📚 Simple Harmonic Motion – AQA A Level Physics Revision | A-Level AQA 物理：简谐运动 考点精讲
Simple harmonic motion (SHM) is a fundamental topic in AQA A Level Physics. It describes a repetitive back-and-forth movement about an equilibrium position, governed by a restoring force proportional to displacement. Mastering the key definitions, equations, and energy changes is essential for exam success. This article provides a focused, bilingual revision guide covering all required aspects of SHM for AQA.
简谐运动（SHM）是AQA A Level物理中的基础主题。它描述了一种围绕平衡位置的重复性往复运动，受正比于位移的回复力支配。掌握关键定义、方程和能量变化对考试成功至关重要。本文提供一份聚焦考点的中英双语复习指南，涵盖AQA考试大纲对简谐运动的所有要求。
1. Defining Simple Harmonic Motion | 简谐运动的定义
Simple harmonic motion occurs when the acceleration of an object is directly proportional to its displacement from a fixed point and is always directed towards that point. The fixed point is the equilibrium position. Mathematically, this is expressed as a = −ω²x, where a is acceleration, x is displacement, and ω is the angular frequency. The negative sign indicates that acceleration opposes the displacement.
简谐运动发生在物体的加速度与它从固定点的位移成正比，并且总是指向该固定点时。该固定点是平衡位置。数学上表示为 a = −ω²x，其中 a 是加速度，x 是位移，ω 是角频率。负号表示加速度与位移方向相反。
In AQA exams, you may need to identify whether a given motion is SHM by checking if the acceleration-displacement graph is a straight line through the origin with a negative gradient. Common examples include a mass on a light spring and a simple pendulum oscillating with small amplitude.
在AQA考试中，你可能需要判断给定运动是否为简谐运动，方法是检查加速度-位移图是否是一条通过原点且斜率为负的直线。常见例子包括轻弹簧上的质量块和小振幅摆动的单摆。
2. Key Quantities: Amplitude, Period, Frequency, and Angular Frequency | 关键物理量：振幅、周期、频率和角频率
Amplitude (A) is the maximum displacement from the equilibrium position. Period (T) is the time taken for one complete oscillation. Frequency (f) is the number of oscillations per second, measured in hertz (Hz). Angular frequency (ω) is related to f and T by ω = 2πf = 2π/T. It has units of rad s⁻¹.
振幅（A）是离平衡位置的最大位移。周期（T）是完成一次全振动所需的时间。频率（f）是每秒钟振动的次数，单位是赫兹（Hz）。角频率（ω）与 f 和 T 的关系为 ω = 2πf = 2π/T，单位为 rad s⁻¹。
Make sure you can convert between these quantities quickly. For example, if T = 0.5 s, then f = 2 Hz and ω = 4π rad s⁻¹. These relationships appear in many SHM problems.
确保你能快速转换这些量。例如，如果 T = 0.5 s，那么 f = 2 Hz，ω = 4π rad s⁻¹。这些关系出现在许多简谐运动问题中。
3. Displacement, Velocity, and Acceleration Equations | 位移、速度和加速度方程
The displacement in SHM as a function of time can be written as x = A cos(ωt) if timing starts at maximum displacement, or x = A sin(ωt) if timing starts at equilibrium moving positively. In AQA, both forms are acceptable, but you must state the starting condition clearly. The velocity is given by v = ± ω √(A² − x²). This shows that maximum speed v_max = ωA occurs at x = 0, and the speed is zero at x = ±A.
简谐运动中位移随时间变化可写为 x = A cos(ωt)（如果从最大位移开始计时），或 x = A sin(ωt)（如果从平衡位置正向运动开始计时）。在AQA中两种形式均可，但必须清楚说明起始条件。速度由 v = ± ω √(A² − x²) 给出。这表明最大速度 v_max = ωA 出现在 x = 0 处，而在 x = ±A 处速度为零。
The acceleration equation a = −ω²x is the defining equation of SHM. By substituting the displacement equation, we also get a = −ω²A cos(ωt) or a = −ω²A sin(ωt). Maximum acceleration magnitude is ω²A, occurring at the extreme positions.
加速度方程 a = −ω²x 是简谐运动的定义方程。通过代入位移方程，我们还可得到 a = −ω²A cos(ωt) 或 a = −ω²A sin(ωt)。最大加速度大小为 ω²A，出现在极端位置。
You must be confident in using these equations to find any variable. Often, the exam will provide some data and expect you to manipulate the equations with the relationships between ω, T, and f.
你必须能熟练使用这些方程求任何变量。考试常会提供一些数据，期望你利用 ω、T 和 f 之间的关系进行方程运算。
4. The Graphical Representation of SHM | 简谐运动的图形表示
Displacement-time, velocity-time, and acceleration-time graphs for SHM are all sinusoidal. For an object starting at maximum positive displacement, the x-t graph is a cosine wave starting at A. The v-t graph is a negative sine wave because velocity leads displacement by 90° (π/2 rad). The a-t graph is a negative cosine wave, which is the mirror of the x-t graph about the time axis, since a = −ω²x.
简谐运动的位移-时间图、速度-时间图和加速度-时间图都是正弦形的。对于从最大正向位移开始的物体，x-t 图是从 A 开始的余弦波。v-t 图是负的正弦波，因为速度超前位移 90°（π/2 rad）。a-t 图是负的余弦波，是 x-t 图关于时间轴的镜像，因为 a = −ω²x。
The phase difference between velocity and displacement is always π/2. The phase difference between acceleration and displacement is π. Recognising these phase relationships helps in sketching graphs and understanding the motion.
速度与位移之间的相位差始终为 π/2。加速度与位移之间的相位差为 π。识别这些相位关系有助于绘制草图并理解运动。
You should also be able to interpret graphs of kinetic energy, potential energy, and total energy against displacement or time. We’ll cover energy separately.
你还应能解读动能、势能和总能量随位移或时间变化的图像。我们稍后会单独讨论能量。
5. The Mass–Spring System | 弹簧振子系统
For a mass m attached to a light spring of spring constant k, the period of oscillation is given by T = 2π √(m/k). This assumes the spring obeys Hooke’s law and the mass is not too large to cause permanent deformation. Notice that T does not depend on the amplitude (isochronous nature), which is a key feature of SHM.
对于连接在劲度系数为 k 的轻弹簧上的质量 m，振动周期由 T = 2π √(m/k) 给出。这假设弹簧遵循胡克定律，且质量不会大到引起永久变形。注意 T 不依赖于振幅（等时性），这是简谐运动的一个关键特征。
Derivation: The restoring force is F = −kx, so acceleration a = F/m = −(k/m)x. Comparing with a = −ω²x gives ω² = k/m, and since T = 2π/ω, we get the formula. You must be able to show this derivation.
推导：回复力为 F = −kx，所以加速度 a = F/m = −(k/m)x。与 a = −ω²x 对比得 ω² = k/m，又因 T = 2π/ω，可得该公式。你必须能展示这一推导过程。
You may encounter questions combining horizontal and vertical spring setups. Remember that vertical oscillation also exhibits SHM about the new equilibrium position, where the extension balances the weight. The period formula remains the same.
你可能遇到结合水平和竖直弹簧装置的问题。记住竖直振动也围绕新的平衡位置作简谐运动，此时伸长量与重力平衡。周期公式保持不变。
6. The Simple Pendulum | 单摆
A simple pendulum consists of a point mass suspended by a light, inextensible string. For small angles (less than about 10°), the motion approximates SHM with period T = 2π √(l/g), where l is the length of the pendulum and g is gravitational field strength. The period is independent of mass and small-angle amplitude.
单摆由用轻质、不可伸长的绳悬挂的质点构成。对于小角度（小于约10°），运动近似为简谐运动，周期 T = 2π √(l/g)，其中 l 是摆长，g 是重力场强度。周期与质量和微小振幅无关。
Derivation: The restoring force tangential to the arc is −mg sinθ. For small θ, sin θ ≈ θ in radians, and the displacement along the arc is x = lθ. This leads to acceleration a = −(g/l)x, giving ω² = g/l. This derivation may be examined.
推导：沿弧切线方向的回复力为 −mg sinθ。对于小角度 θ（以弧度为单位），sin θ ≈ θ，且沿弧的位移 x = lθ。由此得到加速度 a = −(g/l)x，即 ω² = g/l。该推导可能会被考查。
A common exam question asks you to determine g using a pendulum. By measuring T for different lengths and plotting T² against l, the gradient equals 4π²/g.
一个常见的考题是使用单摆测定 g 值。通过测量不同摆长下的周期 T，并绘制 T² 对 l 的图，斜率等于 4π²/g。
7. Energy in Simple Harmonic Motion | 简谐运动中的能量
During SHM, energy continuously interchanges between kinetic energy (KE) and potential energy (PE). The total mechanical energy remains constant if there is no damping. KE = ½ mv², and using v = ω √(A² − x²), we get KE = ½ m ω² (A² − x²). Potential energy in a mass-spring system is PE = ½ kx² = ½ mω²x². Therefore, total energy E_total = KE + PE = ½ m ω² A² = ½ k A².
简谐运动过程中，能量在动能（KE）和势能（PE）之间不断转换。若无阻尼，总机械能保持不变。KE = ½ mv²，利用 v = ω √(A² − x²) 可得 KE = ½ m ω² (A² − x²)。弹簧振子系统的势能为 PE = ½ kx² = ½ mω²x²。因此，总能量 E_total = KE + PE = ½ m ω² A² = ½ k A²。
At the equilibrium position (x = 0), KE is maximum and PE is zero. At the extremes (x = ±A), KE is zero and PE is maximum. The energy-time graphs show that KE and PE both vary between zero and E_total with double the frequency of oscillation, because they depend on x² or v².
在平衡位置（x = 0），动能最大，势能为零。在极端位置（x = ±A），动能为零，势能最大。能量-时间图显示，动能和势能都在零与 E_total 之间变化，且频率为振动频率的两倍，因为它们依赖于 x² 或 v²。
For a pendulum, the potential energy is gravitational: PE = mgh, where h is the vertical height above the lowest point. The same KE formula applies.
对于单摆，势能是重力势能：PE = mgh，其中 h 是最低点以上的竖直高度。动能公式相同。
8. Damping and Its Effects | 阻尼及其影响
Damping occurs when an external resistive force, such as air resistance or friction, removes energy from an oscillating system. Light damping gradually reduces the amplitude over many cycles, while the time period remains almost unchanged. Critical damping brings the system to rest in the minimum time without overshooting. Overdamped systems return to equilibrium more slowly, with no oscillation.
阻尼发生在外部阻力（如空气阻力或摩擦）从振动系统中移除能量时。轻阻尼经过多个周期逐渐减小振幅，而周期几乎不变。临界阻尼使系统在最短时间内回到平衡位置而不发生超调。过阻尼系统更缓慢地回到平衡位置，且无振动发生。
Heavy damping (overdamping) and critical damping are often demonstrated in car suspension systems and measuring instruments. In AQA, you must be able to sketch amplitude-time graphs for different degrees of damping and identify the damping type from a graph or description.
重阻尼（过阻尼）和临界阻尼常见于汽车悬挂系统和测量仪器。在AQA考试中，你必须能绘制不同阻尼程度下的振幅-时间图，并根据图像或描述识别阻尼类型。
Free vibrations occur without external forces after an initial displacement. The natural frequency is the frequency at which a system oscillates when freely vibrating.
自由振动在初始位移后无外力作用时发生。固有频率是系统自由振动时的频率。
9. Forced Vibrations and Resonance | 受迫振动与共振
When a periodic external force is applied to a system, forced vibrations occur at the driving frequency. The amplitude depends on the driving frequency and the amount of damping. When the driving frequency equals the natural frequency of the system, resonance occurs, causing a dramatic increase in amplitude. Graphs of amplitude against driving frequency show sharp peaks at resonance, with the peak height and width depending on damping.
当周期性外力作用于系统时，受迫振动以驱动频率发生。振幅取决于驱动频率和阻尼大小。当驱动频率等于系统的固有频率时，发生共振，导致振幅急剧增大。振幅-驱动频率图在共振处呈现尖峰，峰的高度和宽度取决于阻尼。
Light damping gives a tall, sharp resonance peak, whereas heavier damping produces a lower, broader peak. Examples of resonance include a child on a swing, microwave ovens (water molecule resonance), and the catastrophic collapse of structures like the Tacoma Narrows Bridge. In AQA, you should be able to describe practical examples and interpret resonance curves.
轻阻尼产生高而尖锐的共振峰，而较强阻尼产生较低且较宽的峰。共振的例子包括荡秋千的孩子、微波炉（水分子共振）和塔科马海峡大桥等结构的灾难性倒塌。在AQA中，你应能描述实际例子并解读共振曲线。
You may also need to discuss the risks of resonance in engineering (e.g., designing buildings to avoid resonance with earthquakes) and the benefits (e.g., in musical instruments).
你可能还需讨论共振在工程中的风险（如设计建筑物以避免与地震共振）和益处（如乐器中）。
10. Practical Skills and Experiments | 实验技能与实验
Key practicals for SHM include investigating a mass-spring system to determine k and/or g, and using a simple pendulum to measure g. For the pendulum experiment, you typically vary the length l and measure the period T for small oscillations. Using a light gate or a motion sensor can improve accuracy for a mass-spring system. You must know how to reduce uncertainties, e.g., by timing multiple oscillations to find T accurately.
简谐运动的关键实验包括研究弹簧振子以测定 k 和/或 g，以及使用单摆测量 g。在单摆实验中，通常改变摆长 l 并测量小振幅振动的周期 T。使用光闸或运动传感器可提高弹簧振子实验的准确度。你必须知道如何减小不确定度，例如通过计时多次振动来精确求 T。
AQA often asks you to describe the procedure, state the measurements taken, explain how to plot and interpret a graph (e.g., T² vs l), and use the gradient to find g. Estimating percentage uncertainty and identifying systematic versus random errors are also assessed.
AQA常要求你描述步骤，说明需测量的量，解释如何绘图并解读图像（如 T² 对 l 图），并利用斜率求 g。评估百分不确定度、识别系统误差和随机误差也会被考查。
11. Common Misconceptions and Exam Tips | 常见误解与应试技巧
One common mistake is thinking that the period of SHM depends on amplitude. Remember that for ideal SHM, period is independent of amplitude unless damping is significant or the amplitude is large enough to break the small-angle approximation. Another misconception is confusing the direction of velocity and acceleration: acceleration is always towards the equilibrium position, but velocity can be away or towards equilibrium depending on the phase.
一个常见错误是认为简谐运动的周期依赖于振幅。记住，对于理想的简谐运动，周期与振幅无关，除非阻尼显著或振幅过大破坏了小角近似。另一个误解是混淆速度和加速度的方向：加速度总是指向平衡位置，但速度方向取决于振动相位，可能远离或指向平衡位置。
Always label graphs clearly and show the amplitude and period. Use the correct units. When solving problems, write down the governing equation (a = −ω²x) first, then substitute. Practice deriving the period formulas for spring and pendulum, as these derivations are frequently examined.
务必清晰地标注图像并标出振幅和周期。使用正确的单位。解题时，先写出控制方程（a = −ω²x），然后代入。练习推导弹簧和单摆的周期公式，因为这些推导常被考查。
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