📚 A-Level Physics: Derivation and Application of Simple Harmonic Motion Equations | A-Level 物理：简谐运动方程的推导与运用
In A-Level Physics, Simple Harmonic Motion (SHM) is one of the most frequently tested topics in the mechanics section of CIE examinations. It appears in multiple forms: derivation questions, graph interpretation, energy analysis, and numerical calculations. A thorough understanding of where the SHM equations come from, and how to apply them with confidence, is essential for achieving an A or A* grade. This article walks you through every key derivation step by step, followed by exam-style applications and common pitfalls to avoid.
在 A-Level 物理中，简谐运动（SHM）是 CIE 考试力学部分最常考查的考点之一。它出现在多种题型中：推导题、图像分析题、能量分析题和数值计算题。深入理解 SHM 方程的来源，并能够自信地运用它们，是取得 A 或 A* 成绩的关键。本文将一步一步带你完成每一个关键推导，并配合考试风格的应用题和常见误区分析。
1. What is Simple Harmonic Motion? | 什么是简谐运动？
Simple Harmonic Motion is a special type of periodic oscillation in which the restoring force is directly proportional to the displacement from equilibrium and is always directed toward the equilibrium position. Mathematically, this requires the acceleration to satisfy the condition: acceleration is proportional to negative displacement. Examples include a mass oscillating on an ideal spring, a simple pendulum swinging through small angles, and the vibrating molecules in a solid lattice.
简谐运动是一种特殊的周期性振动，其回复力与偏离平衡位置的位移成正比，且始终指向平衡位置。从数学上，这要求加速度满足条件：加速度与负位移成正比。典型例子包括：理想弹簧上的振子、小角度摆动的单摆，以及固体晶格中振动的分子。
Not every periodic motion is SHM. For example, a ball bouncing between two walls is periodic but not simple harmonic, because the force is not proportional to displacement during the motion. It is crucial to recognise that SHM requires a linear restoring force — this is exactly what produces the sinusoidal forms of displacement, velocity, and acceleration.
并非所有周期运动都是简谐运动。例如，在两墙之间弹跳的小球是周期性的，但不是简谐运动，因为运动过程中力与位移不成正比。必须认识到，SHM 要求线性回复力——正是这一点产生了位移、速度和加速度的正弦形式。
2. The Defining Equation: a = -ω²x | 定义方程：a = -ω²x
The most compact way to define SHM is through its defining differential equation. If a particle’s displacement from equilibrium is x, then the condition for SHM is:
简谐运动最简洁的表述方式是其微分定义方程。若质点偏离平衡位置的位移为 x，则 SHM 的条件为：
a = -ω²x
Here, a is the acceleration, x is the displacement from the equilibrium position, and ω (omega) is the angular frequency measured in rad s⁻¹. The negative sign indicates that acceleration always opposes displacement — when the particle is to the right (x positive), the acceleration is directed to the left (a negative), pulling it back toward equilibrium.
其中，a 是加速度，x 是偏离平衡位置的位移，ω（欧米伽）是角频率，单位为 rad s⁻¹。负号表示加速度始终与位移方向相反——当质点位于右侧（x 为正）时，加速度指向左侧（a 为负），将其拉回平衡位置。
The angular frequency ω is related to the period T and the frequency f by: ω = 2π/T = 2πf. It is crucial to note that ω is not a speed; it measures the rate of phase change in radians per second. When the system is displaced by a larger distance, the restoring acceleration also becomes proportionally larger — this is the essence of SHM.
角频率 ω 与周期 T 和频率 f 的关系为：ω = 2π/T = 2πf。必须注意，ω 不是速度，它表示相位以弧度每秒为单位的變化速率。当系统被位移到更远处时，回复加速度也成比例地增大——这就是 SHM 的本质。
3. Derivation from the Reference Circle | 参考圆推导法
The cleanest way to derive the SHM displacement, velocity, and acceleration equations is to use the reference circle method. Imagine a particle P moving with constant angular speed ω in a circle of radius A, centred at O. A second particle Q is defined as the projection of P onto a diameter (say, the horizontal axis). As P completes one full revolution, Q oscillates back and forth along the diameter between x = +A and x = -A.
推导 SHM 位移、速度和加速度方程最清晰的方法是参考圆法。设想一个质点 P 以恒定角速度 ω 在半径为 A、圆心为 O 的圆周上运动。定义另一个质点 Q 为 P 在某一直径（例如水平轴）上的投影。当 P 完成一整圈圆周运动时，Q 沿直径在 x = +A 和 x = -A 之间来回振动。
Suppose that at time t = 0, the radius OP makes an angle φ with the horizontal axis. After time t, the angular position of P is ωt + φ. The horizontal displacement of Q is therefore the horizontal component of OP:
假设在 t = 0 时刻，半径 OP 与水平轴的夹角为 φ。经过时间 t 后，P 的角位置为 ωt + φ。因此 Q 的水平位移就是 OP 的水平分量：
x = A sin(ωt + φ)
This is the general solution of the SHM differential equation a = -ω²x. The constant A is the amplitude (maximum displacement from equilibrium), and φ is the phase constant (or initial phase), which depends on where the oscillator was at t = 0. Different starting positions simply shift the sine curve along the time axis.
这就是 SHM 微分方程 a = -ω²x 的通解。常数 A 是振幅（偏离平衡位置的最大位移），φ 是初相位（也称为相位常数），它取决于振子在 t = 0 时的初始位置。不同的起始位置只是将正弦曲线沿时间轴平移。
It is worth verifying that this x(t) satisfies a = -ω²x. Taking the second derivative of A sin(ωt + φ) with respect to time gives -ω²A sin(ωt + φ) = -ω²x, confirming that the reference circle construction indeed produces SHM. Conversely, any solution of a = -ω²x must have this sinusoidal form — this is a standard result from solving second-order linear differential equations.
值得验证 x(t) 是否满足 a = -ω²x。对 A sin(ωt + φ) 关于时间求二阶导数，得到 -ω²A sin(ωt + φ) = -ω²x，确认参考圆构造确实产生了简谐运动。反过来，a = -ω²x 的任何解都必须具有这种正弦形式——这是二阶线性微分方程的标准结论。
4. The Displacement Equation | 位移方程
The displacement of an SHM oscillator can be written in two equivalent forms, depending on the initial conditions:
简谐运动振子的位移可以写成两种等价的形式，具体取决于初始条件：
x = A sin(ωt + φ) or x = A cos(ωt + φ’)
Since sin(θ + π/2) = cos θ, the two forms differ only by a phase shift of π/2. If the oscillator starts at equilibrium and moves in the positive direction, then x = A sin(ωt) is the natural choice (φ = 0). If the oscillator starts at maximum positive displacement, then x = A cos(ωt) is more convenient (φ’ = 0). In solving problems, always choose the form that makes the initial condition simplest.
由于 sin(θ + π/2) = cos θ，两种形式仅相差 π/2 的相位。如果振子从平衡位置开始沿正方向运动，则 x = A sin(ωt) 是自然选择（φ = 0）。如果振子从最大正位移处开始，则 x = A cos(ωt) 更方便（φ’ = 0）。解题时，总是选择使初始条件最简单的形式。
The displacement-time graph of SHM is a sine (or cosine) wave. Key features to identify on the graph include: the amplitude A (peak height), the period T (horizontal distance between successive peaks), and the phase constant φ (horizontal shift). CIE examiners frequently ask you to sketch this graph or read values from it, so be precise with labelling axes and marking maximum and minimum points.
位移-时间图像是一条正弦（或余弦）曲线。图像上需要识别的关键特征包括：振幅 A（峰值高度）、周期 T（相邻波峰之间的水平距离）以及初相位 φ（水平平移量）。CIE 考官经常要求你画出此图或从中读数，因此标注坐标轴和标记最值点时要精确。
5. Deriving the Velocity Equation | 速度方程的推导
To obtain the velocity of an SHM oscillator, we differentiate the displacement equation with respect to time. Starting from x = A sin(ωt + φ):
为了得到简谐运动振子的速度，我们对位移方程关于时间求导。从 x = A sin(ωt + φ) 出发：
v = dx/dt = Aω cos(ωt + φ)
Using the identity cos²θ + sin²θ = 1, we can eliminate the time variable and express velocity directly in terms of displacement x. Since sin(ωt + φ) = x/A and cos(ωt + φ) = v/(Aω):
利用恒等式 cos²θ + sin²θ = 1，我们可以消去时间变量，直接用位移 x 表示速度。由于 sin(ωt + φ) = x/A，cos(ωt + φ) = v/(Aω)：
(v/(Aω))² + (x/A)² = 1 → v = ±ω√(A² – x²)
The ± sign indicates direction: the oscillator moves in either the positive or negative direction depending on which side of equilibrium it is on and the stage of its cycle. The magnitude of velocity is greatest when x = 0 (passing through equilibrium), giving v_max = Aω. At the turning points x = ±A, the velocity is zero — the oscillator momentarily comes to rest before reversing direction.
± 符号表示方向：振子根据其处于平衡位置哪一侧以及处于振动周期的哪个阶段，沿正方向或负方向运动。速度的大小在 x = 0（经过平衡位置）时最大，即 v_max = Aω。在转折点 x = ±A 处，速度为零——振子瞬间静止，然后反向运动。
This velocity-displacement relationship is often tested in CIE data analysis questions. If you are given a v-x graph, it has the shape of an ellipse (or a circle if the axes are scaled appropriately), and the maximum velocity occurs at zero displacement. The gradient of the x-t graph at any instant gives the instantaneous velocity — this is worth checking when analysing graphs.
这种速度-位移关系在 CIE 数据分析题中经常出现。如果给你一个 v-x 图像，其形状是椭圆（如果坐标轴按适当比例缩放，则为圆形），最大速度出现在零位移处。x-t 图像在任何时刻的切线斜率给出瞬时速度——分析图像时值得注意这一点。
6. Deriving the Acceleration Equation | 加速度方程的推导
Differentiating the velocity equation v = Aω cos(ωt + φ) with respect to time gives the acceleration:
对速度方程 v = Aω cos(ωt + φ) 关于时间求导，得到加速度：
a = dv/dt = -Aω² sin(ωt + φ) = -ω²x
This confirms that the acceleration is directly proportional to the negative displacement, which is precisely the defining equation of SHM we started with. The maximum acceleration occurs at the extreme positions x = ±A, where a_max = ω²A. At the equilibrium position (x = 0), the acceleration is zero, but this is where the velocity is greatest.
这证实了加速度与负位移成正比，这正是我们一开始给出的 SHM 定义方程。最大加速度出现在极端位置 x = ±A 处，即 a_max = ω²A。在平衡位置（x = 0）处，加速度为零，但此处速度最大。
It is important to keep the three graphs (x-t, v-t, a-t) consistent. The v-t graph is the gradient of the x-t graph, and the a-t graph is the gradient of the v-t graph. In CIE exams, you may be asked to deduce one graph from another, or to compare phase relationships: displacement and acceleration are in antiphase (a is a maximum when x is a minimum), while velocity leads displacement by π/2 radians (i.e., 90°).
保持三条图像（x-t、v-t、a-t）的一致性非常重要。v-t 图是 x-t 图的斜率，a-t 图是 v-t 图的斜率。在 CIE 考试中，可能会要求你根据一幅图像推断另一幅图像，或比较相位关系：位移与加速度反相（a 最大时 x 最小），而速度领先位移 π/2 弧度（即 90°）。
7. Energy in Simple Harmonic Motion | 简谐运动中的能量
An ideal SHM oscillator exchanges energy between kinetic and potential forms, and in the absence of damping, the total mechanical energy remains constant. At maximum displacement (x = ±A), all energy is stored as potential energy; at equilibrium (x = 0), all energy is kinetic.
理想的简谐运动振子在动能和势能之间交换能量，在无阻尼的情况下，总机械能保持不变。在最大位移处（x = ±A），所有能量以势能形式储存；在平衡位置处（x = 0），所有能量均为动能。
The restoring force for SHM is F = -mω²x = -kx, where k = mω² is the equivalent force constant (for a mass-spring system, k is the spring constant). The potential energy is the work done to bring the mass from equilibrium to displacement x:
SHM 的回复力为 F = -mω²x = -kx，其中 k = mω² 是等效力常数（对于弹簧振子系统，k 就是弹簧刚度系数）。势能是将质量从平衡位置移动到位移 x 处所做的功：
PE = ½kx² = ½mω²x²
The kinetic energy is KE = ½mv² = ½mω²(A² – x²), obtained by substituting v² = ω²(A² – x²). Therefore, the total energy is:
动能为 KE = ½mv² = ½mω²(A² – x²)，这是通过代入 v² = ω²(A² – x²) 得到的。因此总能量为：
E_total = KE + PE = ½mω²A² = ½kA²
Notice that the total energy is independent of x — it depends only on the amplitude A and the system parameters (mass and angular frequency). This means that if the amplitude is doubled, the total energy increases by a factor of four. This proportional reasoning appears frequently in multiple-choice questions, so keep it in mind.
注意总能量与 x 无关——它只取决于振幅 A 以及系统参数（质量和角频率）。这意味着如果振幅加倍，总能量变为原来的四倍。这种比例推理经常出现在选择题中，务必牢记。
8. Period of a Mass-Spring System | 弹簧振子的周期
For a mass m attached to a spring of force constant k, Newton’s second law gives F = -kx = ma. Rearranging: a = -(k/m)x. Comparing this with the defining equation a = -ω²x, we identify:
对于连接在力常数为 k 的弹簧上的质量 m，牛顿第二定律给出 F = -kx = ma。整理得：a = -(k/m)x。与定义方程 a = -ω²x 比较，我们得到：
ω² = k/m → ω = √(k/m)
Since ω = 2π/T, the period of a mass-spring system is:
由于 ω = 2π/T，弹簧振子系统的周期为：
T = 2π√(m/k)
This equation tells us that a stiffer spring (larger k) produces a shorter period (faster oscillation), while a larger mass produces a longer period. The period does not depend on the amplitude — this is an important property known as isochronism, which holds exactly for ideal mass-spring oscillators.
该方程告诉我们：弹簧越硬（k 越大），周期越短（振动越快）；质量越大，周期越长。周期与振幅无关——这是一个称为等时性的重要性质，对于理想的弹簧振子精确成立。
When a mass-spring system is placed vertically, gravity shifts the equilibrium position but does not change the period. The weight mg stretches the spring by a static extension x₀ = mg/k, and the oscillation occurs about this new equilibrium point. CIE examiners often use this setup to test whether you understand that g only affects the equilibrium position, not the oscillation frequency.
当弹簧振子在竖直方向放置时，重力会改变平衡位置，但不会改变周期。重力 mg 使弹簧拉伸一个静态伸长量 x₀ = mg/k，振动发生在这个新的平衡点附近。CIE 考官常常利用这种设置来测试你是否理解 g 只影响平衡位置，而不影响振动频率。
9. Period of a Simple Pendulum | 单摆的周期
For a simple pendulum of length L with a bob of mass m, the restoring force when displaced by a small angle θ is F = -mg sin θ. For small angles (θ less than about 10°), sin θ ≈ θ ≈ x/L, where x is the horizontal displacement. Hence:
对于长度为 L、摆锤质量为 m 的单摆，当偏离小角度 θ 时，回复力为 F = -mg sin θ。对于小角度（θ 小于约 10°），sin θ ≈ θ ≈ x/L，其中 x 是水平位移。因此：
F ≈ -mg(x/L) = -(mg/L)x → a = -(g/L)x
Comparing with a = -ω²x gives ω² = g/L, so the period of a simple pendulum is:
与 a = -ω²x 比较，得 ω² = g/L，因此单摆的周期为：
T = 2π√(L/g)
The period of a simple pendulum depends only on its length and the gravitational field strength — not on the mass of the bob or the amplitude (for small angles). This is why pendulums are useful in clocks: their period is highly predictable as long as the length is kept constant.
单摆的周期仅取决于摆长和重力场强度——与摆锤质量和振幅（小角度范围内）无关。这就是摆钟使用单摆的原因：只要长度保持不变，其周期就高度可预测。
If you are asked to determine g using a pendulum in the laboratory, plot T² against L. The gradient of the graph is 4π²/g, from which g can be calculated. Or, if only one measurement is taken, use g = 4π²L/T². CIE practical questions often involve measuring T for different L values and analysing the T²-L graph, so practise drawing straight-line graphs and calculating gradients precisely.
如果要求在实验室中用单摆测定 g，应绘制 T² 关于 L 的图像。图像斜率为 4π²/g，由此可算出 g。或者，如果只进行一次测量，使用 g = 4π²L/T²。CIE 实验题通常涉及测量不同 L 对应的 T，并分析 T²-L 图像，因此要练习绘制直线图和精确计算斜率。
10. Worked Example: Exam-Style Problem | 例题：考试风格题目
Consider the following CIE-style question. A particle of mass 0.20 kg oscillates with simple harmonic motion. The period is 0.80 s and the amplitude is 3.0 cm. Calculate: (a) the angular frequency, (b) the maximum speed, (c) the speed when the displacement is 1.5 cm, and (d) the total energy of the system.
考虑以下 CIE 风格题目。一个质量为 0.20 kg 的质点做简谐运动，周期为 0.
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