωt) = −ω²x
The maximum speed (when x=0) is vmax = Aω. The maximum acceleration (when x = ±A) is amax = Aω². These amplitude values are frequently tested, as is the fact that v=0 at the extreme points.
最大速率（当 x=0 时）为 vmax = Aω。最大加速度（当 x = ±A 时）为 amax = Aω²。这些振幅值经常被考察，同时端点处 v=0 也是常考点。
An extremely useful relationship eliminates time and links speed directly to displacement:
v² = ω²(A² − x²)
This follows from the trigonometric identity sin²θ + cos²θ = 1 and is often used to find the speed at a specific displacement or to determine the amplitude from given speed and position data.
一个非常实用的关系式消去了时间，直接将速率与位移联系起来：
v² = ω²(A² − x²)
它由 sin²θ + cos²θ = 1 推导而来，常用于求给定位移时的速率，或根据已知速率和位置数据求振幅。
Always be careful to take the correct sign for velocity; the expression v = ± ω√(A² − x²) gives the magnitude, and the direction depends on whether the object is moving towards or away from the equilibrium position. A common exam mistake is to forget the direction sign.
注意速度的符号；表达式 v = ± ω√(A² − x²) 给出的是大小，方向取决于物体是朝平衡位置还是远离平衡位置运动。考生常犯的一个错误是忘记方向符号。
4. Graphical Representations | 图像表示
Displacement-time, velocity-time and acceleration-time graphs are all sinusoidal for SHM, but with important phase differences. If x = A cos(ωt), then v = −Aω sin(ωt) and a = −Aω² cos(ωt). This means:
简谐运动的位移-时间、速度-时间和加速度-时间图像均为正弦型，但存在重要的相位差。若 x = A cos(ωt)，则 v = −Aω sin(ωt)，a = −Aω² cos(ωt)。这意味着：
The velocity graph leads the displacement graph by π/2 radians (or a quarter of a period). The acceleration graph leads the velocity graph by another π/2, and hence is in antiphase (π out of phase) with the displacement graph. On a sketch, when x is at a maximum, a is at its most negative maximum.
速度图相位领先位移图 π/2 弧度（即四分之一周期）。加速度图又领先速度图 π/2，因此与位移图反相（相差 π）。在草图中，当 x 处于最大值时，a 处于负最大值。
You must be able to read amplitudes, period, and phase relationships from such graphs. Mark the equilibrium positions clearly, and show that the velocity is zero where the displacement gradient is zero, and maximum where the gradient is steepest. For IB exams, you might need to sketch velocity and acceleration given a displacement graph.
你必须能从这类图像中读出振幅、周期和相位关系。要明确标出平衡位置，并展示出速度在位移梯度为零处为零，在梯度最陡处最大。在 IB 考试中，可能会要求根据位移图画出速度和加速度图。
5. Time Period and Frequency | 周期与频率
Angular frequency ω links the period T and frequency f of the oscillation:
ω = 2π
Published by TutorHao | IB Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导