Simple Harmonic Motion: IB & OCR Revision Guide | 简谐运动考点精讲

📚 Simple Harmonic Motion: IB & OCR Revision Guide | 简谐运动考点精讲

Simple harmonic motion (SHM) is a cornerstone of both IB and OCR A-Level physics, describing the oscillatory behaviour of systems where the restoring force is proportional to displacement from equilibrium. Understanding SHM not only unlocks problems on springs and pendulums but also lays the groundwork for wave theory and alternating current. This guide breaks down the essential definitions, mathematical models, energy transformations, and exam-style reasoning you need to excel.

简谐运动（SHM）是 IB 和 OCR A-Level 物理的核心内容，描述的是恢复力与偏离平衡位置的位移成正比的系统的振荡行为。理解简谐运动不仅能解决弹簧振子和单摆问题，还为波动理论和交流电的学习奠定基础。本指南将系统梳理关键定义、数学模型、能量转化以及应试推理方法，助你稳稳拿下高分。

1. Defining Simple Harmonic Motion | 简谐运动的定义

Simple harmonic motion is defined as an oscillation in which the acceleration of the body is directly proportional to its displacement from a fixed equilibrium position and is always directed towards that equilibrium. Mathematically, the condition can be written as a ∝ −x, or more commonly a = −ω²x, where ω is the angular frequency of the motion.

简谐运动定义为一种振荡：物体的加速度与其相对于固定平衡位置的位移成正比，且加速度方向始终指向该平衡位置。数学上，这一条件可写作 a ∝ −x，更常见的形式是 a = −ω²x，其中 ω 是运动的角频率。

The negative sign is crucial; it indicates that the acceleration and displacement are in opposite directions. Whenever a system satisfies this equation for small displacements, it undergoes SHM, regardless of whether it is a mass on a spring, a simple pendulum, or a floating object bobbing in water.

负号至关重要，它表明加速度与位移方向相反。只要一个系统在小位移下满足该方程，它就会做简谐运动，无论是一个弹簧上的物块、一个单摆，还是漂浮在水面上上下振动的物体。

In IB and OCR exams, you must be able to identify SHM from a graph or an equation. If a graph of acceleration against displacement yields a straight line passing through the origin with a negative gradient, the motion is simple harmonic.

在 IB 和 OCR 考试中，你必须能从图像或方程中识别 SHM。如果加速度-位移图是一条通过原点且斜率为负的直线，那么该运动就是简谐运动。

2. The SHM Differential Equation | 简谐运动的微分方程

The defining equation a = −ω²x can be expressed in differential form as d²x/dt² = −ω²x. This second-order linear differential equation is the hallmark of SHM. Its general solution can be written using sine or cosine functions, depending on initial conditions.

定义方程 a = −ω²x 可以写成微分形式 d²x/dt² = −ω²x。这个二阶线性微分方程是 SHM 的标志。根据初始条件，其通解可用正弦或余弦函数表示。

For IB Higher Level and OCR, you should be comfortable showing that x = A sin(ωt + φ) and x = A cos(ωt + φ) satisfy this differential equation. The choice between sine and cosine is simply a matter of where you set t = 0. If the oscillation starts from maximum displacement, a cosine form is convenient; if it starts from equilibrium moving in the positive direction, a sine form works best.

对于 IB 高水平和 OCR，你应该能熟练证明 x = A sin(ωt + φ) 和 x = A cos(ωt + φ) 都是该微分方程的解。选择正弦还是余弦仅取决于如何设置时间零点。如果振荡从最大位移处开始，余弦形式更方便；如果从平衡位置向正方向启动，正弦形式最合适。

Understanding the differential equation also helps when dealing with energy and deriving the expressions for velocity and period.

理解这个微分方程还有助于处理能量问题，并推导速度和周期表达式。

3. Displacement, Velocity, and Acceleration Equations | 位移、速度和加速度方程

If we choose the displacement-time relation as x = x₀ sin(ωt), where x₀ is the amplitude, then velocity v is the first derivative: v = dx/dt = ωx₀ cos(ωt). The acceleration a is the second derivative: a = dv/dt = −ω²x₀ sin(ωt) = −ω²x.

若选用位移-时间关系 x = x₀ sin(ωt)，其中 x₀ 为振幅，那么速度 v 是一阶导数：v = dx/dt = ωx₀ cos(ωt)。加速度 a 是二阶导数：a = dv/dt = −ω²x₀ sin(ωt) = −ω²x。

Maximum speed occurs when cos(ωt) = ±1, i.e. as the oscillator passes through equilibrium. Hence vₘₐₓ = ωx₀. Maximum acceleration occurs at the extreme positions where displacement is largest: aₘₐₓ = ω²x₀. These relationships are frequently tested in multiple-choice and structured questions.

最大速度出现在 cos(ωt) = ±1 时，即振荡体通过平衡位置时，因此 vₘₐₓ = ωx₀。最大加速度出现在位移最大的端点处：aₘₐₓ = ω²x₀。这些关系经常出现在选择题和结构题中。

For a general phase constant φ in x = x₀ sin(ωt + φ), the velocity is v = ωx₀ cos(ωt + φ), and acceleration remains −ω²x. IB often expects you to extract x₀, ω, and φ from a given equation or graph.

对于带有一般初相位 φ 的表达式 x = x₀ sin(ωt + φ)，速度是 v = ωx₀ cos(ωt + φ)，加速度仍为 −ω²x。IB 经常要求从给出的方程或图像中提取 x₀、ω 和 φ。

4. Period and Frequency | 周期与频率

The angular frequency ω is related to the period T and frequency f by ω = 2πf = 2π/T. The period of SHM is the time for one complete oscillation. For a spring–mass system, T = 2π√(m/k); for a simple pendulum, T = 2π√(L/g). These expressions are derived from the defining SHM equation and do not depend on amplitude, a feature called isochronism.

角频率 ω 与周期 T 及频率 f 的关系为 ω = 2πf = 2π/T。简谐运动的周期是一次完整振荡所需的时间。对于弹簧–质量系统，T = 2π√(m/k)；对于单摆，T = 2π√(L/g)。这些表达式均源自 SHM 的定义方程，且与振幅无关，这一特性称为等时性。

OCR typically asks you to recall these period formulas and explain why amplitude does not affect the period (for small angles in the case of the pendulum). IB may require you to derive T = 2π/ω from the solution of the differential equation.

OCR 通常要求记住这些周期公式，并解释为何振幅不影响周期（单摆需在小角度条件下）。IB 可能要求通过微分方程的解推导出 T = 2π/ω。

It is important to note that the period of a pendulum is independent of mass, a fact often investigated in practical assessments. For a spring, the period depends on mass and spring constant but not on gravitational field strength.

需要注意的是，单摆周期与质量无关，这一事实常在实验评估中考查。而对弹簧摆而言，周期取决于质量和劲度系数，而与重力场强度无关。

5. Energy in Simple Harmonic Motion | 简谐运动中的能量

During SHM, energy continuously alternates between kinetic and potential forms, but the total mechanical energy remains constant if there is no damping. The kinetic energy is Eₖ = ½ m v² = ½ m ω² (x₀² − x²). The potential energy stored in the spring or due to gravity in a pendulum is Eₚ = ½ m ω² x² (for a horizontal spring, or with appropriate modifications for a pendulum).

在 SHM 过程中，能量在动能和势能之间持续转换，但无阻尼时总机械能保持恒定。动能 Eₖ = ½ m v² = ½ m ω² (x₀² − x²)。弹簧的弹性势能或单摆的重力势能可表示为 Eₚ = ½ m ω² x²（适用于水平弹簧振子，单摆则有适当形式）。

Thus, the total energy E_total = Eₖ + Eₚ = ½ m ω² x₀² = ½ k x₀². This maximum energy is proportional to the square of the amplitude. A common exam question requires you to sketch energy–displacement or energy–time graphs, showing Eₖ and Eₚ as parabolas or sine-squared curves.

因此，总能量 E_total = Eₖ + Eₚ = ½ m ω² x₀² = ½ k x₀²。该最大能量与振幅的平方成正比。常见的考试题要求绘制能量-位移或能量-时间图像，表现出 Eₖ 和 Eₚ 的抛物线或正弦平方形状。

At the equilibrium position, Eₖ is maximum and Eₚ is zero; at the extremes, the reverse holds. Understanding this transfer is vital for solving problems involving maximum speed or spring compression.

在平衡位置，动能最大，势能为零；在端点处情况相反。理解这种能量转换对于求解最大速度或弹簧压缩量等问题至关重要。

6. The Horizontal Spring–Mass System | 水平弹簧–质量系统

A mass attached to a spring on a frictionless surface is the simplest SHM example. The restoring force is F = −kx, and using Newton’s second law, ma = −kx so a = −(k/m)x. Comparing with a = −ω²x gives ω² = k/m, leading to T = 2π√(m/k).

一个连接在水平无摩擦表面弹簧上的质量块是最简单的 SHM 实例。恢复力为 F = −kx，运用牛顿第二定律得 ma = −kx，因此 a = −(k/m)x。与 a = −ω²x 对比可得 ω² = k/m，从而 T = 2π√(m/k)。

The displacement is measured from the unstretched equilibrium position. When the mass is displaced and released, it oscillates symmetrically about equilibrium. The maximum speed is at equilibrium: vₘₐₓ = x₀√(k/m).

位移是从弹簧原长时的平衡位置开始测量的。当质量块被拉离并释放后，它将围绕平衡位置对称振动。最大速度出现在平衡位置：vₘₐₓ = x₀√(k/m)。

Vertically loaded springs undergo SHM with a new equilibrium where the spring stretches by mg/k. The period remains unchanged because the net restoring force is still proportional to displacement from that new equilibrium.

竖直悬挂的弹簧振子同样做 SHM，其新平衡位置是弹簧伸长 mg/k 处。由于净恢复力仍与新平衡位置的位移成正比，周期保持不变。

7. The Simple Pendulum | 单摆

For a simple pendulum of length L and bob mass m, the restoring force along the arc is −mg sinθ. For small angles (less than about 10°), sinθ ≈ θ in radians, and the displacement along the arc is s = Lθ. The tangential acceleration is a = −gθ = −(g/L)s. This matches a = −ω²s with ω² = g/L, yielding T = 2π√(L/g).

对于摆长为 L、摆球质量为 m 的单摆，沿弧线的恢复力为 −mg sinθ。在小角度下（约小于 10°），sinθ ≈ θ（弧度制），弧线上的位移 s = Lθ。切向加速度 a = −gθ = −(g/L)s。这与 a = −ω²s 形式一致，故 ω² = g/L，得到 T = 2π√(L/g)。

The approximation sinθ ≈ θ is essential for SHM. If the angle is large, the motion is periodic but no longer simple harmonic because the acceleration is not proportional to displacement. IB and OCR expect you to state the small-angle condition and explain its significance.

近似 sinθ ≈ θ 对于 SHM 至关重要。若角度过大，运动虽然依然周期性但不再是简谐运动，因为加速度与位移不成正比。IB 和 OCR 要求能说出小角度条件并解释其重要性。

Pendulum period depends on L and g, but not on mass or small amplitude. This makes it an excellent tool to measure gravitational acceleration, a classic experiment in both syllabuses.

单摆周期取决于 L 和 g，但与质量及小振幅无关。这使得它成为测量重力加速度的绝佳工具，也是两个课程大纲中的经典实验。

8. Phase, Phase Difference, and Initial Conditions | 相位、相位差与初始条件

The argument (ωt + φ) is called the phase of the motion. The constant φ is the initial phase, which sets the starting position and velocity. Two oscillations may be in phase (Δφ = 0, 2π, 4π…), antiphase (Δφ = π, 3π…), or have a phase difference expressed in radians or degrees.

表达式 (ωt + φ) 称为运动的相位。常量 φ 是初相位，决定了起始位置和速度。两个振荡可能同相（Δφ = 0, 2π, 4π…）、反相（Δφ = π, 3π…），或存在用弧度或度表示的相位差。

In the IB syllabus, you may be given two oscillators with different initial phases and asked to find the displacement, velocity, or acceleration at a certain time. OCR often links phase difference to wave superposition.

在 IB 课程中，可能给出两个初相位不同的振子，要求计算某一时刻的位移、速度或加速度。OCR 常将相位差与波的叠加联系起来。

A handy relationship is that a phase difference of π/2 rad corresponds to one quarter of a cycle. This helps when comparing displacement and velocity graphs: velocity leads displacement by π/2, and acceleration leads velocity by another π/2, meaning acceleration is π out of phase with displacement.

一个实用的关系是 π/2 rad 的相位差对应四分之一个周期。这在比较位移和速度图像时很有帮助：速度比位移超前 π/2，加速度又比速度超前 π/2，因此加速度与位移反相（相位差 π）。

9. Damped Harmonic Motion | 阻尼简谐运动

In reality, oscillators lose energy to friction or air resistance, causing the amplitude to decrease over time – this is damping. The motion is no longer truly simple harmonic because the energy is not constant, but it can be modelled as a decaying sinusoidal variation. Light damping results in a gradual decrease in amplitude; critical damping brings the system back to equilibrium in the shortest time without oscillating; heavy damping produces a very slow return.

现实中，振子会因摩擦或空气阻力而损失能量，导致振幅随时间减小，这就是阻尼。此时运动不再是严格意义上的简谐运动，因为能量不守恒，但可模型化为衰减的正弦振荡。轻阻尼使振幅缓慢减小；临界阻尼使系统在最短时间内回到平衡且不振荡；重阻尼则导致非常缓慢地返回。

IB HL and OCR both explore light, critical, and heavy damping. You should be able to sketch displacement–time graphs for each. The exponential envelope for light damping is characterised by a exponential decay of amplitude: x₀(t) = x₀₀ e⁻ᵞᵗ, where γ is the damping coefficient.

IB 高水平和 OCR 都会探讨轻阻尼、临界阻尼和重阻尼。你应能分别画出它们的位移-时间图像。轻阻尼的指数包络特征是振幅的指数衰减：x₀(t) = x₀₀ e⁻ᵞᵗ，其中 γ 为阻尼系数。

Questions may ask you to identify the damping type from a graph or to explain why a car suspension uses critical damping. Remember that in light damping the period remains approximately constant, but in heavy damping there is no oscillation.

试题可能要求从图像中识别阻尼类型，或解释汽车悬架为何采用临界阻尼。记住，轻阻尼时周期近似不变，而重阻尼状态下没有振荡。

10. Forced Oscillations and Resonance | 受迫振动与共振

When a periodic external force is applied to an oscillator, the system vibrates at the driving frequency. The amplitude of the forced oscillation depends on how close the driving frequency is to the natural frequency f₀ of the system. Resonance occurs when the driving frequency equals the natural frequency, producing a very large amplitude.

当对一个振子施加周期性外力时，系统会以外力的驱动频率振动。受迫振动的振幅取决于驱动频率与系统固有频率 f₀ 的接近程度。当驱动频率等于固有频率时，会发生共振，产生极大的振幅。

The sharpness of the resonance peak is related to damping: light damping gives a tall, narrow peak; heavier damping broadens the peak and reduces the maximum amplitude. This is a popular IB question: draw and interpret resonance curves for different damping levels.

共振曲线的尖锐程度与阻尼有关：轻阻尼产生高而窄的峰；阻尼较大则使曲线变宽，最大振幅降低。这是 IB 的热门题型：画出并解读不同阻尼下的共振曲线。

Real-world examples include the oscillation of bridges (Tacoma Narrows), tuning a radio, and microwave heating. OCR often sets questions on Barton’s pendulum, a classic demonstration of resonance.

现实例子包括桥梁振荡（塔科马海峡大桥）、调谐收音机以及微波加热。OCR 时常以巴顿摆作为共振的经典演示进行设问。

11. Graphical Analysis and Problem-solving Tips | 图像分析与解题技巧

SHM questions often involve interpreting x–t, v–t, and a–t graphs. Recall that gradient of displacement gives velocity, and gradient of velocity gives acceleration. At extreme positions, gradient (velocity) is zero; at equilibrium, gradient is steepest, corresponding to vₘₐₓ.

简谐运动问题常涉及解读 x–t、v–t 和 a–t 图像。记住位移的斜率给出速度，速度的斜率给出加速度。在端点位置，斜率（速度）为零；在平衡位置，斜率最陡，对应 vₘₐₓ。

A common IB exam technique is to use the relationship v = ± ω √(x₀² − x²) to find velocity at a specific position without needing time. Similarly, a = −ω²x gives acceleration directly from displacement. Mastering these magnitude forms will save time.

IB 考试的一个常用技巧是利用关系式 v = ± ω √(x₀² − x²) 求特定位置的速度而无需计算时间。类似地，a = −ω²x 由位移直接给加速度。熟练掌握这些大小形式能节省时间。

OCR structured questions may ask for a description of energy changes over a quarter cycle, or to calculate maximum kinetic energy given amplitude and period. Always check units: ω in rad s⁻¹, T in s, f in Hz.

OCR 结构题可能会要求描述四分之一周期内的能量变化，或根据振幅和周期计算最大动能。记得检查单位：ω 单位 rad s⁻¹，T 单位 s，f 单位 Hz。

12. Key Equations and Summary | 关键公式与总结

Defining equation: a = −ω²x

Displacement: x = x₀ sin(ωt + φ) or x = x₀ cos(ωt + φ)

Velocity: v = ωx₀ cos(ωt + φ) ; vₘₐₓ = ωx₀ ; v = ± ω √(x₀² − x²)

Period: T = 2π/ω = 1/f ; Spring: T = 2π√(m/k) ; Pendulum: T = 2π√(L/g)

Energy: E_total = ½ m ω² x₀² = ½ k x₀² ; Eₖ = ½ m ω² (x₀² − x²) ; Eₚ = ½ m ω² x²

Damping envelope: x₀(t) = x₀₀ e⁻ᵞᵗ

Master these relationships, practise blending algebraic and graphical reasoning, and you will be well prepared for any SHM question. Keep a close eye on small-angle approximations, phase shifts, and the distinction between angular frequency ω and linear frequency f.

掌握这些关系，练习将代数推理与图像分析相结合，你就能轻松应对任何 SHM 题目。请格外注意小角度近似、相位移以及角频率 ω 和频率 f 的区别。

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Simple Harmonic Motion: IB & OCR Revision Guide | 简谐运动 考点精讲