📚 Simple Harmonic Motion: Key Concepts for Edexcel Physics | Edexcel 物理：简谐运动考点精讲
Simple harmonic motion (SHM) forms the backbone of oscillation topics in the Edexcel A Level Physics specification. Mastery of SHM requires a clear understanding of the defining equation, graphical relationships between displacement, velocity and acceleration, energy transformation, and the behaviour of real systems such as mass-spring oscillators and simple pendulums. This article unpacks each of these concepts with precise bilingual explanations, helping you build confidence for both multiple-choice and extended-response exam questions.
简谐运动是 Edexcel A Level 物理大纲中振动部分的核心。要真正掌握 SHM，需要透彻理解其定义方程、位移–速度–加速度之间的图像关系、能量转化以及弹簧振子和单摆等实际系统的行为。本文用精准的双语解析逐一拆解这些概念，帮助你从容应对选择题和论述题。
1. What is Simple Harmonic Motion? | 什么是简谐运动？
Simple harmonic motion is a special type of periodic oscillation where the restoring force is directly proportional to the displacement from equilibrium and always acts towards the equilibrium position. Mathematically, this leads to the key relation: acceleration a is proportional to negative displacement x.
简谐运动是一种特殊的周期振动，其回复力与偏离平衡位置的位移成正比，并且始终指向平衡位置。数学上，这导出核心关系：加速度 a 与位移 x 成正比且方向相反。
The defining equation of SHM is
SHM 的定义方程为
a = -ω²x
where ω is the angular frequency of the motion. Any system that obeys this equation is said to be executing simple harmonic motion. In Edexcel exams, you must be able to recognise this relationship and use it to deduce whether a given motion is SHM.
其中 ω 是振动的角频率。任何满足此方程的系统都在做简谐运动。在 Edexcel 考试中，你必须能够识别这一关系，并利用它判断某个运动是否为 SHM。
2. Conditions for SHM | 简谐运动的条件
For an object to undergo true SHM, two conditions must be met simultaneously. First, the resultant force F acting on the object must always be directed towards the equilibrium point. Second, the magnitude of this force must be directly proportional to the displacement from equilibrium.
一个物体要真正做简谐运动，必须同时满足两个条件。第一，作用在物体上的合力 F 必须始终指向平衡位置。第二，该力的大小必须与偏离平衡位置的位移成正比。
These conditions can be summarised by Hooke's law for a spring: F = -kx, where the negative sign indicates the restoring direction. However, SHM is not limited to springs; any system where the restoring force obeys this linear law will exhibit SHM for small displacements. A common exam pitfall is to assume all oscillatory motion is SHM; graphically, the displacement-time graph must be a pure sine or cosine function.
这些条件可以用弹簧的胡克定律概括：F = -kx，其中负号表示回复方向。然而，SHM 不仅限于弹簧；任何在小位移下回复力满足这一线性关系的系统都会表现出 SHM。考试中常见的错误是认为所有振动都是 SHM；从图像上看，位移–时间图必须是一条纯粹的正弦或余弦曲线。
3. Displacement, Velocity and Acceleration Equations | 位移、速度和加速度方程
For an oscillation starting at maximum positive displacement A (amplitude), the displacement x as a function of time t is given by
对于从最大正位移 A（振幅）开始的振动，位移 x 随时间 t 变化的函数为
x = A cos(ωt)
If the motion starts from equilibrium at t = 0, the equation becomes x = A sin(ωt). The velocity v can be derived by differentiating displacement with respect to time, yielding
如果在 t = 0 时从平衡位置开始运动，方程变为 x = A sin(ωt)。通过对位移求导，可得到速度 v
v = -ωA sin(ωt) = ±ω√A² - x²
The maximum speed is vmax = ωA and occurs at the equilibrium position. Acceleration is obtained by further differentiation:
最大速率为 vmax = ωA，出现在平衡位置。加速度通过进一步求导得到：
a = -ω²A cos(ωt) = -ω²x
Thus the acceleration is always directed opposite to displacement, with maximum magnitude amax = ω²A at the extremes of motion.
因此加速度始终与位移反向，在运动端点达到最大量值 amax = ω²A。
4. Graphical Representations | 图形表示
Edexcel examiners often test your ability to sketch and interpret displacement-time, velocity-time and acceleration-time graphs for SHM. The displacement x-t graph is a sine or cosine wave of amplitude A and period T. The velocity v-t graph is also sinusoidal but shifted by a quarter of a period, with amplitude ωA.
Edexcel 考官经常考查你绘制和解读 SHM 的位移–时间、速度–时间和加速度–时间图像的能力。位移 x-t 图是振幅为 A、周期为 T 的正弦或余弦波。速度 v-t 图同样是正弦波，但相位移动了四分之一周期，振幅为 ωA。
The acceleration a-t graph is exactly the mirror of the x-t graph about the time axis, because a = -ω²x. A crucial point to remember: when displacement is zero, velocity is maximum and acceleration is zero; when displacement is maximum, velocity is zero and acceleration is maximum in the opposite direction.
加速度 a-t 图则恰好是 x-t 图关于时间轴的镜像，因为 a = -ω²x。一个关键点要记住：位移为零时，速度最大而加速度为零；位移最大时，速度为零而加速度达到最大值且方向相反。
A summary table helps visualise these phases:
下面表格有助于直观理解这些相位关系：
| Position | Displacement x | Velocity v | Acceleration a |
|---|---|---|---|
| Equilibrium | 0 | ± max | 0 |
| Extreme (+A) | +A | 0 | -ω²A |
| Extreme (-A) | -A | 0 | +ω²A |
5. Energy in SHM | 简谐运动中的能量
In an ideal SHM system there is no energy loss, so the total mechanical energy remains constant. The total energy Etotal is the sum of kinetic energy KE and potential energy PE. Using the SHM equations, these can be expressed in terms of displacement:
在理想 SHM 系统中没有能量损失，因此总机械能保持不变。总能量 Etotal 是动能 KE 与势能 PE 之和。利用 SHM 方程，这些能量可以用位移表示为：
KE = ½mv² = ½mω²(A² - x²)
PE = ½mω²x²
Etotal = ½mω²A²
Kinetic energy is maximum at equilibrium and zero at the extremes, while potential energy is zero at equilibrium (taking the equilibrium as the zero-potential reference) and maximum at the extremes. The energy–displacement graphs are parabolas: KE is an inverted parabola, and PE is a parabola opening upwards.
动能在平衡位置最大，在端点为零；势能在平衡位置为零（以平衡位置为零势能参考点），在端点最大。能量–位移图均为抛物线：动能是开口向下的抛物线，势能是开口向上的抛物线。
For a mass-spring system, the potential energy is stored as elastic potential energy in the spring. For a simple pendulum, it is gravitational potential energy. In examinations, you may be asked to determine the speed at a given displacement or to calculate the fraction of energy that is kinetic.
对于弹簧振子系统，势能以弹性势能的形式储存在弹簧中；对于单摆，势能则是重力势能。考试可能要求你计算给定位移处的速率或动能占比。
6. Mass-Spring System | 弹簧振子系统
A mass attached to an ideal, massless spring is a classic example of SHM. The restoring force is provided by Hooke's law: F = -kx. Comparing with F = ma and a = -ω²x, we obtain the angular frequency and period:
连接在理想轻质弹簧上的物块是 SHM 的经典实例。回复力由胡克定律提供：F = -kx。将其与 F = ma 及 a = -ω²x 对比，可得到角频率和周期：
ω = √k/m
T = 2π √m/k
Notice that the period does not depend on amplitude, which is a key feature of SHM (isochronism). For a vertical mass-spring system, the equilibrium position shifts but the SHM characteristics remain unchanged because the net restoring force is still linearly proportional to the displacement from the new equilibrium.
注意周期与振幅无关，这是 SHM 的一个关键特征（等时性）。对于竖直弹簧振子，平衡位置会移动，但因为从新平衡位置量起的回复力仍然与位移成线性关系，SHM 的特性保持不变。
Exam questions often involve messy combinations of series and parallel springs. Remember: for springs in parallel, effective spring constant keff = k₁ + k₂; for springs in series, 1/keff = 1/k₁ + 1/k₂. These combinations alter the period accordingly.
考试中常出现弹簧串并联的复杂组合。记住：弹簧并联时有效劲度系数 keff = k₁ + k₂；串联时 1/keff = 1/k₁ + 1/k₂。这些组合会相应改变周期。
7. The Simple Pendulum | 单摆
A simple pendulum consists of a point mass suspended from a light, inextensible string. For small angular displacements (less than about 10°), the motion approximates SHM. The restoring force is a component of the weight, leading to a period independent of mass:
单摆由一个悬挂在轻质且不可伸长的细线上的质点构成。当角位移较小（约小于 10°）时，运动近似为 SHM。回复力是重力的一个分量，由此得出周期与质量无关：
T = 2π √l/g
where l is the length of the pendulum and g proximations involved: sinθ ≈ θ for small angles, which makes the restoring torque linear.
其中 l 是摆长，g 是重力加速度。Edexcel 要求你理解其中涉及的近似：小角度下 sinθ ≈ θ，这使得回复力矩成为线性。
Common pitfalls include forgetting that the period depends only on length and gravitational field strength, not on amplitude or mass. You may also be asked to calculate g from a pendulum experiment by measuring T for different lengths and plotting T² against l.
常见错误包括忘记周期仅取决于摆长和重力场强度，而与振幅或质量无关。你还可能遇到要求通过测量不同摆长的周期 T，并绘制 T²–l 图来测定 g 的考题。
8. Damped Oscillations | 阻尼振动
Real oscillating systems lose energy over time due to resistive forces such as air resistance or internal friction. This energy loss causes the amplitude to decay exponentially—a phenomenon called damping. Edexcel distinguishes three types of damping: light (underdamped), critical, and heavy (overdamped).
真实的振动系统会因空气阻力或内摩擦等耗散力而损失能量。这种能量损失导致振幅指数衰减——这一现象称为阻尼。Edexcel 将阻尼分为三类：弱阻尼（欠阻尼）、临界阻尼和强阻尼（过阻尼）。
Light damping results in oscillations whose amplitude decreases gradually, with the period remaining roughly constant. Critical damping brings the system back to equilibrium in the shortest possible time without overshooting. Heavy damping returns the system to equilibrium slowly with no oscillation. Graphical interpretation of these cases is a frequent multiple-choice topic.
弱阻尼导致振幅逐渐减小，但周期大致保持不变。临界阻尼使系统在最短时间内返回平衡位置且不发生过冲。强阻尼则使系统缓慢回到平衡位置，无振荡。对这些情况的图像解读是常见的选择题考点。
9. Forced Oscillations and Resonance | 受迫振动与共振
When a periodic external force drives an oscillatory system, the system vibrates at the driving frequency. As the driving frequency approaches the natural frequency of the system, the amplitude increases dramatically—this is resonance. In the ideal case with no damping, the amplitude would become infinite, but real systems always have some damping, limiting the amplitude to a finite maximum.
当周期性外力驱动一个振动系统时，系统将以驱动频率振动。当驱动频率接近系统固有频率时，振幅急剧增大——这就是共振。在无阻尼的理想情况下振幅会趋向无穷大，但实际系统总存在阻尼，将振幅限制在一个有限最大值。
The resonance curve (amplitude versus driving frequency) shows a sharp peak for lightly damped systems and a broader, lower peak for heavily damped systems. The phase relationship between displacement and driving force also changes: at low frequencies they oscillate nearly in phase; at resonance there is a 90° (π/2 rad) phase lag; at high frequencies they are almost in anti-phase (180° out of phase).
共振曲线（振幅–驱动频率图）在弱阻尼时显示尖锐的峰值，强阻尼时则显示较宽较低的峰值。位移与驱动力之间的相位关系也会变化：低频时几乎同相；共振时存在 90°（π/2 弧度）相位滞后；高频时几乎反相（相差 180°）。
Examples such as the Tacoma Narrows Bridge collapse, microwave heating of food, and MRI scanners demonstrate the practical importance of resonance. Edexcel may ask you to describe both beneficial and destructive effects.
塔科马海峡大桥坍塌、微波炉食物加热、核磁共振成像等实例展示了共振的实际重要性。Edexcel 可能会要求你描述共振的有益和破坏性影响。
10. Phase and Phase Difference | 相位与相位差
Phase describes the position and direction of motion of an oscillator within one cycle. Two objects undergoing SHM can be compared by their phase difference, measured in radians or degrees. In Edexcel questions, you might need to calculate phase difference from displacement or velocity equations, or to predict when two pendulums will next be in phase.
相位描述了一个周期内振子所处的位置和运动方向。两个做 SHM 的物体可以通过相位差进行比较，相位差以弧度或度为单位。在 Edexcel 考题中，你可能需要从位移或速度方程计算相位差，或预测两个单摆何时再次同相。
For example, if oscillator A follows x₁ = A sin(ωt) and oscillator B follows x₂ = A sin(ωt + φ) they are in phase; when φ = π they are exactly out of phase. Understanding the meaning of 'lead' and 'lag' is essential for interpreting oscilloscope traces or wavefront diagrams.
例如，若振子 A 遵从 x₁ = A sin(ωt)，振子 B 遵从 x₂ = A sin(ωt + φ)，则相位差为 φ。当 φ = 0 时它们同相；当 φ = π 时它们完全反相。理解“超前”和“滞后”的含义对于解读示波器轨迹或波前图至关重要。
11. Mathematical Derivation and Key Relationships | 数学推导与关键关系
A solid grasp of the calculus linking displacement, velocity and acceleration is highly beneficial for A Level Physics. Starting from x = A cos(ωt), velocity is the first derivative: v = dx/dt = -ωA sin(ωt). Acceleration is the second derivative: a = d²x/dt² = -ω²A cos(ωt) = -ω²x. This elegantly demonstrates that SHM arises from a second-order linear differential equation.
扎实掌握联系位移、速度和加速度的微积分知识对 A Level 物理大有裨益。从 x = A cos(ωt) 出发，速度是一阶导数：v = dx/dt = -ωA sin(ωt)。加速度是二阶导数：a = d²x/dt² = -ω²A cos(ωt) = -ω²x。这优雅地表明 SHM 源于一个二阶线性微分方程。
Additionally, the relationship v = ±ω√A² - x² is not in the formula booklet but frequently appears in problems. It can be derived from the energy equations or from trigonometric identities. Practise using it to find velocity at any displacement without knowing time, and vice versa.
此外，v = ±ω√A² - x² 这个关系虽不在公式表中，但经常出现在题目中。它可以从能量方程或三角恒等式推导出来。练习在未知时间的情况下用该式求任意位移下的速度，反之亦然。
12. Exam Technique and Common Misconceptions | 考试技巧与常见误区
fy that the restoring force is proportional to displacement (and opposite in direction) before using SHM formulas. Another frequent mistake is forgetting the square in a = -ω²x, leading to an incorrect value for ω.
许多学生因混淆振幅与峰－峰位移，或将 SHM 方程应用于非 SHM 情景而失分。在使用 SHM 公式之前，始终要验证回复力是否与位移成正比且方向相反。另一个常见错误是忘记 a = -ω²x 中的平方，导致 ω 值计算错误。
When tackling multi-step problems, break the motion into phases. Use the energy method for speed at a known displacement, and the derivative method when given a displacement-time function. Always check whether the oscillation starts from equilibrium or from an extreme, as this determines whether you use sine or cosine for displacement. Pay close attention to units and significant figures as per Edexcel requirements.
处理多步问题时，将运动分解为不同阶段。对于已知位移求速率，使用能量方法；对于给定位移–时间函数的问题，使用求导方法。始终检查振动是从平衡位置还是从端点开始，这决定了位移函数用正弦还是余弦。严格按照 Edexcel 的要求注意单位与有效数字。
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