📚 GCSE WJEC Mathematics: Simple Harmonic Motion – Key Points | GCSE WJEC 数学：简谐运动 考点精讲
Simple harmonic motion (SHM) is not just a physics topic; in WJEC GCSE Mathematics, it appears as an application of trigonometric functions. You will be expected to interpret and manipulate equations of the form x = A sin(ωt) or x = A cos(ωt) to describe oscillating motion. This article covers every crucial point you need to master for the exam, from understanding amplitude and period to solving for displacement and time using sine and cosine graphs.
简谐运动 (SHM) 不仅是物理课题，在 WJEC GCSE 数学中，它作为三角函数的应用出现。你需要解释并处理形如 x = A sin(ωt) 或 x = A cos(ωt) 的方程，以描述振动运动。本文涵盖你考试需要掌握的每一个关键点，从理解振幅和周期，到利用正弦和余弦图像求解位移和时间。
1. Introduction to Simple Harmonic Motion in GCSE Maths | 简谐运动在 GCSE 数学中的引入
At GCSE level, simple harmonic motion is modelled using sine or cosine functions. A particle moving back and forth about a central equilibrium point can have its displacement x from the centre expressed as a function of time t. This mathematical model allows you to predict the position at any moment.
在 GCSE 阶段，简谐运动用正弦或余弦函数来建模。一个围绕中心平衡点来回运动的质点，其相对中心的位移 x 可以表示为时间 t 的函数。这个数学模型让你能够预测任意时刻的位置。
The core idea is that the acceleration is proportional to the negative of displacement, but you will not be required to derive this. Instead, you work directly with the displacement-time equations and their graphs.
核心思想是加速度与位移的负值成正比，但你不需要推导这个关系。相反，你直接使用位移-时间方程及其图像。
2. The Basic Equation: x = A sin(ω t) and x = A cos(ω t) | 基本方程：x = A sin(ω t) 与 x = A cos(ω t)
The two most common forms are x = A sin(ω t) and x = A cos(ω t), where x is the displacement from the equilibrium position, A is the amplitude, ω is the angular frequency, and t is time. The choice between sine and cosine depends on the starting position of the oscillation.
最常见的两种形式是 x = A sin(ω t) 和 x = A cos(ω t)，其中 x 是相对于平衡位置的位移，A 是振幅，ω 是角频率，t 是时间。选择正弦还是余弦取决于振动的起始位置。
If the motion starts at the equilibrium position at t = 0, the sine form is used: x = A sin(ω t). If it starts at maximum positive displacement, the cosine form is appropriate: x = A cos(ω t).
如果运动在 t = 0 时从平衡位置开始，则使用正弦形式：x = A sin(ω t)。如果从最大正位移开始，余弦形式则合适：x = A cos(ω t)。
3. Understanding Amplitude (A) | 理解振幅 (A)
The amplitude A is the maximum displacement from the equilibrium position. It is always a positive value and is measured in metres, centimetres, or any suitable unit of length. In the equation x = 5 sin(2 t), the amplitude is 5, meaning the particle moves between +5 and -5.
振幅 A 是离平衡位置的最大位移。它总是一个正值，以米、厘米或任何适当的长度单位计量。在方程 x = 5 sin(2 t) 中，振幅为 5，意味着质点在 +5 和 -5 之间运动。
On a displacement-time graph, the amplitude is the vertical distance from the midline (t-axis) to a peak or trough. Always check the coefficient in front of the trigonometric term to identify A.
在位移-时间图像上，振幅是从中线（t 轴）到波峰或波谷的垂直距离。始终检查三角函数项前面的系数来确定 A。
4. Angular Frequency (ω) and Its Relation to Period and Frequency | 角频率 (ω) 及其与周期和频率的关系
Angular frequency ω (omega) is measured in radians per second (rad/s). It determines how rapidly the oscillation occurs. The relationship between ω, period T, and frequency f is given by:
角频率 ω（欧米伽）的单位是弧度每秒 (rad/s)。它决定了振动的快慢。ω、周期 T 和频率 f 之间的关系如下：
ω = 2π f = 2π / T
Here, T is the time for one complete cycle, and f is the number of cycles per second (Hertz). Rearranging these formulas is a key skill; for instance, T = 2π / ω and f = ω / 2π.
这里，T 是完成一个完整周期所需的时间，f 是每秒的周期数（赫兹）。重排这些公式是一项关键技能；例如，T = 2π / ω，f = ω / 2π。
Understanding ω is vital because when you see x = A sin(3t), ω = 3 rad/s, so the period is T = 2π / 3 seconds.
理解 ω 至关重要，因为当你看到 x = A sin(3t) 时，ω = 3 rad/s，所以周期为 T = 2π / 3 秒。
5. Period (T) and Frequency (f) | 周期 (T) 与频率 (f)
The period T is the time taken for one full oscillation. It is always positive and can be found from the graph as the horizontal distance between two successive peaks or two successive troughs.
周期 T 是一次完整振动所需的时间。它总为正值，并可以从图像上找到，即两个连续波峰或两个连续波谷之间的水平距离。
Frequency f is the reciprocal of the period: f = 1 / T. If a particle completes 4 oscillations in 2 seconds, f = 2 Hz and T = 0.5 s. In exam questions, you may be asked to calculate T or f directly from the equation by extracting ω.
频率 f 是周期的倒数：f = 1 / T。如果一个质点在 2 秒内完成 4 次振动，则 f = 2 Hz，T = 0.5 s。在考题中，你可能被要求通过提取 ω 直接从方程计算 T 或 f。
6. Graphing SHM: Sine and Cosine Waves | 简谐运动的图像：正弦波与余弦波
Displacement-time graphs for SHM are sine or cosine curves. The sine graph starts at 0, rises to A, returns to 0, goes to -A, and returns to 0 over one period. The cosine graph starts at A, drops to 0, goes to -A, and returns to A.
简谐运动的位移-时间图像是正弦或余弦曲线。正弦图从 0 开始，升至 A，回到 0，到 -A，一个周期后回到 0。余弦图从 A 开始，降至 0，到 -A，再回到 A。
You need to be able to sketch these graphs for given A and T, label key points, and identify the equation from a graph. Pay attention to the starting displacement and the direction of motion.
你需要能够对给定的 A 和 T 绘制这些图像的草图，标注关键点，并且能够根据图像识别方程。注意起始位移和运动方向。
7. Finding Displacement at a Given Time | 求特定时间的位移
To find the displacement x at a chosen time t, simply substitute the value of t into the equation. Ensure your calculator is in the correct angle mode. If ω is given in rad/s, use radian mode; if the question uses degrees explicitly for the argument, switch to degree mode.
要求出在选定时间 t 的位移 x，只需将 t 的值代入方程。确保你的计算器处于正确的角度模式。如果 ω 的单位是 rad/s，则使用弧度模式；如果问题明确对角度参数使用度数，则切换到角度模式。
Example: For x = 8 cos(π t), at t = 0.5 s, x = 8 cos(π × 0.5) = 8 cos(π/2) = 0. This means the particle passes through equilibrium at t = 0.5 s.
示例：对于 x = 8 cos(π t)，在 t = 0.5 s 时，x = 8 cos(π × 0.5) = 8 cos(π/2) = 0。这意味着质点在 t = 0.5 s 时通过平衡位置。
8. Solving for Time When Displacement is Given | 由位移求时间
Given a specific displacement x, you often need to find the corresponding time t. Set up the equation x = A sin(ω t) and solve for t. This involves inverse trigonometric functions: t = arcsin(x/A) / ω.
给定一个特定的位移 x，你通常需要求相应的时间 t。建立方程 x = A sin(ω t) 并求解 t。这涉及反三角函数：t = arcsin(x/A) / ω。
Remember that the sine function has multiple solutions within one period. For instance, sin(ω t) = 0.5 gives principal solutions ω t = π/6 and ω t = 5π/6. You must consider the context of the motion to choose the correct t value.
记住，在一个周期内正弦函数有多个解。例如，sin(ω t) = 0.5 给出的主解为 ω t = π/6 和 ω t = 5π/6。你必须考虑运动的情境来选择正确的 t 值。
Exam tip: always draw a quick sketch of the sine or cosine graph and mark the given displacement. This helps you find all possible times within a specified interval.
考试技巧：始终快速绘制正弦或余弦图像的草图，并标出给定的位移。这有助于你在指定区间内找出所有可能的时间。
9. Phase Difference and Cosine Form | 相位差与余弦形式
A phase shift allows you to convert between sine and cosine forms. The identity sin(θ + π/2) = cos θ means that x = A sin(ω t + π/2) is identical to x = A cos(ω t). Similarly, x = A cos(ω t – π/2) = A sin(ω t).
相位偏移允许你在正弦和余弦形式之间转换。恒等式 sin(θ + π/2) = cos θ 意味着 x = A sin(ω t + π/2) 等同于 x = A cos(ω t)。类似地，x = A cos(ω t – π/2) = A sin(ω t)。
WJEC questions may ask you to rewrite an expression from sine to cosine form or vice versa, or to interpret the starting condition. Always check the phase constant carefully.
WJEC 考题可能会要求你将表达式从正弦形式改写为余弦形式，或反之，或者解释起始条件。始终仔细检查相位常数。
10. Practical Examples and Exam-style Questions | 实例与考试型题目
Let’s work through a typical question: “A particle moves with SHM according to x = 6 sin(2t), where x is in cm and t in seconds. Find (a) the amplitude, (b) the period, (c) the displacement at t = π/3 seconds, and (d) the time when x = 3 cm for the first time.”
我们来看一个典型问题：“一个质点按照 x = 6 sin(2t) 做简谐运动，其中 x 的单位是 cm，t 为秒。求 (a) 振幅，(b) 周期，(c) t = π/3 秒时的位移，以及 (d) 第一次到达 x = 3 cm 的时间。”
Solutions: (a) Amplitude A = 6 cm. (b) ω = 2 rad/s, so T = 2π/2 = π seconds. (c) x = 6 sin(2 × π/3) = 6 sin(2π/3) = 6 × (√3/2) = 3√3 ≈ 5.20 cm. (d) Set 3 = 6 sin(2t) → sin(2t) = 0.5 → 2t = π/6 → t = π/12 seconds (first occurrence).
解答：(a) 振幅 A = 6 cm。(b) ω = 2 rad/s，所以 T = 2π/2 = π 秒。(c) x = 6 sin(2 × π/3) = 6 sin(2π/3) = 6 × (√3/2) = 3√3 ≈ 5.20 cm。(d) 设 3 = 6 sin(2t) → sin(2t) = 0.5 → 2t = π/6 → t = π/12 秒（第一次出现）。
Practise with both sine and cosine versions, and always check your calculator mode. Also be prepared for questions involving the derivative of displacement to find velocity, although at GCSE this is usually given as a new function or simply interpreted from gradient.
练习时使用正弦和余弦两种版本，并始终检查计算器模式。也要准备涉及位移导数求速度的问题，不过在 GCSE 中通常直接给出速度函数或通过斜率来解释。
11. Common Mistakes to Avoid | 常见错误与避坑指南
Mistake 1: Confusing degrees and radians. If ω contains π, radian mode is almost certainly required. Always check the context.
错误 1：混淆角度制和弧度制。如果 ω 中含有 π，几乎肯定需要弧度模式。始终根据上下文确认。
Mistake 2: Forgetting that amplitude is a positive number, even if the coefficient appears negative (e.g., x = -4 sin(t) still has amplitude 4).
错误 2：忘记振幅是正数，即使系数带有负号（例如 x = -4 sin(t) 的振幅仍是 4）。
Mistake 3: Misidentifying the period from a graph: use the horizontal distance between equivalent points, not just where the curve crosses the axis.
错误 3：从图像上错误判断周期：使用等价点之间的水平距离，而不仅仅是曲线与轴的交点。
Mistake 4: When solving for t, only giving the principal value. Always consider the symmetry of the trigonometric graph to find all solutions within the required domain.
错误 4：求解 t 时只给出主值。总要考虑三角函数图像的对称性，以找到所需定义域内的所有解。
12. Summary and Key Formulas | 总结与核心公式
Here is a quick reference of the essential formulas you must memorise for WJEC GCSE Mathematics SHM questions:
以下是你必须记住的 WJEC GCSE 数学简谐运动问题核心公式快速参考：
| Formula / 公式 | Description / 描述 |
|---|---|
| x = A sin(ω t) or x = A cos(ω t) | Displacement-time equation / 位移-时间方程 |
| ω = 2π f = 2π / T | Angular frequency relations / 角频率关系式 |
| T = 1 / f | Period–frequency link / 周期与频率的联系 |
| t = arcsin(x/A) / ω | Finding time from displacement (principal value) / 由位移求时间（主值） |
Approach every problem methodically: identify A and ω, determine if sine or cosine, sketch the graph if needed, and solve using inverse trig functions with careful attention to angle units. With practice, SHM questions become a reliable source of marks.
系统性地处理每个问题：识别 A 和 ω，确定是正弦还是余弦，必要时绘制图形草图，并用反三角函数求解，特别注意角度单位。通过练习，简谐运动题目将成为可靠的得分点。
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