Simple Harmonic Motion for IGCSE AQA Physics | IGCSE AQA 物理：简谐运动考点精讲

📚 Simple Harmonic Motion for IGCSE AQA Physics | IGCSE AQA 物理：简谐运动考点精讲

Simple harmonic motion (SHM) is a special type of oscillatory motion that appears again and again in physics, from a mass bouncing on a spring to the swing of a pendulum. In the AQA IGCSE specification, you need to grasp the defining features of SHM, interpret displacement–time and velocity–time graphs, and link acceleration to displacement. Mastering SHM not only helps you tackle exam questions but also builds a foundation for understanding waves and alternating current. This article walks you through every tested concept with clear English explanations followed by accurate Chinese translations, all organised into bite-sized sections.

简谐运动（SHM）是物理中反复出现的一种特殊振动形式，从弹簧上跳动的小球到摆动的单摆都属于它的范畴。在 AQA IGCSE 的考试大纲中，你需要掌握简谐运动的定义特征、理解位移-时间图和速度-时间图，并建立加速度与位移之间的联系。掌握简谐运动不仅能帮你搞定考题，也为理解波动和交流电打下基础。本文将以清晰的英文讲解配以准确的中文翻译，分小节带你梳理每一个必考概念。

1. What is Simple Harmonic Motion? | 什么是简谐运动？

Simple harmonic motion is defined as a repetitive back-and-forth movement about an equilibrium position, where the acceleration is directly proportional to the displacement from the equilibrium position and is always directed towards that equilibrium point. In other words, the further the object moves from the middle, the stronger the restoring force pulling it back. The acceleration a is proportional to negative displacement x, giving the hallmark equation a ∝ −x.

简谐运动的定义是：物体围绕平衡位置做往复运动，其加速度与相对平衡位置的位移成正比，且方向始终指向平衡位置。换句话说，物体离开平衡位置越远，把它往回拉的回复力就越强。加速度 a 正比于负的位移 x，构成了标志性的数学关系 a ∝ −x。

For an object to perform SHM, the restoring force must obey Hooke’s law type behaviour (within the elastic limit) or the equivalent restoring torque for a pendulum. The motion is isochronous, meaning the period T remains constant regardless of amplitude, provided the amplitude is small enough (for a pendulum, typically angular displacement < 10°). This isochronism makes SHM an excellent timekeeping mechanism.

物体做简谐运动时，回复力必须满足胡克定律式的线性关系（在弹性限度内），或者对单摆而言满足相应的回复力矩关系。该运动是等时的，即只要振幅足够小（单摆通常要求角位移< 10°），周期 T 与振幅无关。正是这种等时性让 SHM 成为理想的计时机制。

In IGCSE exams, you may be asked to identify SHM from a description or a graph. Remember the two essential conditions: acceleration proportional to displacement, and acceleration opposite in direction to displacement.

在 IGCSE 考试中，你可能会被要求根据文字描述或图像判断是否为简谐运动。务必牢记两个核心条件：加速度与位移成正比，且加速度的方向始终与位移相反。

2. Key quantities: Amplitude, Period and Frequency | 关键物理量：振幅、周期和频率

Amplitude (A) is the maximum displacement from the equilibrium position. It is always a positive scalar measured in metres. The total distance travelled in one complete oscillation is 4A, because the object goes from equilibrium to a maximum on one side, back through equilibrium to the other maximum, and returns to equilibrium.

振幅（A）是物体偏离平衡位置的最大位移，是一个正的标量，单位是米。完成一次全振动所经过的总路程为 4A，因为物体从平衡位置运动到一侧的最大位移处，再回到平衡位置，又运动到另一侧的最大位移处，最后返回平衡位置。

Period (T) is the time taken for one complete oscillation. It is measured in seconds. Frequency (f) is the number of complete oscillations per second, measured in hertz (Hz). The relationship is f = 1/T and T = 1/f. In IGCSE, you may need to calculate period from a graph by measuring the time between two successive identical points, such as peak to peak or trough to trough.

周期（T）是完成一次全振动所需的时间，单位是秒。频率（f）是每秒钟完成的全振动次数，单位是赫兹（Hz）。两者的关系为 f = 1/T 和 T = 1/f。在 IGCSE 中，你可能需要从图像上测量两个相邻的同一状态点（如波峰到波峰或波谷到波谷）之间的时间来求周期。

Angular frequency ω (omega) is not always required at IGCSE but appears in some extension material. It relates to frequency by ω = 2πf and to period by ω = 2π/T. Remember these relationships when analysing sinusoidal graphs of SHM.

角频率 ω（omega）在 IGCSE 阶段不一定必考，但会出现在拓展内容中。它与频率的关系为 ω = 2πf，与周期的关系为 ω = 2π/T。在分析简谐运动的正弦图形时，记住这些关系。

f = 1/T T = 1/f ω = 2πf = 2π/T

3. Displacement–time graphs for SHM | 简谐运动的位移-时间图

The displacement–time (x–t) graph for an object starting at equilibrium, moving in one direction, is a sine curve. If the object is released from maximum displacement at t = 0, the graph is a cosine curve. Both shapes are sinusoidal and show the characteristic smooth reversal at the extremes.

如果物体从平衡位置开始向一个方向运动，它的位移-时间（x–t）图像是正弦曲线。如果物体在 t = 0 时从最大位移处释放，图像则是余弦曲线。两种形状都是正弦型的，并在两端展现出平滑的反向过程。

From the x–t graph you can read the amplitude A directly as the maximum displacement from the time axis. The period T is the time for one complete cycle. The graph also shows where the object is momentarily at rest (gradient zero at maxima and minima) and where it moves fastest (steepest gradient, passing through equilibrium).

从 x–t 图中可以直接读取振幅 A，即时间轴上方或下方的最大位移。周期 T 是一个完整循环所用的时间。图像还可以看出物体在何处瞬间静止（极大值和极小值处斜率为零），以及何处速度最大（通过平衡位置时斜率最陡）。

Exam tip: When asked to sketch a displacement–time graph, label the axes clearly: displacement on the vertical axis, time on the horizontal axis. Mark amplitude and period. If the question states the initial conditions (e.g., starting at equilibrium moving right), start your sine curve accordingly.

考试技巧：当题目要求绘制位移-时间图时，要清晰标注坐标轴：纵轴为位移，横轴为时间。标出振幅和周期。如果题目给出了初始条件（例如，从平衡位置开始向右运动），你的正弦曲线也要相应地从零点向上画起。

4. Velocity in SHM and velocity–time graphs | 简谐运动的速度与速度-时间图

The velocity of an object in SHM is not constant. Maximum speed vₘₐₓ occurs at the equilibrium position, where displacement is zero and the restoring force has accelerated the object through the midpoint. At the extreme displacements, the velocity is zero because the object changes direction. The velocity–time (v–t) graph is also sinusoidal but shifted relative to the displacement graph.

简谐运动中的速度不是恒定的。最大速度 vₘₐₓ 出现在平衡位置，此时位移为零，回复力已将物体加速到最快。在最大位移处，速度为零，因为物体正要改变运动方向。速度-时间（v–t）图也是正弦型曲线，但相对于位移曲线有相位偏移。

If the displacement is a sine function starting at zero, the velocity will be a cosine function, reaching maximum when x = 0 and zero when x = ±A. Thus, velocity leads displacement by a quarter of a cycle (π/2 radians). You do not need to quote phase differences in radian at IGCSE, but understanding the lag/lead helps with graph interpretation.

如果位移是从零点开始的正弦函数，速度就是余弦函数，在 x = 0 时达到最大值，在 x = ±A 时为零。因此，速度超前位移四分之一周期（π/2 弧度）。IGCSE 不要求说出弧度制的相位差，但理解这种超前/滞后关系有助于解读图像。

The maximum speed vₘₐₓ can be expressed as vₘₐₓ = ωA, where ω = 2πf. This formula is useful when you are given frequency and amplitude. Practise calculating vₘₐₓ for a mass–spring system or a simple pendulum.

最大速度 vₘₐₓ 可表示为 vₘₐₓ = ωA，其中 ω = 2πf。当已知频率和振幅时，这个公式很有用。练习计算弹簧振子或单摆的最大速度。

vₘₐₓ = ωA = 2πfA

5. Acceleration and restoring force | 加速度与回复力

The acceleration a of an object in SHM is given by a = −ω²x, where x is the displacement from equilibrium. The minus sign indicates that acceleration always acts towards the equilibrium position. This is a direct consequence of the defining relation a ∝ −x. The constant of proportionality is ω², so the acceleration is proportional to displacement and opposite in direction.

简谐运动中物体的加速度 a 由公式 a = −ω²x 给出，其中 x 是相对平衡位置的位移。负号表示加速度始终指向平衡位置。这是定义关系 a ∝ −x 的直接体现。比例常数为 ω²，所以加速度大小与位移成正比，方向相反。

At the extreme displacement x = A, the acceleration is maximum in magnitude: aₘₐₓ = −ω²A. At the equilibrium position x = 0, acceleration is zero. This matches the forces: the restoring force F = ma = −mω²x. For a horizontal spring, F = −kx, so we identify k = mω², which leads to ω = √(k/m) and T = 2π √(m/k).

在最大位移 x = A 处，加速度的大小达到最大值：aₘₐₓ = −ω²A。在平衡位置 x = 0 处，加速度为零。这与受力情况一致：回复力 F = ma = −mω²x。对于水平弹簧振子，F = −kx，因此可以得到 k = mω²，进而推出 ω = √(k/m) 和 T = 2π √(m/k)。

The acceleration–displacement (a–x) graph for SHM is a straight line passing through the origin with a negative slope. This is a quick check for SHM: if a graph of a against x is a straight line with negative gradient, the motion is simple harmonic.

简谐运动的加速度-位移（a–x）图是一条通过原点、斜率为负的直线。这是检验 SHM 的快捷方法：如果 a 随 x 变化的图像是一条具有负斜率的直线，该运动就是简谐运动。

6. Simple pendulum: period and factors affecting it | 单摆：周期及其影响因素

A simple pendulum consists of a point mass (the bob) suspended from a fixed point by a light, inextensible string. When displaced by a small angle (less than about 10°), the bob performs SHM. The period T of a simple pendulum is given by T = 2π √(L/g), where L is the length of the pendulum and g is the acceleration due to gravity.

单摆由一个质点（摆锤）通过轻质、不可伸长的细线悬挂在固定点组成。当摆角很小（一般小于 10°）时，摆锤会做简谐运动。单摆的周期 T 由公式 T = 2π √(L/g) 给出，其中 L 是摆长，g 是重力加速度。

Notice that the period does not depend on the mass of the bob nor on the amplitude (as long as the small-angle approximation holds). This is an important experimental fact: you can test it by keeping L constant and varying the mass or initial displacement; the period remains the same. The only variable that changes the period is the length L (and the local g).

注意，周期并不取决于摆锤的质量，也与振幅无关（只要小角度近似成立）。这是一个重要的实验事实：你可以通过保持 L 不变，改变质量或起始位移来验证，周期始终保持不变。唯一改变周期的变量是摆长 L（以及当地的 g 值）。

The relationship T² ∝ L is often tested. A graph of T² against L yields a straight line through the origin, and the gradient can be used to determine g: gradient = 4π²/g ⇒ g = 4π²/gradient. IGCSE practical questions may ask you to plan or analyse such an experiment.

常考的关系是 T² ∝ L。绘制 T² 对 L 的图像会得到一条过原点的直线，利用斜率可以测定重力加速度：斜率 = 4π²/g ⇒ g = 4π²/斜率。IGCSE 的实验题可能会要求你设计或分析这一实验。

Factor	Effect on period T	因素	对周期 T 的影响
Length L	Larger L → larger T	摆长 L	L 增大 → T 增大
Mass of bob	No effect	摆锤质量	无影响
Amplitude (small)	No effect	振幅（小角度）	无影响
Gravitational field strength g	Larger g → smaller T	重力场强度 g	g 增大 → T 减小

7. Mass–spring system: dynamics and period | 弹簧振子：动力学与周期

A mass attached to a horizontal spring on a frictionless surface exhibits SHM when displaced from equilibrium. The restoring force follows Hooke’s law: F = −kx, where k is the spring constant (stiffness). The minus sign shows the force is always opposite to displacement. Using Newton’s second law, F = ma, we obtain a = −(k/m)x, confirming a ∝ −x.

将一质量块连接在水平弹簧上并置于无摩擦表面，当偏离平衡位置后，系统会做简谐运动。回复力遵循胡克定律：F = −kx，其中 k 是弹簧劲度系数（刚度）。负号表示力的方向始终与位移相反。利用牛顿第二定律 F = ma，可得 a = −(k/m)x，确认了 a ∝ −x 关系。

The period of oscillation for a mass–spring system is T = 2π √(m/k). A stiffer spring (larger k) produces a shorter period, meaning faster oscillations. A larger mass (larger m) increases the inertia and thus lengthens the period. Unlike the pendulum, the mass does affect the period here.

弹簧振子的周期为 T = 2π √(m/k)。弹簧越硬（k 越大），周期越短，即振动越快。质量越大（m 越大），惯性增大，周期变长。与单摆不同，这里的质量确实会影响周期。

When analysing vertical mass–spring systems, the equilibrium position shifts due to gravity, but the motion about the new equilibrium is still SHM with the same period T = 2π √(m/k), because the gravitational force adds a constant offset that does not affect the restoring force’s dependence on displacement.

分析竖直弹簧振子时，平衡位置会因为重力而下移，但围绕新平衡位置的运动仍然是简谐运动，周期同样为 T = 2π √(m/k)。这是因为重力只叠加了一个恒定的偏移量，不影响回复力与位移之间的线性关系。

T = 2π √(m/k) for mass–spring system

8. Energy transformations in SHM | 简谐运动中的能量转化

In an ideal undamped SHM system, total mechanical energy is conserved and continuously transforms between kinetic energy (KE) and potential energy (PE). At the equilibrium position, KE is maximum (because speed is maximum) and PE is minimum (often taken as zero for a horizontal spring). At the extreme positions, the velocity is zero so KE = 0, and all energy is stored as PE.

在理想无阻尼的简谐运动系统中，总机械能守恒，并在动能（KE）和势能（PE）之间不断转换。在平衡位置，动能最大（因为速度最大），势能最小（水平弹簧常取为零）。在最大位移处，速度为零，因此 KE = 0，全部能量以势能形式储存。

For a horizontal mass–spring system, the elastic potential energy is PE = ½ kx². At maximum displacement x = A, total energy E = ½ kA². At any point, KE = E − PE = ½ k(A² − x²). This relationship can be plotted against displacement, showing parabolic curves for KE and PE, which sum to a constant horizontal line for total energy.

对于水平弹簧振子，弹性势能为 PE = ½ kx²。在最大位移 x = A 处，总能量 E = ½ kA²。在任意位置，动能为 KE = E − PE = ½ k(A² − x²)。这个关系可以绘制成位移的函数图像，显示出 KE 和 PE 的抛物线形状，总和为一条水平直线（总能量恒定）。

For a simple pendulum, the potential energy is gravitational: PE = mgh, where h is the height relative to the lowest point. Again, energy swaps back and forth. Understanding energy conservation helps explain why the amplitude remains constant in the absence of damping and why the system continues oscillating indefinitely in theory.

对于单摆，势能是重力势能：PE = mgh，其中 h 是相对最低点的高度。能量同样来回转换。理解能量守恒可以解释为何在没有阻尼的情况下振幅保持不变，以及理论上系统会永远振荡下去。

Total energy E = ½ kA² (spring) or E = mgHₘₐₓ (pendulum)

9. Damping and its effects | 阻尼及其影响

In real-world oscillations, energy is gradually lost to the surroundings due to friction, air resistance or other resistive forces. This loss causes the amplitude of oscillation to decrease over time, a phenomenon called damping. The frequency usually remains nearly unchanged for light damping, but the period may lengthen slightly as amplitude decays.

在现实振动中，由于摩擦、空气阻力或其他阻力的存在，能量会逐渐耗散到周围环境中。这种能量损失导致振幅随时间逐渐减小，这称为阻尼。轻阻尼时频率几乎保持不变，但随着振幅衰减，周期可能会略微变长。

Light damping: the system completes many oscillations before coming to rest. The amplitude decreases exponentially. Critical damping: the system returns to equilibrium in the shortest possible time without oscillating (used in car suspension and door closers). Heavy damping (overdamping): the system returns to equilibrium slowly without oscillation.

轻阻尼：系统在停止前会经历许多次振荡，振幅呈指数衰减。临界阻尼：系统以最短时间回到平衡位置而不发生振荡（应用于汽车悬挂和闭门器）。重阻尼（过阻尼）：系统不经振荡缓慢回到平衡位置。

Exam questions may ask you to sketch amplitude–time graphs for different degrees of damping or to identify damping from a given trace. Remember that damping does not change the period significantly unless the damping is very heavy.

考题可能要求你画出不同阻尼程度下的振幅-时间图，或根据给定波形判断阻尼类型。记住，除极大阻尼外，阻尼不会显著改变周期。

10. Resonance and forced oscillations | 共振与受迫振动

When a periodic force is applied to a system capable of SHM, the system vibrates at the driving frequency. This is called forced oscillation. If the driving frequency matches the system’s natural frequency, resonance occurs: the amplitude of oscillation increases dramatically, often to dangerous levels. Bridges, buildings and machines must be designed to avoid resonance.

当对能够做简谐运动的系统施加周期性外力时，系统会以外加驱动频率进行振动，称为受迫振动。如果驱动频率恰好等于系统的固有频率，就会发生共振：振幅急剧增大，常常达到危险的程度。桥梁、建筑和机器设计时必须避免共振。

The natural frequency f₀ of a simple pendulum is f₀ = 1/(2π) √(g/L); for a mass–spring system, f₀ = 1/(2π) √(k/m). At resonance, the energy transfer from the driver to the system is most efficient. A small periodic driving force can produce a large-amplitude oscillation if the frequencies match.

单摆的固有频率 f₀ = 1/(2π) √(g/L)；弹簧振子的固有频率 f₀ = 1/(2π) √(k/m)。共振时，驱动系统对振动系统的能量传递效率最高。一个小小的周期驱动力，如果频率匹配，就能引发大振幅振荡。

Resonance curves show amplitude against driving frequency. The peak is sharp for light damping and broader for heavier damping. In IGCSE, you might be asked to interpret a resonance graph or give an example of resonance, such as a swing being pushed at the right moment, a wine glass shattering from a singer’s voice, or the Tacoma Narrows Bridge collapse (though that was actually aeroelastic flutter).

共振曲线展示振幅随驱动频率的变化。轻阻尼时共振峰尖锐，阻尼增大时峰变宽。在 IGCSE 中，你可能需要解读共振图或举出共振的例子，如秋千在恰当时机被推高，歌声震碎酒杯，或塔科马海峡大桥崩塌（尽管那实际上是气动弹性颤振）。

11. Graphical analysis: linking x–t, v–t and a–t | 图像分析：关联 x–t、v–t 和 a–t 图

One powerful skill for IGCSE is to sketch and compare the three motion graphs for SHM on the same time axis. Starting from equilibrium at t=0, displacement is a sine wave; velocity is a cosine wave (leading by ¼ cycle); acceleration is a negative sine wave (leading displacement by ½ cycle, i.e., opposite direction). All three have the same period.

IGCSE 的一项重要技能是在同一时间轴上绘制并比较简谐运动的三种运动图像。若 t=0 时从平衡位置开始，位移是正弦波；速度是余弦波（超前 ¼ 周期）；加速度是负正弦波（超前位移 ½ 周期，即方向相反）。三者周期相同。

You can deduce velocity from the gradient of the x–t graph, and acceleration from the gradient of the v–t graph. Where x is zero and gradient is steepest, v is max; where v is max and its gradient is zero, a is zero – exactly at equilibrium. At extremes, x is max, gradient of x–t is zero so v = 0; gradient of v–t is steepest so a is max (and points back to equilibrium).

你可以通过 x–t 图的斜率推断速度，通过 v–t 图的斜率推断加速度。当 x 为零且斜率最陡时，v 最大；当 v 最大且其斜率为零时，a 为零——正好在平衡位置。在极限位置，x 最大，x–t 斜率为零所以 v = 0；v–t 斜率最陡所以 a 最大（并指向平衡位置）。

Be able to annotate these graphs with amplitude A, period T, and to mark points of max velocity and max acceleration. Understand that the a–x graph is a straight line through the origin, which is a defining test for SHM.

要能在这些图上标注振幅 A、周期 T，并标出最大速度和最大加速度的位置。理解 a–x 图是一条过原点的直线，这是判断 SHM 的决定性检验。

12. Common exam mistakes and tips | 常见考试错误与应对技巧

Many students lose marks by confusing amplitude with distance travelled in one cycle (4A not 2A). Another common error is assuming the period depends on amplitude for small oscillations – it does not. When using the pendulum formula, ensure L is measured to the centre of the bob. For spring problems, check whether the question refers to horizontal or vertical configuration, but remember the period formula T = 2π √(m/k) works for both.

许多学生混淆振幅与一个周期内的路程（4A 而非 2A），导致失分。另一个常见错误是以为小角度振荡时周期与振幅有关——实则无关。使用单摆公式时，确保 L 量到摆锤中心。在弹簧问题中，要看清题目描述的是水平还是竖直放置，但记住周期公式 T = 2π √(m/k) 两者都适用。

Always show working steps: write the correct formula, substitute values, and give the answer with units. When describing energy changes, specify the points (equilibrium, extreme) and what forms of energy are involved. For graph sketching, label axes, use a ruler for straight lines, and draw smooth curves for sinusoidal shapes.

务必展示解题步骤：写出正确的公式，代入数值，并给出带单位的答案。在描述能量转换时，要指明位置（平衡点、极限点）以及涉及的能量形式。画图时，标注坐标轴，画直线用尺，画正弦曲线要平滑。

Finally, practise past paper questions on SHM, especially those combining pendulum period, spring constant, and energy. Being comfortable with interpreting graphs and applying formulas will make this topic a straightforward marks-earner.

最后，多练习历年真题中与 SHM 相关的题目，尤其是结合单摆周期、弹簧劲度系数和能量的综合题。熟练解读图像和应用公式，会让你在这一专题上轻松得分。

Published by TutorHao | Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Simple Harmonic Motion for IGCSE AQA Physics | IGCSE AQA 物理：简谐运动 考点精讲