Simple Harmonic Motion: A Comprehensive Guide for AQA A-Level Physics | 简谐运动：AQA A-Level 物理完整指南

📚 Simple Harmonic Motion: A Comprehensive Guide for AQA A-Level Physics | 简谐运动：AQA A-Level 物理完整指南

Simple harmonic motion (SHM) is one of the most important topics in AQA A-Level Physics because it links mechanics, waves, energy and circular motion. Mastering SHM gives you a strong foundation for answering both short-answer questions and multi-step calculations.

简谐运动（SHM）是 AQA A-Level 物理中最重要的主题之一，因为它将力学、波动、能量与圆周运动联系在一起。掌握简谐运动，能为你回答简答题和多步骤计算题打下坚实基础。

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

Simple harmonic motion is a periodic oscillation in which the acceleration of an object is directly proportional to its displacement from a fixed equilibrium position, and is always directed towards that equilibrium position.

简谐运动是一种周期性振荡，物体离开固定平衡位置的位移越大，其加速度就越大；并且加速度方向始终指向平衡位置。

a ∝ -x 或 a = -ω²x

The negative sign shows that the acceleration and displacement are in opposite directions. This condition leads to a sinusoidal oscillation in time.

负号表示加速度与位移方向相反。这一条件导致物体随时间做正弦规律振荡。

2. Key Conditions for SHM | 简谐运动的关键条件

For a system to perform true SHM, two conditions must be satisfied:

要让一个系统真正做简谐运动，必须满足两个条件：

Restoring force: The net force must be proportional to displacement from equilibrium.
Opposite direction: The force must always point back towards equilibrium.

回复力：合外力必须与偏离平衡位置的位移成正比。
方向相反：回复力必须始终指向平衡位置。

In AQA exams, you may be asked to “show that” a system undergoes SHM. The best method is to write the acceleration in terms of displacement and identify the constant ω².

在 AQA 考试中，题目可能会要求你“证明”某系统做简谐运动。最常用的方法是写出加速度与位移的关系式，并找出常数 ω²。

3. Mathematical Description of SHM | 简谐运动的数学描述

The displacement of an object undergoing SHM can be expressed as a cosine or sine function of time:

做简谐运动的物体的位移可以表示为时间 t 的余弦或正弦函数：

x = A cos(ωt) 或 x = A sin(ωt)

Here, A is the amplitude (maximum displacement from equilibrium), and ω is the angular frequency, measured in radians per second (rad s⁻¹).

其中 A 是振幅（离开平衡位置的最大位移），ω 是角频率，单位是弧度每秒（rad s⁻¹）。

The period T and frequency f are related to ω by:

周期 T 和频率 f 与 ω 的关系是：

ω = 2πf = 2π / T

Because AQA often uses the cosine form, remember that at t = 0, x = A when the motion starts from maximum displacement.

由于 AQA 常使用余弦形式，请记住：当运动从最大位移开始时，t = 0 时 x = A。

4. Velocity and Acceleration in SHM | 简谐运动中的速度与加速度

Differentiating the displacement equation gives the velocity:

对位移方程求导，可以得到速度：

v = -Aω sin(ωt)

The maximum speed occurs as the object passes through equilibrium:

最大速度出现在物体经过平衡位置时：

v_max = Aω

Differentiating again gives acceleration:

再次求导得到加速度：

a = -Aω² cos(ωt) = -ω²x

This equation is central to SHM. At the amplitude extremes, the acceleration is maximum (a_max = Aω²), while at equilibrium the acceleration is zero.

这个方程是简谐运动的核心。在振幅最大处，加速度最大（a_max = Aω²）；而在平衡位置，加速度为零。

A useful link between velocity and displacement is:

速度与位移之间的一个重要关系是：

v² = ω²(A² – x²)

5. Graphs of SHM | 简谐运动的图像

AQA frequently asks you to sketch or interpret displacement-time, velocity-time and acceleration-time graphs.

AQA 经常要求你画出或解读位移-时间图、速度-时间图和加速度-时间图。

The displacement-time graph is a cosine or sine curve.
The velocity-time graph is shifted by a quarter of a period (90°).
The acceleration-time graph is an inverted version of the displacement graph.

位移-时间图是余弦或正弦曲线。
速度-时间图比位移图提前四分之一周期（90°）。
加速度-时间图是位移图的反相版本。

Quantity	Equation	Maximum	Zero at
Displacement \| 位移	x = A cos(ωt)	±A	equilibrium \| 平衡位置
Velocity \| 速度	v = -Aω sin(ωt)	±Aω	amplitude extremes \| 振幅两端
Acceleration \| 加速度	a = -ω²x	±Aω²	equilibrium \| 平衡位置

6. Phase and Phasors | 相位与相量

In SHM, displacement, velocity and acceleration all have different phases. The phase difference is measured in radians or degrees.

在简谐运动中，位移、速度和加速度的相位各不相同。相位差用弧度或度表示。

Velocity leads displacement by π/2 rad (90°), while acceleration leads displacement by π rad (180°).

速度比位移超前 π/2 弧度（90°），而加速度比位移超前 π 弧度（180°）。

A phasor is a rotating vector whose projection gives the displacement. The idea of phasors helps you convert between SHM and circular motion.

相量是一个旋转矢量，它的投影给出位移。相量的概念帮助你理解简谐运动与圆周运动之间的相互转换。

Phase difference = (time shift / T) × 2π

Notice that displacement and acceleration are said to be in antiphase because their phase difference is exactly π.

注意：位移与加速度被称为反相，因为它们的相位差正好为 π。

7. Energy Changes in SHM | 简谐运动中的能量变化

During SHM, energy constantly shifts between kinetic energy and potential energy. The total mechanical energy remains constant if there is no damping.

在简谐运动过程中，能量不断在动能与势能之间转换。如果没有阻尼，总机械能保持不变。

E_total = ½ k A²

At equilibrium, all energy is kinetic:

在平衡位置，所有能量都为动能：

E_k, max = ½ m v_max² = ½ mω²A²

At maximum displacement, all energy is potential:

在最大位移处，所有能量都为势能：

E_p, max = ½ k A²

For an oscillating mass-spring system, ½ kx² is the elastic potential energy. For a pendulum, the potential energy is gravitational, E_p = mgh.

对于弹簧振子系统，½ kx² 是弹性势能；对于单摆，势能为重力势能，E_p = mgh。

AQA examiners expect you to sketch energy-time graphs and energy-displacement graphs, including the constant total energy line.

AQA 考官希望你能画出能量-时间图和能量-位移图，并标出恒定的总能量线。

8. Mass-Spring System | 质量-弹簧系统

A horizontal mass-spring system is a classic example of SHM. The restoring force is given by Hooke’s law:

水平弹簧振子系统是简谐运动的经典例子。回复力由胡克定律给出：

F = -kx

Combining this with Newton’s second law (F = ma) gives:

将其与牛顿第二定律（F = ma）结合，得到：

a = -(k/m)x，所以 ω² = k/m

Therefore the period is:

因此周期为：

T = 2π√(m/k)

In AQA practical work, you may investigate how the period changes with mass. Plotting T² against m should give a straight line through the origin.

在 AQA 实验题中，你可能会研究周期如何随质量变化。作出 T² 随 m 变化的图像，应得到一条过原点的直线。

9. Simple Pendulum | 单摆

A simple pendulum consists of a small bob suspended from a light string. For small angular displacements, the motion approximates SHM.

单摆由一根轻绳悬挂一个小球构成。对于小角度摆动，其运动近似为简谐运动。

The restoring force is the component of weight tangentially: F = -mg sinθ. For small θ, sinθ ≈ θ, so the motion is SHM.

回复力是重力沿切线方向的分量：F = -mg sinθ。在小角度时，sinθ ≈ θ，因此运动可视为简谐运动。

For a pendulum, the angular frequency and period are:

单摆的角频率和周期为：

ω = √(g/l)，T = 2π√(l/g)

The period is independent of the mass of the bob and, for small angles, independent of the amplitude. This is called isochronism.

单摆周期与摆球质量无关，且在小角度下与振幅无关。这被称为等时性。

10. Damping | 阻尼

Damping occurs when an external force, such as air resistance or friction, removes energy from the oscillating system.

当空气阻力或摩擦力等外力不断从振荡系统中带走能量时，就会产生阻尼。

Light damping: The amplitude gradually decreases, but the period remains approximately constant.
Heavy damping: The system returns slowly to equilibrium without oscillating.
Critical damping: The system returns to equilibrium in the shortest possible time without overshooting.

轻阻尼：振幅逐渐减小，但周期近似保持不变。
重阻尼：系统缓慢回到平衡位置，不发生振荡。
临界阻尼：系统以最短时间回到平衡位置，且不越过平衡位置。

In the displacement-time graph for light damping, the envelope of the peaks decays exponentially:

在轻阻尼的位移-时间图中，峰值包络线按指数规律衰减：

A = A₀ e^(-bt)

Critical damping is very important in real life, for example in car suspension systems and automatic door closers.

临界阻尼在实际生活中非常重要，例如汽车悬挂系统和自动关门装置中都有应用。

11. Resonance and Forced Vibrations | 共振与受迫振动

When an external periodic force is applied to an oscillator, the oscillator performs forced vibration. The amplitude depends on how close the driving frequency is to the natural frequency.

当一个周期性外力作用在振荡器上时，振荡器做受迫振动。振幅取决于驱动频率与固有频率的接近程度。

Resonance occurs when the driving frequency equals the natural frequency of the system. At resonance, energy is transferred most efficiently, producing a maximum amplitude.

当驱动频率等于系统固有频率时，就会发生共振。共振时能量传递效率最高，振幅达到最大。

f_driver = f_natural → amplitude is maximum

In the AQA specification, you should also understand the effect of damping on resonance:

在 AQA 大纲中，你还需要理解阻尼对共振的影响：

With light damping, the resonance peak is high and narrow.
With heavy damping, the resonance peak is low and broad.

轻阻尼时，共振峰又高又窄。
重阻尼时，共振峰又低又宽。

Examples of resonance include the Tacoma Narrows Bridge collapse, microwave ovens, and musical instruments. Applications include MRI scanning and quartz watches.

共振的例子包括塔科马海峡大桥坍塌、微波炉和乐器。应用包括核磁共振成像和石英钟表。

12. Exam Tips and Common Pitfalls | 考试技巧与常见误区

Many AQA students lose marks on SHM questions because they confuse amplitude, frequency and phase. Here are the key points to remember:

许多 AQA 学生在简谐运动题目上丢分，是因为混淆了振幅、频率和相位。以下是要记住的关键点：

Always state whether x is measured from equilibrium. A common mistake is measuring from the starting point instead.
Check units. Angular frequency ω is in rad s⁻¹, not Hz.
Label your graphs properly. Mark maximum values and the equilibrium position.
Use v² = ω²(A² – x²) when the question gives displacement instead of time.

始终说明 x 是从平衡位置测量的。常见错误是从起始点测量。
检查单位。角频率 ω 的单位是 rad s⁻¹，而不是 Hz。
正确标注图像。标出最大值和平衡位置。
当题目给出位移而不是时间时，使用 v² = ω²(A² – x²)。

Also, remember that the total energy of an undamped oscillator is constant. When drawing energy graphs, the total energy line must stay horizontal.

另外，请记住无阻尼振荡器的总能量恒定。在绘制能量图时，总能量线必须是水平的。

Finally, if you are asked to describe a resonance experiment, mention a vibration generator, a variable frequency oscillator, and how you detect the maximum amplitude.

最后，如果题目要求你描述共振实验，要提到振动发生器、可调频率振荡器，以及如何检测最大振幅。

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Simple Harmonic Motion: A Comprehensive Guide for AQA A-Level Physics | 简谐运动：AQA A-Level 物理完整指南