Simple Harmonic Motion | 简谐运动

📚 Simple Harmonic Motion | 简谐运动

Simple harmonic motion (SHM) forms a cornerstone of A-Level Physics, linking oscillations, waves, and mechanical systems. In the WJEC specification, you are expected to model SHM mathematically, interpret its defining equation, analyse energy interchanges, and apply the theory to real-world pendulums and mass–spring systems. This revision guide breaks down every essential concept, from the kinematic equations and graphical representations to damping, resonance, and phase.

简谐运动（SHM）是 A-Level 物理的基石，它将振动、波动和力学系统连接在一起。在 WJEC 考试大纲中，你不仅要能推导简谐运动的定义方程，还要会画位移、速度和加速度随时间变化的图像，分析能量转换，并把理论应用到单摆和弹簧振子等实际系统中。这篇考点精讲将带你从定义出发，逐步攻克运动方程、能量特征、阻尼振动和共振，确保你在考场上游刃有余。

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

Simple harmonic motion is an oscillation in which the acceleration of the particle is directly proportional to its displacement from a fixed equilibrium position and is always directed towards that equilibrium. Mathematically, this condition is expressed as a = –ω²x, where a is acceleration, x is displacement, and ω is the angular frequency. The minus sign indicates that acceleration opposes displacement, always pointing back towards the centre. For an object moving along a straight line with amplitude A, the motion is SHM if and only if the restoring force obeys Hooke’s law and no other forces interfere.

简谐运动是一种加速度与位移成正比且方向始终指向平衡位置的振动。数学上表示为 a = –ω²x，其中 a 是加速度，x 是位移，ω 是角频率。负号表明加速度总是与位移反向，指向中心。如果一物体沿直线运动，振幅为 A，那么当且仅当回复力符合胡克定律且无其他外力干扰时，该物体的运动就是简谐运动。这个条件也是判断一个系统是否做 SHM 的根本出发点。

2. The Reference Circle and Angular Frequency | 参考圆与角频率

SHM can be understood as the projection of uniform circular motion onto a diameter. If a particle moves around a circle of radius A with constant angular speed ω, its projection on the x-axis executes SHM. The angular frequency ω is related to the period T by ω = 2π/T and to the frequency f by ω = 2πf. In the WJEC exam, you must be able to identify ω from a graph, calculate it from T, and use it to find maximum velocity and acceleration. Remember: ω has units rad s⁻¹, but SHM can be described without explicitly mentioning radians when using trigonometric functions.

简谐运动可以看作匀速圆周运动在直径上的投影。设想一个质点以角速度 ω 沿半径 A 的圆运动，那么它在 x 轴上的投影就做简谐运动。角频率 ω 与周期 T 的关系是 ω = 2π/T，与频率 f 的关系是 ω = 2πf。在 WJEC 考试中，你既要能从图像上识别 ω，也要会根据周期计算 ω，并用它求出最大速度和最大加速度。需要注意的是，ω 的单位是 rad s⁻¹，但在使用三角函数描述 SHM 时，不一定显式地写出弧度。

3. Displacement, Velocity and Acceleration Equations | 位移、速度与加速度方程

If timing starts when the particle is at maximum positive displacement, the displacement–time relation is x = A cos(ωt). Velocity v is the first derivative: v = –Aω sin(ωt). Acceleration a is the second derivative: a = –Aω² cos(ωt) = –ω²x. The maximum speed occurs as the particle passes through equilibrium: vₘₐₓ = ωA. Maximum acceleration occurs at the extreme positions: aₘₐₓ = ω²A. An alternative expression linking speed and displacement without time is v = ±ω√(A² – x²). This form is particularly useful for energy calculations and for checking the speed at a known displacement.

若计时起点选在最大正向位移处，则位移时间关系为 x = A cos(ωt)。速度 v 是位移的一阶导数：v = –Aω sin(ωt)；加速度 a 是二阶导数：a = –Aω² cos(ωt) = –ω²x。质点经过平衡位置时速度最大：vₘₐₓ = ωA；而在极端位置处加速度最大：aₘₐₓ = ω²A。不显含时间的速度位移关系 v = ±ω√(A² – x²) 在能量计算和给定位移求速度时非常方便。WJEC 题目经常要求你从 x-t 图推导 v-t 和 a-t 图，务必熟练掌握。

4. Graphical Representations of SHM | 简谐运动的图像表示

Three graphs are fundamental: displacement–time (x–t), velocity–time (v–t), and acceleration–time (a–t). For x = A cos(ωt), the v–t graph is a negative sine wave shifted by a quarter period, and the a–t graph is an inverted cosine wave. Key features to label: peak velocity at t = T/4, 3T/4 where x = 0; peak acceleration at t = 0, T/2 where x = ±A; zero acceleration at equilibrium. In WJEC questions, you may be asked to sketch these graphs for a given phase constant or to interpret the phase difference between them. Remember that velocity leads displacement by π/2 rad, and acceleration leads velocity by another π/2 rad, meaning acceleration is always π rad out of phase with displacement.

你需要掌握三张基本图像：位移–时间图（x–t）、速度–时间图（v–t）和加速度–时间图（a–t）。以 x = A cos(ωt) 为例，v–t 是一条相位延迟四分之一周期的负正弦曲线，a–t 则是一条倒置的余弦曲线。标出关键特征：在 t = T/4 和 3T/4 处 x = 0，速度最大；在 t = 0 和 T/2 处 x = ±A，加速度最大；平衡位置处加速度为零。WJEC 考题可能要求你根据给定的初相画出这些图像，或者解读它们之间的相位差。记住：速度超前位移 π/2 弧度，加速度又超前速度 π/2 弧度，因此加速度与位移始终保持 π 弧度的相位差。

5. The Mass–Spring System | 弹簧振子系统

For a body of mass m attached to a light spring of spring constant k, the restoring force is F = –kx. Using Newton’s second law, m a = –kx, so a = –(k/m) x. Comparing with a = –ω²x gives ω² = k/m, hence the period T = 2π√(m/k). This is independent of amplitude, provided the spring obeys Hooke’s law and the mass of the spring is negligible. WJEC may ask you to investigate the effect of changing m or k, or to analyse a combination of springs in series and parallel. For springs in parallel, the effective spring constant kₚₐᵣₐₗₗₑₗ = k₁ + k₂; for springs in series, 1/kₛₑᵣᵢₑₛ = 1/k₁ + 1/k₂.

质量为 m 的物体连接在一根劲度系数为 k 的轻弹簧上，回复力为 F = –kx。由牛顿第二定律 m a = –kx，得 a = –(k/m) x。与 a = –ω²x 对比可知 ω² = k/m，因此周期 T = 2π√(m/k)。只要弹簧满足胡克定律且质量可忽略，该周期与振幅无关。WJEC 考试可能会让你讨论 m 或 k 改变时周期的变化，或者分析弹簧串联与并联的组合。注意：并联时等效劲度系数 kₚₐᵣₐₗₗₑₗ = k₁ + k₂，串联时满足 1/kₛₑᵣᵢₑₛ = 1/k₁ + 1/k₂。

6. The Simple Pendulum | 单摆

A simple pendulum consists of a point mass suspended by a light, inextensible string of length l. For small angular displacements (θ < 10° approximately), the restoring force is proportional to –θ, leading to SHM. The tangential acceleration is given by a = –(g/l) x, where x is the arc length displacement. Thus ω² = g/l and the period is T = 2π√(l/g). This expression is independent of mass and amplitude (for small angles). WJEC experiments often involve measuring T as l is varied, plotting T² against l, and using the gradient to determine g. Make sure you can explain why the small-angle approximation is necessary: without it, the motion is no longer simple harmonic.

单摆由一根轻质不可伸长的细绳悬挂一个质点构成，绳长为 l。当角位移很小（通常 θ < 10°）时，回复力与 –θ 成正比，摆球的运动为简谐运动。切向加速度 a = –(g/l) x，其中 x 是弧长位移。因此 ω² = g/l，周期 T = 2π√(l/g)。该周期与摆球质量和小振幅下的振幅无关。WJEC 实验题经常要求你改变 l 测量 T，然后绘制 T²–l 图像，利用斜率求重力加速度 g。务必清楚为什么需要小角度近似：只有当 sinθ ≈ θ 时，运动才满足 a = –恒定值×位移 的形式，否则就是非简谐运动。

7. Energy in Simple Harmonic Motion | 简谐运动的能量

In SHM, energy continuously oscillates between kinetic and potential forms. The kinetic energy is Eₖ = ½ m v². Using v = ±ω√(A² – x²), we obtain Eₖ = ½ m ω² (A² – x²). For a spring system, the elastic potential energy is Eₚ = ½ k x² = ½ m ω² x². The total mechanical energy remains constant: Eₜₒₜₐₗ = Eₖ + Eₚ = ½ m ω² A² = ½ k A². At equilibrium, energy is entirely kinetic; at the extremes, it is entirely potential. WJEC questions often ask you to sketch energy–displacement or energy–time graphs, or to calculate speed from energy conservation.

在简谐运动过程中，动能和势能不断相互转化。动能 Eₖ = ½ m v²，代入 v 与位移的关系得 Eₖ = ½ m ω² (A² – x²)。对于弹簧振子，弹性势能 Eₚ = ½ k x² = ½ m ω² x²。系统的总机械能守恒：Eₜₒₜₐₗ = Eₖ + Eₚ = ½ m ω² A² = ½ k A²。在平衡位置，能量全部为动能；在最大位移处，能量全部为势能。WJEC 常让你画出能量随位移或时间变化的图像，或者根据能量守恒计算某一位置的速度。注意：总能量与振幅的平方成正比，这是 SHM 的一个重要结论。

8. Damping and Its Effects | 阻尼及其影响

Real oscillators lose energy to resistive forces, causing the amplitude to decrease over time — this is called damping. Light damping results in a gradual decrease in amplitude while the period stays nearly unchanged. Critical damping brings the system to equilibrium in the shortest possible time without overshooting; this is important in car suspensions and galvanometer design. Heavy damping returns the system to equilibrium slowly without oscillating. The WJEC syllabus expects you to recognise these three types from displacement–time graphs and to explain how damping affects the amplitude and frequency. The natural frequency of an oscillator does not change, but the frequency of the damped motion may decrease slightly under heavy damping.

实际振动系统总会因阻力而损失能量，振幅随时间减小，这就是阻尼。轻阻尼下振幅逐渐减小，但周期几乎不变。临界阻尼使系统在最短时间内返回平衡位置且不发生超调，这种特性在汽车悬挂和电流计设计中至关重要。过阻尼则让系统缓慢回到平衡，无振荡。WJEC 要求你能够从位移–时间图像中识别这三种阻尼类型，并解释阻尼如何影响振幅和频率。系统的固有频率不变，但在重阻尼下，振动的频率会略微降低。

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

When a periodic driving force is applied to an oscillator, the system vibrates at the frequency of the driver, not its natural frequency. As the driving frequency approaches the natural frequency, the amplitude of oscillation increases dramatically — this is resonance. At resonance, energy transfer from the driver to the oscillator is most efficient, and the phase difference between displacement and driving force becomes π/2. The sharpness of the resonance peak is described by the quality factor Q. WJEC questions may ask you to plot amplitude against driving frequency for different damping levels: light damping gives a sharp, high peak, while heavy damping broadens and lowers the peak. You must also be able to discuss examples such as Barton’s pendulums, microwave heating, and bridge collapse.

当周期性驱动力作用于振子上时，系统会以驱动力的频率振动，而非其固有频率。当驱动频率接近固有频率时，振幅急剧增大，这种现象称为共振。在共振状态下，能量从驱动力到振子的传输效率最高，位移与驱动力之间的相位差为 π/2。共振峰的尖锐程度用品质因子 Q 描述。WJEC 可能让你绘制不同阻尼下振幅随驱动频率变化的图像：轻阻尼下共振峰尖而高，重阻尼下峰变宽变低。你还需要能够讨论巴顿摆、微波加热和桥梁坍塌等实例。

10. Phase Relationships in SHM | 简谐运动中的相位关系

Phase describes the stage an oscillator has reached in its cycle, usually expressed in radians. Two oscillators can be ‘in phase’ (phase difference 0 or 2π), meaning they cross equilibrium in the same direction simultaneously. A phase difference of π radians puts them ‘in antiphase’. In WJEC, you will be asked to calculate the phase difference between two points on a wave, or between displacement, velocity, and acceleration. For SHM starting with x = A cos(ωt), the velocity leads displacement by π/2, acceleration leads displacement by π, and velocity lags acceleration by π/2. You should be able to express displacement as x = A sin(ωt + φ) and determine φ from initial conditions.

相位描述振子在振动周期中所处的阶段，通常以弧度表示。两个振子若同时同向通过平衡位置，则称它们同相（相位差为 0 或 2π）；相位差为 π 时为反相。WJEC 考试会要求你计算波上两点间的相位差，或者分析位移、速度和加速度之间的相位关系。对于以 x = A cos(ωt) 开始的 SHM，速度超前位移 π/2，加速度超前位移 π，因此加速度与位移反相。你还要能够将位移写为 x = A sin(ωt + φ) 的形式，并根据初始条件确定初相 φ。

11. Experimental Determination of g Using a Pendulum | 用单摆测定重力加速度 g

A core practical in WJEC is to determine g using a simple pendulum. Measure the period T for several lengths l, using small amplitudes and timing multiple swings to reduce uncertainty. Plot T² vs l; the gradient is 4π²/g, so g = 4π² / gradient. Sources of uncertainty include reaction time, measurement of length to the centre of the bob, and the small-angle approximation. You must know how to minimise these, for example, by using a fiducial marker and a large number of oscillations. The experiment also reinforces your understanding of the T² ∝ l relationship.

WJEC 的一个重要实验是用单摆测定重力加速度 g。在保持小角度的前提下，改变摆长 l，测量对应的周期 T（通常计时多个完整周期以减少随机误差）。绘制 T²–l 图像，其斜率等于 4π²/g，因此 g = 4π² / 斜率。不确定度的主要来源包括反应时间、摆长测量（从悬挂点到摆球中心）以及小角度近似偏差。你需要知道如何通过使用参照标记和增加摆动次数来减小误差。这个实验也强化了你对 T² ∝ l 关系的理解。

12. Summary Checklist for WJEC SHM | WJEC 简谐运动考点清单

• Define SHM and use a = –ω²x.
• 定义 SHM 并运用 a = –ω²x。
• Derive and apply x = A cos(ωt), v = –Aω sin(ωt), a = –Aω² cos(ωt).
• 推导并应用 x = A cos(ωt)、v = –Aω sin(ωt)、a = –Aω² cos(ωt)。
• Use v = ±ω√(A² – x²) for speed at displacement x.
• 使用 v = ±ω√(A² – x²) 求任意位移处的速率。
• Calculate T for mass–spring (2π√(m/k)) and pendulum (2π√(l/g)).
• 计算 弹簧振子周期 T = 2π√(m/k) 和单摆周期 T = 2π√(l/g)。
• Analyse energy: Eₖ = ½ m ω² (A² – x²), Eₚ = ½ k x², Eₜₒₜₐₗ = ½ k A².
• 分析能量：Eₖ = ½ m ω² (A² – x²)，Eₚ = ½ k x²，Eₜₒₜₐₗ = ½ k A²。
• Sketch and interpret x–t, v–t, a–t graphs, noting phase differences.
• 绘制并解读 x–t、v–t、a–t 图像，注意相位差。
• Recognise light, critical, heavy damping from graphs.
• 从图像识别 轻阻尼、临界阻尼和过阻尼。
• Describe resonance, draw amplitude–frequency curves, and link to damping.
• 描述共振，画出振幅–频率曲线，并与阻尼建立联系。
• Perform and evaluate the pendulum experiment to find g.
• 完成并评估 用单摆测 g 的实验。

Published by TutorHao | Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）