📚 Simple Harmonic Motion: Essential Exam Concepts | 简谐运动考点精讲
Simple Harmonic Motion (SHM) is a fundamental topic in both IB and CIE Mathematics, linking calculus, trigonometry, and mechanics. Understanding its defining equation, graphical representations, and energy transformations is crucial for tackling examination questions. This revision guide covers all essential concepts, common pitfalls, and effective problem-solving strategies.
简谐运动 (SHM) 是 IB 与 CIE 数学中的核心主题，连接微积分、三角学与力学。掌握其定义方程、图像表示和能量转化是攻克相关考题的关键。本文梳理全部核心概念、常见错误与高效解题策略，帮助你在考场上从容应对。
1. Defining Simple Harmonic Motion | 简谐运动的定义
A particle performs simple harmonic motion if its acceleration is directly proportional to its displacement from a fixed equilibrium position and is always directed towards that point. The constant of proportionality involves the angular frequency ω, giving the hallmark equation a = –ω²x.
当物体的加速度与其相对于平衡位置的位移成正比且方向相反（总指向平衡点）时，该物体在作简谐运动。比例常数关联角频率 ω，得标志性方程 a = –ω²x。
a = −ω²x
The negative sign indicates that acceleration opposes the displacement. In exam problems, you must identify this relationship to confirm that a motion is simple harmonic.
负号表示加速度与位移方向相反。考题中，你需要正确识别这一关系以判断运动是否为简谐运动。
2. Differential Equation of SHM | 简谐运动的微分方程
Since acceleration is the second derivative of displacement, the SHM condition can be written as a second-order linear differential equation: d²x/dt² = –ω²x. This is the standard form both IB and CIE expect you to recognise and solve.
因为加速度是位移的二阶导数，简谐运动条件可写为二阶线性微分方程：d²x/dt² = –ω²x。这是 IB 与 CIE 都要求识别并求解的标准形式。
d²x/dt² + ω²x = 0
The solution to this equation involves sine or cosine functions, with two arbitrary constants determined by initial conditions. Being able to set up this equation from a physical description is a key skill.
该方程的解包含正弦或余弦函数，两个任意常数由初始条件决定。能够根据物理情境列出这一方程是一项关键能力。
3. Displacement–Time Functions | 位移–时间函数
The general displacement function for SHM can be expressed as x = A sin(ωt + φ) or x = A cos(ωt + φ), where A is the amplitude, φ is the phase constant, and ω is the angular frequency. Both forms are equivalent, differing only by a phase shift.
简谐运动的一般位移函数可表示为 x = A sin(ωt + φ) 或 x = A cos(ωt + φ)，其中 A 为振幅，φ 为初相位，ω 为角频率。这两种形式等价，仅相差一个相位移。
x = A sin(ωt + φ)
Choosing which form to use depends on the initial conditions given. For instance, if the particle starts at the equilibrium position with positive velocity, the sine form is more convenient.
选用哪种形式取决于给出的初始条件。例如，若物体从平衡位置以正速度开始运动，采用正弦形式更方便。
4. Velocity and Acceleration Equations | 速度与加速度方程
The velocity in SHM can be derived by differentiating displacement: v = dx/dt = Aω cos(ωt + φ). However, a more useful expression linking v to x is v = ± ω√(A² – x²). This shows that speed is maximum at the equilibrium point (x=0) and zero at the extremes (x=±A).
对位移求导可得速度：v = dx/dt = Aω cos(ωt + φ)。但更实用的是联系 v 与 x 的公式 v = ± ω√(A² – x²)。由此可见，速度在平衡位置 (x=0) 最大，在两端 (x=±A) 为零。
v = ± ω√(A² − x²)
Acceleration is given by a = dv/dt = –Aω² sin(ωt + φ), which simplifies to a = –ω²x. This confirms the defining property: magnitude of acceleration grows linearly with distance from centre, directed towards it.
加速度可由 a = dv/dt = –Aω² sin(ωt + φ) 给出，并可简化为 a = –ω²x。这印证了定义特征：加速度大小随距中心距离线性增长，方向指向中心。
5. Period and Frequency | 周期与频率
The period T is the time taken to complete one full oscillation. It is related to angular frequency by T = 2π/ω. Frequency f = 1/T, and the number of oscillations per second equals f hertz. Angular frequency ω is measured in rad/s.
周期 T 是完成一次全振动所需的时间，与角频率的关系为 T = 2π/ω。频率 f = 1/T，每秒振动次数等于 f 赫兹。角频率 ω 的单位为 rad/s。
T = 2π/ω f = 1/T
These relationships are independent of amplitude, which is a key characteristic of SHM. In exam questions, you may need to find ω from a given period or vice versa.
这些关系与振幅无关，这是简谐运动的重要特征。考试中常需根据给定周期求 ω，或由 ω 算周期。
6. Phase Angle and Initial Conditions | 相位角与初始条件
The phase constant φ determines the state of the oscillator at t = 0. By substituting t=0 into displacement and velocity equations, you can solve for A and φ using given initial displacement x₀ and initial velocity v₀.
初相常数 φ 决定了 t=0 时振子的状态。将 t=0 代入位移和速度方程，结合已知的初位移 x₀ 和初速度 v₀，便可解出振幅 A 与初相 φ。
x₀ = A sin φ v₀ = Aω cos φ
From these, A = √(x₀² + (v₀/ω)²) and tan φ = (ω x₀)/v₀. Understanding how to shift between sine and cosine forms using phase differences is also examined regularly.
由此可得 A = √(x₀² + (v₀/ω)²)，tan φ = (ω x₀)/v₀。如何使用相位差在正弦和余弦形式间进行转换也是常见考点。
7. The Spring–Block System | 弹簧–物块系统
For a mass m attached to a light spring of stiffness k, Newton’s second law gives ma = –kx, hence a = –(k/m)x. Comparing with a = –ω²x yields ω = √(k/m) and T = 2π √(m/k). This system is a standard SHM example.
对于连接到劲度系数为 k 的轻弹簧上的质量 m，牛顿第二定律给出 ma = –kx，故 a = –(k/m)x。与 a = –ω²x 对比得 ω = √(k/m)，T = 2π √(m/k)。这是标准的简谐运动实例。
T = 2π √(m/k)
Note that the period depends only on mass and spring stiffness, not on amplitude. When a spring is horizontal or vertical, the equilibrium position shifts, but the SHM equations remain valid as long as the restoring force obeys Hooke’s law.
注意，周期只取决于质量和弹簧劲度系数，与振幅无关。无论弹簧水平放置还是竖直悬挂，只要恢复力遵循胡克定律，平衡位置虽有移动，但简谐运动方程仍然成立。
8. The Simple Pendulum | 单摆
A simple pendulum consists of a point mass m suspended by a light string of length L. For small angular displacements (θ < 10°), the motion approximates SHM with ω = √(g/L) and T = 2π √(L/g).
单摆由轻绳悬挂的质点 m 构成。角位移较小（θ < 10°）时，运动近似为简谐运动，角频率 ω = √(g/L)，周期 T = 2π √(L/g)。
T = 2π √(L/g)
Here the period is independent of mass and depends only on length and gravitational field strength. You may be asked to derive these relations from the tangential component of weight or from energy considerations.
此处周期与质量无关，仅取决于摆长和重力场强度。考题可能要求你从重力的切向分量或能量角度推导这些关系。
9. Energy in Simple Harmonic Motion | 简谐运动的能量
In SHM, the total mechanical energy is conserved and equals the sum of kinetic energy (K = ½mv²) and potential energy. For a spring–mass system, potential energy is U = ½kx². The total energy can be expressed as E = ½kA² = ½mω²A².
简谐运动中总机械能守恒，为动能 (K = ½mv²) 与势能之和。对于弹簧–质量系统，势能为 U = ½kx²。总能量可表示为 E = ½kA² = ½mω²A²。
E = ½kA² = ½mω²A²
At equilibrium, potential energy is zero and kinetic energy is maximum. At extremes, kinetic energy is zero and potential energy is maximum. Graphs of K and U against displacement are parabolic.
在平衡位置，势能为零，动能最大；在两端，动能为零，势能最大。动能和势能随位移变化的图为抛物线。
10. Solving the SHM Differential Equation | 微分方程求解 (IB HL)
For IB Higher Level and some CIE further problems, you need to solve d²x/dt² + ω²x = 0 analytically. The auxiliary equation is m² + ω² = 0, giving complex roots ±iω, so the general solution is x = A cos ωt + B sin ωt.
在 IB 高水平及部分 CIE 进阶题中，需要解析求解 d²x/dt² + ω²x = 0。辅助方程为 m² + ω² = 0，得共轭复根 ±iω，因此通解为 x = A cos ωt + B sin ωt。
x = A cos ωt + B sin ωt
Using trigonometric identities, this can be rewritten as x = R sin(ωt + φ) or R cos(ωt − α). Applying initial conditions allows you to determine the constants A and B (or R and φ).
利用三角恒等式，上式可改写为 x = R sin(ωt + φ) 或 R cos(ωt − α)。代入初始条件即可确定常数 A 和 B（或 R 和 φ）。
11. Graphical Representations | 图像分析
Exam questions frequently ask you to sketch or interpret x–t, v–t, and a–t graphs. All are sinusoidal with the same period. Velocity leads displacement by 90° (π/2 rad), and acceleration leads velocity by another 90°, being perfectly out of phase with displacement.
考题常要求绘制或解读 x–t, v–t 和 a–t 图像。三者均为正弦曲线，周期相同；速度超前位移 90° (π/2 rad)，加速度再超前速度 90°，与位移恰好反相。
Plotting energy–time or energy–displacement graphs is also common. The total energy appears as a horizontal line, while kinetic and potential energies exchange in complementary parabolas.
能量–时间或能量–位移图的绘制也属常见题型。总能量表现为水平直线，而动能与势能则呈互补的抛物线交替变化。
12. Common Pitfalls and Exam Tips | 常见错误与应考提示
Common mistakes include forgetting the negative sign in a = –ω²x, mixing up v = ±ω√(A²–x²) with v = ωx, and using degrees instead of radians in trigonometric differentiation. Always convert angles to radians when using calculus.
常见错误包括遗漏 a = –ω²x 中的负号、混淆 v = ±ω√(A²–x²) 与 v = ωx，以及在求导时将角度单位用度而不是弧度。涉及微积分时必须使用弧度制。
Also, be careful when identifying the equilibrium position from a force balance, especially in vertical spring or pendulum problems. Read the wording carefully to distinguish between ‘speed’ and ‘velocity’, and between ‘displacement’ and ‘distance travelled’.
另外，对竖直弹簧或单摆等情形，要通过受力平衡正确找出平衡位置。仔细审题，区分‘速率’与‘速度’，以及‘位移’与‘路程’。
Finally, in SHM differential equation problems, always express the second derivative term with the coefficient 1 before ω²x. Show clear steps when equating coefficients to extract ω.
最后，在微分方程类题目中，务必使二阶导数项系数为 1，再对比 ω²x。写出清晰的系数对比过程，以正确提取角频率 ω。
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