Simple Harmonic Motion: Edexcel A-Level Maths Key Points | Edexcel A-Level 数学：简谐运动考点精讲

📚 Simple Harmonic Motion: Edexcel A-Level Maths Key Points | Edexcel A-Level 数学：简谐运动考点精讲

Simple Harmonic Motion (SHM) is a core topic in Mechanics within the Edexcel A-Level Mathematics syllabus. It models oscillatory systems such as springs and pendulums, where the restoring force is proportional to the displacement from equilibrium. Mastering SHM requires fluency in differential equations, trigonometric solutions, and energy methods.

简谐运动 (SHM) 是 Edexcel A-Level 数学力学模块的核心考点，用于描述弹簧振子、单摆等恢复力与位移成正比的周期运动。掌握 SHM 需要熟练运用微分方程、三角解法和能量分析。

1. Understanding Simple Harmonic Motion | 理解简谐运动

An object performs SHM when it moves to and fro about a fixed point (the equilibrium position) and its acceleration is directly proportional to its displacement from that point, always directed towards it.

当物体在固定点（平衡位置）附近来回运动，且加速度始终指向平衡点并与位移成正比时，该物体就在做简谐运动。

The defining characteristic is that the restoring force, and hence acceleration, is always opposite in direction to the displacement. Mathematically this is expressed as a = -kx, but for SHM we set k = ω².

简谐运动的本质特征是恢复力（从而加速度）始终与位移方向相反。数学上可写为 a = -kx，在标准 SHM 中我们令 k = ω²。

Common examples include a mass on a light spring obeying Hooke’s Law, and a simple pendulum oscillating through a small angle. In the Edexcel specification, both horizontal spring systems and the simple pendulum are examined.

常见实例包括满足胡克定律的弹簧振子和小角度摆动的单摆。Edexcel 考纲同时考查水平弹簧振子和单摆模型。

2. The Fundamental Equation a = -ω²x | 基本方程 a = -ω²x

The hallmark of SHM is the linear relationship between acceleration and displacement: a = -ω²x. Here x is the displacement from the equilibrium position, ω is the angular frequency (constant for a given system), and the negative sign indicates that a and x are in opposite directions.

SHM 的标志性方程是加速度与位移的线性关系：a = -ω²x。其中 x 为相对平衡位置的位移，ω 为角频率（特定系统的常数），负号表示加速度与位移反向。

Since acceleration is the second derivative of displacement with respect to time, this can be written as the second‑order differential equation d²x/dt² = -ω²x. Students are expected to recognise this form and be able to solve it.

因为加速度是位移对时间的二阶导数，该关系可写为二阶微分方程 d²x/dt² = -ω²x。考生需要能识别该方程并求解。

a = -ω²x or d²x/dt² + ω²x = 0

a = -ω²x 或 d²x/dt² + ω²x = 0

3. Solving the Differential Equation | 解微分方程

The general solution to d²x/dt² = -ω²x involves sine and cosine functions. The two most common forms are x = A sin(ωt) and x = A cos(ωt), where A is the amplitude (maximum displacement). A full solution can include a phase constant φ, written as x = A sin(ωt + φ) or x = A cos(ωt + φ).

微分方程 d²x/dt² = -ω²x 的通解包含正弦和余弦函数。最常见的两种形式为 x = A sin(ωt) 和 x = A cos(ωt)，其中 A 为振幅（最大位移）。完整的通解可包含初相 φ，写为 x = A sin(ωt + φ) 或 x = A cos(ωt + φ)。

The choice between sine and cosine depends on the initial conditions. If the motion starts at the equilibrium position with maximum velocity, use sine. If it starts at maximum displacement from rest, use cosine.

选用正弦还是余弦取决于初始条件。若从平衡位置以最大速度开始运动，用正弦；若从最大位移处静止释放，用余弦。

To verify that x = A sin(ωt) satisfies the equation, differentiate twice. You obtain a = -Aω² sin(ωt) = -ω²x, confirming the result.

为验证 x = A sin(ωt) 满足方程，对其求二阶导数可得 a = -Aω² sin(ωt) = -ω²x，确认成立。

4. Displacement as a Function of Time | 位移作为时间的函数

The displacement can be expressed as:

位移可表示为：

x = A sin(ωt) or x = A cos(ωt) or x = A sin(ωt + φ)

In Edexcel problems, you are often given a specific form, e.g. x = 0.3 cos(2t). You must be able to identify the amplitude (0.3 m) and the angular frequency (2 rad s⁻¹) immediately.

在 Edexcel 考题中，通常会给出具体形式，如 x = 0.3 cos(2t)。必须能立即识别出振幅（0.3 m）和角频率（2 rad s⁻¹）。

The equilibrium position corresponds to x = 0. The greatest positive displacement is x = +A and the greatest negative displacement is x = -A. The motion is symmetric about the equilibrium point.

平衡位置对应 x = 0。最大正向位移为 x = +A，最大负向位移为 x = -A。运动关于平衡点对称。

When plotting x against t, a cosine curve starts at A when t = 0; a sine curve starts at 0. The amplitude A determines the height of the peaks and troughs.

绘制位移-时间图时，余弦曲线从 t=0 时的 x=A 开始；正弦曲线从 x=0 开始。振幅 A 决定波峰与波谷的高度。

5. Velocity and Acceleration in SHM | 简谐运动的速度与加速度

Velocity is obtained by differentiating displacement with respect to time. For x = A sin(ωt), v = dx/dt = Aω cos(ωt). For x = A cos(ωt), v = -Aω sin(ωt).

速度由位移对时间求导得到。若 x = A sin(ωt)，则 v = Aω cos(ωt)；若 x = A cos(ωt)，则 v = -Aω sin(ωt)。

Acceleration is the second derivative: a = dv/dt = -Aω² sin(ωt) = -ω²x. The sign of a is always opposite to the sign of x.

加速度为二阶导数：a = dv/dt = -Aω² sin(ωt) = -ω²x。加速度的符号始终与位移符号相反。

It is crucial to remember that maximum speed occurs as the particle passes through the equilibrium position (x = 0), and acceleration is zero there. Conversely, at the extreme points (x = ±A), speed is zero but acceleration magnitude is maximum (|a| = ω²A).

务必记住，物体通过平衡位置 (x=0) 时速度最大，加速度为零；在两端极限位置 (x=±A) 时速度为零，加速度大小最大 (|a| = ω²A)。

6. The Velocity-Displacement Formula | 速度-位移公式

A very powerful equation that links speed directly to displacement without involving time is v = ± ω √(A² – x²). This is derived from the identity cos²θ + sin²θ = 1, eliminating t.

一条极为有用的方程直接将速率与位移联系起来而无需时间变量：v = ± ω √(A² – x²)。该公式通过三角恒等式 cos²θ + sin²θ = 1 消去 t 推导得出。

The ± sign indicates direction of motion. Typically you take the positive root for speed (magnitude) and add a sign according to the direction of travel.

± 号表示运动方向。通常取正平方根作为速率大小，再根据运动方向添加正负号。

v = ± ω √(A² – x²)

This equation is particularly useful for finding speed at a given displacement, or finding displacement when a given speed is reached. It is frequently tested in Edexcel exams.

该公式在已知位移求速度、或已知速度求位移时特别重要，是 Edexcel 考试的常考点。

7. Period and Frequency | 周期与频率

The period T is the time taken for one complete oscillation. Since the sine and cosine functions have period 2π, the motion repeats when ωT = 2π, giving T = 2π/ω.

周期 T 是完成一次全振动所需的时间。由于正弦和余弦函数的周期为 2π，当 ωT = 2π 时运动重复，因此 T = 2π/ω。

The frequency f is the number of oscillations per unit time: f = 1/T = ω/(2π). Angular frequency ω is related to f by ω = 2πf. The unit of f is hertz (Hz), T is in seconds (s), ω in rad s⁻¹.

频率 f 是单位时间内的振动次数：f = 1/T = ω/(2π)。角频率 ω 与 f 的关系为 ω = 2πf。f 的单位是赫兹 (Hz)，T 为秒 (s)，ω 为弧度每秒 (rad s⁻¹)。

T = 2π/ω f = ω/(2π) ω = 2πf

In many problems you will be given T or f and must first find ω before applying other SHM formulas.

很多问题会给出 T 或 f，需要先求出 ω，再应用其他 SHM 公式。

8. Maximum Values of Displacement, Velocity and Acceleration | 位移、速度与加速度的最大值

Quantity	Maximum value	Occurs at
Displacement x	A (amplitude)	Extreme positions
Speed \|v\|	ωA	x = 0 (equilibrium)
Acceleration \|a\|	ω²A	x = ±A (extremes)

v_max = ωA a_max = ω²A

Knowing these relations is essential for sketching graphs and for solving problems where extreme values are given. For example, if you are told the maximum speed is 5 m s⁻¹ and the amplitude is 0.2 m, you can find ω = v_max / A = 25 rad s⁻¹.

掌握这些关系对绘制图像和解答极值类问题至关重要。例如，已知最大速率为 5 m s⁻¹，振幅为 0.2 m，便可求得 ω = v_max / A = 25 rad s⁻¹。

9. Energy Considerations | 能量分析

For a horizontal spring-mass system undergoing SHM, the total mechanical energy is conserved and continuously exchanges between kinetic energy (KE) and elastic potential energy (PE). At any instant:

对于水平弹簧振子的 SHM，总机械能守恒，并在动能 (KE) 和弹性势能 (PE) 之间不断转化。任意时刻有：

KE = ½ m v² = ½ m ω² (A² – x²)

PE = ½ m ω² x²

Total Energy T.E. = ½ m ω² A²

Note that the total energy depends on the square of the amplitude and the square of the angular frequency. It remains constant provided no external damping forces act.

总能量与振幅的平方和角频率的平方成正比。若无阻尼外力作用，总能量将保持恒定。

At the equilibrium position, all energy is kinetic; at the extremes, all energy is potential. This principle can be used to find speed without calculus when x is known.

在平衡位置，能量全部为动能；在极限位置，能量全部为势能。利用此原理可在已知 x 时直接求速率而无需微积分。

10. Horizontal Spring-Mass System | 水平弹簧振子系统

A mass m attached to a light spring of stiffness k oscillates horizontally on a smooth surface. The restoring force is Hooke’s Law: F = -kx. Using Newton’s second law, m a = -kx, leading to a = -(k/m) x. Comparing with a = -ω²x gives ω² = k/m.

质量为 m 的物体连接在劲度系数为 k 的轻弹簧上，在光滑水平面上振动。恢复力为胡克定律：F = -kx。由牛顿第二定律 ma = -kx，得 a = -(k/m)x。与 a = -ω²x 对比，得 ω² = k/m。

Hence the period of oscillation is T = 2π √(m/k). This is independent of amplitude, making the system isochronous for small amplitudes where Hooke’s Law holds perfectly.

因此振动周期为 T = 2π √(m/k)。周期与振幅无关，只要在胡克定律严格成立的微小振幅范围内，系统就具有等时性。

ω = √(k/m) T = 2π √(m/k)

Remember to convert all quantities to SI units: k in N m⁻¹, m in kg. When the spring is hung vertically, the equilibrium position shifts but the SHM parameters ω and T remain the same.

注意所有物理量均使用国际单位：k 为 N m⁻¹，m 为 kg。若弹簧竖直悬挂，平衡位置虽会移动，但 SHM 的 ω 和 T 不变。

11. Simple Pendulum (Small Angle Approximation) | 单摆（小角度近似）

For a simple pendulum of length l and a bob of mass m, the restoring force for small angular displacements θ (θ ≈ sin θ, in radians) is proportional to displacement along the arc. The resulting equation is d²θ/dt² = -(g/l) θ, which is SHM in terms of angular displacement.

对于摆长为 l、摆球质量为 m 的单摆，在小角度位移（θ ≈ sin θ，θ 以弧度为单位）下，沿弧线的恢复力与位移成正比，运动方程为 d²θ/dt² = -(g/l) θ，即角位移的 SHM。

Hence ω = √(g/l) and the period T = 2π √(l/g). The period depends only on length and gravitational field strength, not on mass or amplitude (for small angles).

由此可得 ω = √(g/l)，周期 T = 2π √(l/g)。周期只取决于摆长和重力场强度，与质量及（小角度下）振幅无关。

ω = √(g/l) T = 2π √(l/g)

In Edexcel questions, you may be given the period of a pendulum on the Moon or an unknown planet to determine g. Always use θ ≤ 10° for the small-angle approximation to be valid.

在 Edexcel 考题中，可能会给出月球或未知行星上单摆的周期，要求测定 g。确保角度 θ ≤ 10° 才能使用小角度近似。

12. Typical Exam Questions and Strategy | 典型考题与策略

Edexcel exam questions on SHM typically involve a mix of calculation, graph interpretation, and proof. A common task is: given a displacement equation x = 0.4 cos(5t), find (a) amplitude and period, (b) velocity at t = 2 s, (c) maximum acceleration, (d) speed when x = 0.1 m.

Edexcel 关于 SHM 的考题通常综合计算、图像解读与证明。常见的题目为：已知位移方程 x = 0.4 cos(5t)，求 (a) 振幅和周期，(b) t=2 s 时的速度，(c) 最大加速度，(d) 当 x=0.1 m 时的速率。

Step-by-step strategy: (1) Extract ω and A from the given function. (2) Use differentiation to obtain v and a as functions of t. (3) Apply v = ± ω √(A² – x²) for speed at a specific displacement. (4) For energy questions, write KE or PE in terms of x and use conservation. (5) For pendulum or spring problems, identify the correct ω expression first.

解题策略：(1) 从给定函数提取 ω 与 A；(2) 求导得到 v 和 a 关于 t 的函数；(3) 用 v = ± ω √(A² – x²) 求特定位移下的速率；(4) 能量问题将 KE 或 PE 表示为 x 的函数并用守恒原理；(5) 单摆或弹簧类问题先确定正确的 ω 表达式。

Also be prepared to show from first principles that a particular force law leads to SHM with a stated period. This requires equating m a to the net restoring force and rearranging into the form d²x/dt² + ω²x = 0.

此外，准备从基本原理出发证明某一力的规律导致 SHM 并给出周期。这需要令 m a 等于净恢复力，整理为 d²x/dt² + ω²x = 0 的形式。

Always include units in your final answers, and when sketching graphs, label amplitude, period, and axes clearly.

最终答案务必包含单位；绘制图像时须清晰标注振幅、周期及坐标轴。

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Simple Harmonic Motion: Edexcel A-Level Maths Key Points | Edexcel A-Level 数学：简谐运动 考点精讲