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

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

Welcome to this focused revision guide on Simple Harmonic Motion (SHM) for A-Level WJEC Mathematics. SHM is a cornerstone of mechanics, modelling systems ranging from vibrating springs to pendulums. To excel in WJEC exams, you need fluency in the defining differential equation, the various forms of solutions, the interplay between displacement, velocity and acceleration, and the application of energy principles. This guide unpacks every essential point with bilingual clarity, ensuring you can tackle both computational and proof-style questions with confidence.

欢迎阅读这篇针对A-Level WJEC数学的简谐运动 (SHM) 考点精讲。简谐运动是力学的基石，可模拟从弹簧振动到单摆的各种系统。要在WJEC考试中脱颖而出，你需要熟练掌握其定义微分方程、各类解的形式、位移、速度和加速度之间的变化关系，以及能量原理的应用。本指南以双语清晰讲解每一个核心要点，确保你能自信应对计算题和证明题。

1. Definition and Conditions for SHM | 简谐运动的定义和条件

A particle is said to perform simple harmonic motion if its acceleration is directly proportional to its displacement from a fixed equilibrium position and is always directed towards that position. This means the restoring force (and hence acceleration) opposes the displacement. The defining scalar equation, expressed in terms of a positive constant ω (angular frequency), is:

若质点加速度与它相对固定平衡位置的位移成正比，且始终指向该平衡位置，则质点在做简谐运动。这意味着恢复力（以及因此产生的加速度）与位移方向相反。用一个正常数 ω（角频率）表示的定义方程为：

a = –ω²x

The negative sign is compulsory; it guarantees that when x is positive, acceleration is negative (towards the origin), and vice versa. In WJEC mechanics, SHM is always about one-dimensional rectilinear motion along a fixed axis, typically the x-axis. The equilibrium point is taken as x = 0, and the constant ω² is always positive.

负号是必不可少的；它保证当 x 为正时，加速度为负（指向原点），反之亦然。在WJEC力学中，简谐运动始终是沿固定轴（通常是 x 轴）的一维直线运动。平衡点取为 x = 0，常数 ω² 恒为正值。

2. The Differential Equation d²x/dt² = –ω²x | 微分方程 d²x/dt² = –ω²x

Because acceleration is the second derivative of displacement with respect to time, the condition a = –ω²x is rewritten as the second-order linear ordinary differential equation:

因为加速度是位移对时间的二阶导数，条件 a = –ω²x 可改写为二阶线性常微分方程：

d²x/dt² = –ω²x

This equation is fundamental. Any problem where you derive such a relationship from Newton’s second law or energy considerations indicates SHM, and the coefficient of x immediately gives ω². For instance, if you find d²x/dt² = –25x, you can conclude ω = 5 rad/s. Recognising this structure is a key exam skill.

该方程是基础。无论你从牛顿第二定律还是从能量角度推导出这种关系，只要得到此式就表明简谐运动，x 的系数立即给出 ω²。例如，若你得到 d²x/dt² = –25x，即可断定 ω = 5 rad/s。识别这种结构是一项关键的应试技能。

It is a homogeneous equation with constant coefficients. Its characteristic equation is r² + ω² = 0, giving complex roots r = ±iω. This guarantees oscillatory solutions consisting of sines and cosines.

它是常系数齐次方程。其特征方程为 r² + ω² = 0，给出共轭复根 r = ±iω，这就保证了由正弦和余弦构成的振荡解。

3. General Solutions to the Differential Equation | 微分方程的一般解

The most general solution of d²x/dt² = –ω²x can be written in two equivalent forms. Form A uses a single trigonometric function with a phase angle:

d²x/dt² = –ω²x 的通解可以写成两种等价形式。形式A使用带相位角的单一三角函数：

x = A cos(ω t + φ) or x = A sin(ω t + φ)

Here A is the amplitude (maximum displacement) and φ is the phase constant (or initial phase), both determined by initial conditions. Form B is a linear combination of cosine and sine without an explicit phase shift:

这里 A 是振幅（最大位移），φ 是相位常数（或初相），两者均由初始条件决定。形式B是余弦与正弦的线性组合，不显含相移：

x = C cos(ω t) + D sin(ω t)

Both forms are used in WJEC past papers. The phase-angle form is more intuitive for comparing waveforms, while the C and D form is often easier to handle when initial conditions involve both displacement and velocity at t = 0. You can switch between them using the compound-angle identities.

两种形式在WJEC往年试卷中均有出现。相位角形式更直观，便于波形比较；而当 t = 0 时初始条件同时给出位移和速度时，C 和 D 形式通常更容易处理。你可以利用和角公式在两者之间转换。

4. Expressions for Displacement, Velocity and Acceleration | 位移、速度和加速度表达式

Starting from x = A cos(ω t), differentiation gives the velocity and acceleration functions. Velocity is the first derivative:

从 x = A cos(ω t) 出发，求导得到速度和加速度函数。速度是一阶导数：

v = dx/dt = –A ω sin(ω t)

Acceleration is the second derivative, or equivalently the derivative of velocity:

加速度是二阶导数，或者等价地为速度的导数：

a = dv/dt = –A ω² cos(ω t) = –ω² x

This confirms the defining SHM equation. If the chosen solution is x = A sin(ω t), the corresponding expressions are v = A ω cos(ω t) and a = –A ω² sin(ω t) = –ω² x. The velocity amplitude is Aω, and the acceleration amplitude is Aω².

这验证了简谐运动的定义方程。如果选取的解是 x = A sin(ω t)，相应的表达式为 v = A ω cos(ω t)，a = –A ω² sin(ω t) = –ω² x。速度振幅为 Aω，加速度振幅为 Aω²。

When solving problems, you may be given a specific form such as x = 0.2 cos(4t + π/3). You are expected to differentiate correctly to find velocity and acceleration, and to interpret the maximum values and the instants when the particle passes through equilibrium.

解题时，题目可能给出诸如 x = 0.2 cos(4t + π/3) 的特定形式。你需要正确求导以得到速度和加速度，并能解释最大值以及质点通过平衡位置的时刻。

5. Amplitude, Angular Frequency, Period and Frequency | 振幅、角频率、周期和频率

The amplitude A is simply the maximum distance from equilibrium, always taken as positive. The angular frequency ω (unit: rad/s) determines how rapidly the oscillations occur. It is related to the period T and ordinary frequency f by the standard circular motion relations:

振幅 A 就是离平衡位置的最大距离，始终取正值。角频率 ω（单位：rad/s）决定了振荡的快慢。它通过标准的圆周运动关系与周期 T 和普通频率 f 相联系：

ω = 2π / T and f = 1 / T = ω / (2π)

In WJEC questions, you may need to extract ω from a graph of displacement against time, or from a given equation such as x = 0.5 cos(8t). Here ω = 8 rad/s, so T = 2π/8 = π/4 seconds. Always keep track of units: amplitude in metres (or cm if consistent), time in seconds, and ω in rad/s.

在WJEC考题中，你可能需要从位移-时间图像或给定方程（如 x = 0.5 cos(8t)）中提取 ω。此时 ω = 8 rad/s，所以 T = 2π/8 = π/4 秒。始终注意单位：振幅用米（或若统一则用厘米），时间用秒，ω 用 rad/s。

It is also vital to understand that ω is not the same as frequency f. Confusing the two is a common pitfall. Remember that f is the number of complete cycles per second measured in hertz (Hz), while ω is the angular measure per second.

同样非常重要的是，要理解 ω 与频率 f 不是一回事。混淆二者是常见错误。记住 f 是每秒完整循环的次数，单位为赫兹 (Hz)；而 ω 是每秒的弧度度量。

6. Maximum Velocity and Maximum Acceleration | 最大速度和最大加速度

From the velocity equation v = –A ω sin(ω t + φ) or v = A ω cos(ω t + …), it is clear that the speed |v| reaches its maximum value when the sine or cosine term has magnitude 1. Therefore:

从速度方程 v = –A ω sin(ω t + φ) 或 v = A ω cos(ω t + …) 可清楚看出，当正弦或余弦项的绝对值为1时，速率 |v| 达到最大值。因此：

v_max = ω A

The maximum speed occurs as the particle passes through the equilibrium point x = 0, because at that instant all the energy is kinetic. Similarly, the acceleration magnitude reaches its peak when the cosine or sine term has magnitude 1:

最大速率发生在质点经过平衡点 x = 0 的时刻，因为此时全部能量为动能。类似地，加速度大小在余弦或正弦项绝对值为1时达到峰值：

a_max = ω² A

Maximum acceleration occurs at the extreme positions x = ±A, where the restoring force is greatest and velocity is zero. These two extreme values are frequently required in exam calculations, especially when linking to Newton’s second law to find maximum force or tension in a spring.

最大加速度发生在极端位置 x = ±A，此处恢复力最大而速度为零。这两个极值是考试计算中经常需要求的，尤其是在运用牛顿第二定律求解弹簧中的最大力或张力时。

7. Energy in Simple Harmonic Motion | 简谐运动中的能量

In SHM, energy continually transforms between kinetic and potential forms, but the total mechanical energy remains constant (assuming no damping). The kinetic energy is:

在简谐运动中，能量在动能和势能之间不断转化，但总机械能保持不变（假设无阻尼）。动能为：

KE = ½ m v² = ½ m ω² (A² – x²)

The potential energy associated with the restoring force can be shown to be PE = ½ m ω² x², taking the equilibrium as the zero of potential. Adding them gives the constant total energy:

与恢复力相关的势能可表示为 PE = ½ m ω² x²，取平衡位置为零势能点。两者相加得到恒定的总能量：

E_total = KE + PE = ½ m ω² A²

This expression is extremely useful. For example, you can use it to find speed at any given displacement without tackling trigonometric functions: v = ω √(A² – x²). The derived speed equation is valid regardless of whether the motion starts with a cosine or sine form, as it arises directly from energy conservation.

此式极为有用。例如，你可以利用它求任意给定位移处的速率而无需求解三角函数：v = ω √(A² – x²)。这个导出速率方程与运动是从余弦还是正弦开始无关，因为它直接来自能量守恒。

In WJEC examination contexts, energy considerations often provide a simpler alternative to differentiation. When a question asks for speed at a specific displacement, reach for ½ m v² + ½ m ω² x² = ½ m ω² A² rather than struggling with inverse trig functions and time.

在WJEC考试中，能量分析常常提供一种比求导更简便的方法。当题目要求求特定位移处的速率时，应直接使用 ½ m v² + ½ m ω² x² = ½ m ω² A²，而不是费力处理反三角函数和时间。

8. Graphical Relationships (x-t, v-t, a-t) | 位移-时间、速度-时间、加速度-时间图像关系

Graphing the SHM quantities against time reveals important phase differences. Taking x = A cos(ω t) as a reference:

将简谐运动各物理量对时间作图会显示出重要的相位差。以 x = A cos(ω t) 为参考：

The velocity v = –A ω sin(ω t) can be expressed as A ω cos(ω t + π/2), meaning the velocity leads the displacement by π/2 (quarter of a period). The acceleration a = –A ω² cos(ω t) = A ω² cos(ω t + π), so acceleration leads displacement by π (half a period), i.e., it is always opposite in sign.

速度 v = –A ω sin(ω t) 可表示为 A ω cos(ω t + π/2)，这意味着速度超前位移 π/2（四分之一周期）。加速度 a = –A ω² cos(ω t)= A ω² cos(ω t + π)，所以加速度超前位移 π（半个周期），即加速度总与位移反号。

In a WJEC graph-based question, you might be shown a displacement-time curve and asked to sketch the corresponding velocity-time or acceleration-time graphs on the same axes. Key points to mark: when x is maximum, v = 0 and a is at its minimum (most negative if x is positive); when x = 0, v is at its maximum magnitude and a = 0. Practising these sketches helps solidify your understanding of the derivatives.

在WJEC的读图题中，可能会给出位移-时间曲线，要求在同一坐标轴上画出相应的速度-时间或加速度-时间图像。需标记的关键点：当 x 最大时，v = 0，a 处于最小值（若 x 为正则 a 最负）；当 x = 0 时，v 达到最大幅度，a = 0。练习这些草图绘制有助于巩固你对导数的理解。

9. Applications: Mass-Spring and Simple Pendulum | 应用：弹簧振子和单摆

The SHM framework applies directly to two classic physical models. For a mass m attached to an ideal spring of stiffness k oscillating horizontally on a smooth surface, Newton’s second law gives m a = –k x, so a = –(k/m) x. Comparing with a = –ω² x yields:

简谐运动框架可直接应用于两个经典物理模型。对于在光滑水平面上振动的、连接在劲度系数为 k 的理想弹簧上的质量 m，牛顿第二定律给出 m a = –k x，故 a = –(k/m) x。与 a = –ω² x 比较得到：

ω = √(k / m)

Thus the period of oscillation is T = 2π √(m / k). This holds as long as the spring obeys Hooke’s law and remains within its elastic limit.

因此振荡周期为 T = 2π √(m / k)。只要弹簧遵从胡克定律且处于弹性限度内，此式即成立。

For a simple pendulum consisting of a point mass at the end of a light, inextensible string of length L, the restoring force is –mg sinθ. For small angles (θ less than about 10°, or in radians when sinθ ≈ θ), the motion approximates SHM with angular frequency:

对于由轻质、不可伸长的长度 L 的细绳悬挂质点构成的单摆，恢复力为 –mg sinθ。在小角度下（θ 约小于 10°，或在弧度制下 sinθ ≈ θ），运动近似为简谐运动，角频率为：

ω = √(g / L)

and the period T = 2π √(L / g). In WJEC mechanics, you will often be asked to derive this result from the tangential equation of motion and the small-angle approximation. Be prepared to explain why the motion is not exactly SHM for large angles.

周期 T = 2π √(L / g)。在WJEC力学中，你常会被要求从切向运动方程和小角度近似推导这一结果。准备好解释为什么大角度下的运动并非严格的简谐运动。

10. Common Exam Techniques and Pitfalls | 常见考试技巧与易错点

Approach SHM problems systematically. (1) Identify the equilibrium position and define displacement x from it. (2) Write the resultant force or acceleration in terms of x, aiming for the form a = – (positive constant) × x. This immediately confirms SHM and gives ω². (3) Write the general solution appropriate to the initial conditions — if the body is released from rest at t=0 from x = A, use x = A cos(ω t); if it passes through equilibrium with a positive velocity at t=0, use x = A sin(ω t). (4) Differentiate to obtain v and a, or use energy equations if required. (5) Check your maximum values and ensure they are physically reasonable.

系统化地处理简谐运动问题。(1) 确定平衡位置并定义相对此位置的位移 x。(2) 将合力或加速度用 x 表达，努力写成 a = – (正常数) × x 的形式。这立即确认简谐运动并给出 ω²。(3) 根据初始条件写出合适的一般解——若物体在 t=0 时从 x = A 处由静止释放，用 x = A cos(ω t)；若在 t=0 时以正速度经过平衡点，用 x = A sin(ω t)。(4) 求导获取 v 和 a，或视需要使用能量方程。(5) 检查极值，确保它们在物理上合理。

Watch out for common errors: forgetting the negative sign in a = –

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