Simple Harmonic Motion | 简谐运动考点精讲

📚 Simple Harmonic Motion | 简谐运动考点精讲

Simple harmonic motion (SHM) may sound like a topic reserved for advanced physics, but at GCSE Edexcel Mathematics level, it provides a brilliant context for applying trigonometric functions. You will explore how sine and cosine graphs describe repeated oscillatory motions, such as a pendulum swinging or a mass on a spring, without needing calculus. By recognising amplitude, period, frequency and angular frequency directly from equations like y = A sin(ωt) or y = A cos(ωt + φ), you gain a deeper understanding of trigonometric modelling and are fully prepared for graph interpretation and problem-solving questions on your Edexcel exam.

简谐运动（SHM）听起来可能像是高阶物理课的内容，但在 GCSE Edexcel 数学中，它为三角函数应用提供了极佳的场景。你将学习如何用正弦和余弦图像描述重复的振荡运动，例如单摆的摆动或弹簧上物体的振动，而无需使用微积分。通过从方程 y = A sin(ωt) 或 y = A cos(ωt + φ) 中识别振幅、周期、频率和角频率，你会加深对三角函数建模的理解，并完全准备好应对 Edexcel 考试中的图像解读和问题求解题目。

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

Simple harmonic motion is a type of periodic motion where an object’s displacement from a central equilibrium position follows a sinusoidal pattern in time. In GCSE Mathematics, we model this with sine and cosine functions. The motion repeats itself at regular intervals, making it ideal for connecting real-world oscillations to trigonometric graphs.

简谐运动是一种周期运动，物体偏离中心平衡位置的位移随时间按正弦规律变化。在 GCSE 数学中，我们用正弦和余弦函数来建模。这种运动每隔固定时间重复一次，因此非常适合将现实世界中的振荡现象与三角函数图像联系起来。

Key quantities include: the maximum displacement from the centre, called the amplitude; the time taken for one complete oscillation, known as the period; and the number of complete oscillations per unit time, the frequency. These map directly onto the parameters of a sine or cosine function.

关键量包括：偏离中心的最大位移，称为振幅；完成一次完整振荡所需的时间，即周期；以及单位时间内完整振荡的次数，也就是频率。这些量直接对应正弦或余弦函数中的参数。

2. The Model: Displacement as a Sinusoidal Function | 模型：位移的正弦函数形式

In Edexcel GCSE problems, the displacement x (or y) at time t is often given by:

x = A sin(ωt) or x = A cos(ωt)

where A is the amplitude (maximum displacement) and ω (omega) is the angular frequency, measured in radians per second. The product ωt is in radians, so t must be in consistent time units. If the motion starts when the object is at the centre and moving in the positive direction, the sine function is typical; if it starts at maximum displacement, the cosine function is used.

在 Edexcel GCSE 试题中，位移 x（或 y）随时间 t 的变化通常表示为：

x = A sin(ωt) 或 x = A cos(ωt)

其中 A 是振幅（最大位移），ω（omega）是角频率，单位为弧度每秒。乘积 ωt 以弧度为单位，因此 t 必须使用一致的时间单位。若物体从中心开始向正方向运动，则常用正弦函数；若从最大位移处开始，则使用余弦函数。

The sine and cosine graphs oscillate between +A and −A. The shape is identical except for a horizontal shift of π/2 radians. This relationship is often tested when you must choose the correct equation for a given starting condition.

正弦与余弦图像在 +A 和 −A 之间振荡。除了水平移动 π/2 弧度外，形状完全相同。当需要根据给定的起始条件选择正确的方程时，常会考查这一关系。

3. Amplitude | 振幅

Amplitude is the maximum distance the object moves from its equilibrium position. In the equations x = A sin(ωt) and x = A cos(ωt), the amplitude is simply the absolute value of A. For GCSE, you need to identify the amplitude from a graph (the height of a crest or trough from the midline) or from an equation. Amplitude is always positive.

振幅是物体偏离平衡位置的最大距离。在方程 x = A sin(ωt) 和 x = A cos(ωt) 中，振幅就是 A 的绝对值。对于 GCSE，你需要能够从图像（波峰或波谷到中线的距离）或从方程中识别出振幅。振幅始终为正。

If the equation is x = −5 cos(2t), the amplitude is |−5| = 5. The negative sign indicates a reflection but does not affect the distance. In word problems, marks are often lost by ignoring units; always state amplitude with the correct unit, such as cm or m.

若方程为 x = −5 cos(2t)，则振幅为 |−5| = 5。负号表示图像反射，但不影响距离。在文字题中，常因忽略单位而失分；务必用正确单位表示振幅，如 cm 或 m。

4. Period | 周期

The period T is the time required to complete one full oscillation. For sine and cosine functions of the form sin(ωt) or cos(ωt), the period is calculated using:

T = 2π / ω

If the equation is x = 4 sin(3t), then ω = 3 and the period is 2π/3 seconds. The period is the horizontal length of one complete cycle on the graph.

周期 T 是完成一次完整振荡所需的时间。对于形如 sin(ωt) 或 cos(ωt) 的正弦和余弦函数，周期计算公式为：

T = 2π / ω

若方程为 x = 4 sin(3t)，则 ω = 3，周期为 2π/3 秒。周期就是图像上一个完整循环的水平长度。

Always remember that ω must be in radians per unit time; if ω is given as a number without units, it is understood to be rad/s. In GCSE questions, you might be asked to compare periods of different motions or find how many oscillations occur in a given time by using T.

务必记住 ω 的单位是弧度每单位时间；若 ω 给出的是一个数而没有单位，则隐含单位为 rad/s。在 GCSE 问题中，你可能需要比较不同运动的周期，或通过 T 求在给定时间内发生了多少次振荡。

5. Frequency | 频率

Frequency f is the number of complete oscillations per second (or per unit time). It is the reciprocal of the period:

f = 1 / T and f = ω / (2π)

For example, if T = 0.5 s, then f = 2 oscillations per second, measured in hertz (Hz). You may need to convert between period and frequency in Edexcel questions involving timing of swings or vibrations.

频率 f 是每秒（或每单位时间）完整振荡的次数。它是周期的倒数：

f = 1 / T 且 f = ω / (2π)

例如，若 T = 0.5 s，则 f = 2 次每秒，单位是赫兹（Hz）。在涉及摆动或振动计时的 Edexcel 题目中，你可能需要在周期和频率之间进行转换。

A common error is mixing up f and ω; ω is the angular frequency (approximately 6.28 times larger than f). Ensure you use the correct formula, and never substitute f directly into the equation as ω unless the question specifies the unit is in cycles per radian.

一个常见的错误是混淆 f 和 ω；ω 是角频率（大约比 f 大 6.28 倍）。请确保使用正确的公式，除非题目明确单位是周期每弧度，否则切勿将 f 直接当作 ω 代入方程。

6. Angular Frequency (ω) | 角频率

Angular frequency ω links the mathematical description to the physical rate of oscillation. It is defined as ω = 2πf = 2π/T. In the function x = A sin(ωt), ω determines how many complete cycles fit into a 2π interval. The larger ω is, the more oscillations occur per unit time, making the graph appear ‘squashed’ horizontally.

角频率 ω 将数学描述与物理振荡速率联系起来。其定义为 ω = 2πf = 2π/T。在函数 x = A sin(ωt) 中，ω 决定了在 2π 区间内包含多少个完整循环。ω 越大，单位时间内振荡次数越多，图像在水平方向上显得更‘压缩’。

To find ω from a given graph, identify the period T first, then compute ω = 2π/T. Alternatively, count the number of cycles in a known time span and find f, then multiply by 2π. This skill is directly tested in graph interpretation questions.

要从给定的图像中求 ω，首先确定周期 T，然后计算 ω = 2π/T。或者，统计已知时间跨度内的循环数求出 f，再乘以 2π。这一技能会直接在图像解读题中考查。

7. Interpreting Graphs of SHM | 简谐运动图像解读

The displacement-time graph of SHM is a sinusoidal curve. The midline (usually x = 0) represents the equilibrium position. Peaks are at x = A and troughs at x = −A. The horizontal distance between two successive peaks or troughs equals one period T.

简谐运动的位移-时间图像是一条正弦曲线。中线（通常为 x = 0）代表平衡位置。波峰在 x = A 处，波谷在 x = −A 处。两个相邻波峰或波谷之间的水平距离等于一个周期 T。

At t = 0, if the graph passes through the origin with a positive slope, it matches a sine function; if it starts at a maximum, it corresponds to a cosine function. You should be able to label amplitude, period, maximum and minimum points, and the equation given a sketched graph.

在 t = 0 时，若图像经过原点且斜率为正，则对应正弦函数；若从最大值开始，则对应余弦函数。你应能够根据给出的草图标注振幅、周期、最大与最小值点，并写出方程。

Here is a quick reference table linking graph features to equation parameters:

Graph feature	Symbol	How to read
Maximum displacement	A	Peak height from midline
Time for one cycle	T	Distance between two peaks
Angular frequency	ω	ω = 2π/T
Starting position	φ (phase)	Shift from standard curve

8. Phase Shift and Initial Conditions | 初相与初始条件

Sometimes the oscillation does not conveniently start at x = 0 or x = A. A general form is x = A sin(ωt + φ) or x = A cos(ωt + φ), where φ (phi) is the phase constant or initial phase, measured in radians. φ shifts the graph horizontally. If φ > 0, the curve shifts to the left; if φ < 0, it shifts to the right.

有时振荡并非恰好从 x = 0 或 x = A 开始。更一般的形式为 x = A sin(ωt + φ) 或 x = A cos(ωt + φ)，其中 φ (phi) 是相位常数或初相，以弧度为单位。φ 使图像水平移动。若 φ > 0，曲线向左移；若 φ < 0，曲线向右移。

To determine φ from a graph, find the horizontal shift relative to the standard sine curve. For example, if the curve reaches its first positive peak at t = π/6 instead of at T/4, you can set up an equation to solve for φ. In GCSE problems, you may be given the initial displacement and velocity information to write the equation.

要从图像中确定 φ，找出相对于标准正弦曲线的水平移动量。例如，若曲线在 t = π/6（而非 T/4）处达到第一个正波峰，则可建立方程求解 φ。在 GCSE 问题中，你可能会被给出初始位移和速度信息来写出方程。

9. Using the Equation to Solve Problems | 用方程解题

GCSE Edexcel questions often ask you to substitute a specific time into the displacement equation to find the position of an object, or to find the time when the object is at a particular displacement. For instance, given x = 3 sin(2t), find x when t = 1.5 s. Simply calculate: x = 3 sin(3) ≈ 3 × 0.1411 = 0.423 m. Ensure your calculator is in radian mode!

GCSE Edexcel 题目常要求你将特定时间代入位移方程，求物体的位置，或找出物体处于特定位移时的时间。例如，给定 x = 3 sin(2t)，求 t = 1.5 s 时的 x。只需计算：x = 3 sin(3) ≈ 3 × 0.1411 = 0.423 m。确保计算器处于弧度模式！

To find the first time the object reaches x = 1.5 m when x = 3 sin(2t), set 3 sin(2t) = 1.5 ⇒ sin(2t) = 0.5. The principal solution is 2t = π/6 ⇒ t = π/12 s. Because the sine function gives multiple solutions, you must consider the context (first occurrence, within one period, etc.). Edexcel mark schemes expect you to list all relevant times within the given domain.

若要求物体第一次到达 x = 1.5 m 的时间（已知 x = 3 sin(2t)），则设 3 sin(2t) = 1.5 ⇒ sin(2t) = 0.5。主解为 2t = π/6 ⇒ t = π/12 s。由于正弦函数有多个解，你需要根据情境考虑（首次出现、在一个周期内等）。Edexcel 评分标准要求列出给定范围内所有相关的时间点。

10. Worked Examples | 典型例题

Example 1: A particle moves with displacement x = 5 cos(4t) cm. Find the amplitude, angular frequency, period and frequency.
Solution: Amplitude A = 5 cm; ω = 4 rad/s; period T = 2π/4 = π/2 ≈ 1.57 s; frequency f = 1/T = 2/π ≈ 0.637 Hz.

例1：一质点以位移 x = 5 cos(4t) cm 运动。求振幅、角频率、周期和频率。
解：振幅 A = 5 cm；ω = 4 rad/s；周期 T = 2π/4 = π/2 ≈ 1.57 s；频率 f = 1/T = 2/π ≈ 0.637 Hz。

Example 2: The displacement-time graph of a particle is shown with peaks at t = 0, 3, 6, … seconds and amplitude 2 cm. Write down its displacement equation using a cosine function.
Solution: The particle starts at maximum displacement, so we use cosine. Period T = 3 s, thus ω = 2π/3. The equation is x = 2 cos(2πt/3) cm.

例2：一质点的位移-时间图像显示在 t = 0, 3, 6, … 秒处出现波峰，振幅为 2 cm。用余弦函数写出其位移方程。
解：质点从最大位移开始，因此使用余弦函数。周期 T = 3 s，故 ω = 2π/3。方程为 x = 2 cos(2πt/3) cm。

Always include units and clearly state each step. Using radians is critical; many students lose marks by applying degree mode when the equation contains ωt implicitly in radians.

务必包含单位并清晰写出每一步。使用弧度至关重要；很多学生因在方程中含弧度的 ωt 时使用度数模式而失分。

11. Common Mistakes | 常见错误

Confusing ω with f: Remember ω = 2πf, not f. Using f directly in sin(ft) gives the wrong period.

混淆 ω 与 f：切记 ω = 2πf，而非 f。直接在 sin(ft) 中使用 f 会得出错误的周期。
Forgetting radian mode: In sine and cosine calculations where ωt involves π, your calculator must be in radian mode. A degree mode yields nonsensical results.

忘记弧度模式：在涉及 π 的 ωt 正余弦计算中，计算器必须处于弧度模式。度数模式会产生荒谬的结果。
Misreading amplitude: Always take the absolute value; negative sign only means reflection, not a negative amplitude.

误读振幅：始终取绝对值；负号仅表示图像反射，而非振幅为负。
Omitting units: Answers like ‘A = 4’ without cm or m are incomplete. Always include units.

遗漏单位：如“A = 4”而不带 cm 或 m 的答案为不完整。始终包含单位。
Phase shift errors: Not accounting for φ when it is present leads to incorrect starting position and timing of events.

初相错误：当表达式含有 φ 时未予以考虑，会导致起始位置和事件时间错误。

12. Exam Tips | 考试技巧

Highlight key values: On a graph, mark A, −A, and T clearly. This helps you extract information rapidly and minimises mistakes.

标注关键值：在图像上清楚标出 A、−A 和 T。这有助于快速提取信息并减少错误。
Write down given data: List amplitude, ω, T, f, and φ as you read the question. This structures your approach.

列出已知数据：读题时列出振幅、ω、T、f 和 φ。这有助于构建解题思路。
Check domain: When solving sin(ωt + φ) = k, find all solutions within the requested time interval, often 0 ≤ t ≤ T or for the first few oscillations.

检查定义域：解 sin(ωt + φ) = k 时，找出所要求时间区间内的所有解，常为 0 ≤ t ≤ T 或最初几个振荡周期。
Use graph symmetry: The sine graph’s symmetry about t = T/4, T/2, etc., helps predict multiple solutions without lengthy calculations.

利用图像对称性：正弦图像关于 t = T/4, T/2 等的对称性有助于预测多个解，而无需冗长计算。
Show working: Even if the final answer is partially wrong, method marks in Edexcel are generous when the reasoning is clear and the correct formula is applied.

展示解题过程：即使最终答案部分有误，只要思路清晰且使用了正确公式，Edexcel 的过程分也相当可观。

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Simple Harmonic Motion | 简谐运动 考点精讲