A-Level OCR Maths: Simple Harmonic Motion Revision | A-Level OCR 数学：简谐运动考点精讲

📚 A-Level OCR Maths: Simple Harmonic Motion Revision | 简谐运动考点精讲

Simple Harmonic Motion (SHM) is a fundamental topic in the Mechanics section of OCR A-Level Mathematics. It builds on kinematics, calculus, and energy principles to model oscillations ranging from springs to pendulums. Mastery of the definition, the differential equation, and the standard results for displacement, velocity, and acceleration is essential for exam success. This revision guide walks you through every key point, with worked examples and clear, bilingual explanations.

简谐运动（SHM）是 OCR A-Level 数学力学部分的核心专题。它综合运用了运动学、微积分和能量原理，可以描述从弹簧到单摆的振动。熟练掌握定义、微分方程以及位移、速度和加速度的标准结论，是考试取得高分的关键。本复习指南以中英双语形式，带你逐一攻克所有考点，并配有典型例题精讲。

---

1. Definition of SHM | 简谐运动的定义

An object is moving with simple harmonic motion if its acceleration is directly proportional to its displacement from a fixed point and is always directed towards that point. Mathematically, this is expressed as a ∝ -x, or more precisely:

物体做简谐运动时，其加速度与相对于某固定点（平衡位置）的位移成正比，且方向始终指向该点。数学上可表示为 a ∝ -x，更精确的方程是：

a = -ω²x

where ω is the angular frequency of the motion, a positive constant. The negative sign ensures the acceleration opposes the displacement, which is the restoring condition.

其中 ω 是运动的角频率，为一个正常数。负号保证了加速度与位移方向相反，即回复力条件。

In OCR exams, you may be given a = -kx and asked to deduce that the motion is SHM with ω = √k, so you should be comfortable recognising the form.

在 OCR 考试中，题目可能给出 a = -kx 并要求你判断该运动为简谐运动，并得出 ω = √k，因此你应能熟练识别这种形式。

---

2. Differential Equation and Its Solution | 微分方程及其解

Since acceleration is the second derivative of displacement with respect to time, we obtain the defining differential equation for SHM:

因为加速度是位移对时间的二阶导数，我们得到简谐运动的定义微分方程：

d²x/dt² = -ω²x

This is a second-order linear homogeneous differential equation with constant coefficients. Its general solution can be written in several equivalent forms:

这是一个常系数二阶线性齐次微分方程。它的通解有几种等价形式：

x = A cos(ωt) + B sin(ωt)

x = R cos(ωt + φ)   or   x = R sin(ωt + α)

where A, B, R, φ and α are constants determined by initial conditions. R represents the amplitude of the oscillation. You should be able to convert between the sum-of-sine-cosine form and the compound-angle form using trigonometric identities.

其中 A、B、R、φ 和 α 为由初始条件决定的常数。R 代表振幅。你需要能够利用三角恒等式在正弦余弦和角形式与复合角形式之间进行转换。

---

3. Displacement, Velocity and Acceleration | 位移、速度与加速度

Starting from the form x = R cos(ωt + φ), the velocity and acceleration are obtained by successive differentiation:

从 x = R cos(ωt + φ) 出发，通过逐次求导可得速度和加速度：

v = dx/dt = -ωR sin(ωt + φ)

a = d²x/dt² = -ω²R cos(ωt + φ) = -ω²x

An extremely useful relation linking velocity and displacement without explicit time dependence can be derived by integrating a = v dv/dx:

利用 a = v dv/dx 积分，可以得到一个极为有用的不含时间 t 的速度-位移关系式：

v² = ω²(R² – x²)

This equation is widely used to find speed at a given displacement directly.

该公式广泛用于直接求已知位移时的速率。

---

4. Amplitude and Phase Angle | 振幅与初相

The amplitude R is the maximum distance the particle moves from the equilibrium position. The phase angle φ (or α) determines the state of the motion at t = 0. Given initial conditions x = x₀ and v = v₀ at t = 0, we have:

振幅 R 是质点离平衡位置的最大距离。初相 φ（或 α）决定了 t = 0 时的运动状态。给定 t = 0 时 x = x₀, v = v₀，有：

x₀ = R cos φ   and   v₀ = -ωR sin φ

Squaring and adding gives R = √(x₀² + v₀²/ω²), and dividing gives tan φ = -v₀/(ωx₀). Always check the quadrant of φ based on the signs of x₀ and v₀.

平方相加得 R = √(x₀² + v₀²/ω²)，两式相除得 tan φ = -v₀/(ωx₀)。注意要根据 x₀ 和 v₀ 的符号判断 φ 的象限。

---

5. Period and Frequency | 周期与频率

The motion repeats itself after a certain time interval. The period T is the time for one complete oscillation:

运动每隔一定时间重复。周期 T 是一次完整的振动所需时间：

T = 2π/ω

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

频率 f 是单位时间内的振动次数：f = 1/T = ω/(2π)。角频率 ω 的单位是 rad s⁻¹，频率 f 的单位是赫兹（Hz）。

---

6. Maximum Speed and Maximum Acceleration | 最大速度与最大加速度

From v² = ω²(R² – x²), speed is greatest when x = 0:

由 v² = ω²(R² – x²) 可知，当 x = 0 时速率最大：

v_max = ωR

From a = -ω²x, the magnitude of acceleration is maximum at the extreme points x = ±R:

由 a = -ω²x，加速度大小在端点 x = ±R 处最大：

a_max = ω²R

These maxima are frequently tested alongside energy calculations.

这些最大值常与能量计算一起考查。

---

7. Energy in SHM | 简谐运动中的能量

For an undamped SHM system, total mechanical energy is conserved. For a mass-spring system with spring constant k, kinetic energy (KE) and potential energy (PE) are:

对于无阻尼的简谐运动系统，总机械能守恒。以弹簧振子为例，劲度系数为 k，动能和弹性势能为：

KE = ½ m v²   PE = ½ k x²

Using ω² = k/m and v² = ω²(R² – x²), the total energy simplifies to:

利用 ω² = k/m 和 v² = ω²(R² – x²)，总能量可化简为：

E_total = ½ m ω² R² = ½ k R²

At any instant, KE + PE = constant. When x = 0, all energy is kinetic; when x = ±R, all energy is potential. This energy approach provides a quick route to finding speed at a given position.

在任意时刻，动能与势能之和为常数。x = 0 时全部能量为动能；x = ±R 时全部能量为势能。这种能量方法可以快速求出任一位置处的速率。

---

8. Mass-Spring System | 弹簧振子系统

A block of mass m attached to a light spring of stiffness k on a smooth horizontal surface undergoes SHM about the equilibrium position. Comparing Hooke’s law with the definition a = -kx/m yields:

质量为 m 的物体系于劲度系数为 k 的轻弹簧，置于光滑水平面上，物块将围绕平衡位置做简谐运动。对比胡克定律和定义式 a = -kx/m，可得：

ω = √(k/m)   T = 2π√(m/k)

If the spring is vertical, the equilibrium extension is e = mg/k, and the motion is still SHM about that new equilibrium. In all cases, measure displacement x from the equilibrium position, not from the natural length.

若弹簧竖直悬挂，平衡伸长量为 e = mg/k，运动仍关于该新平衡位置做简谐运动。务必注意：任何情况下位移 x 都应从平衡位置量起，而非从原长位置量起。

---

9. Simple Pendulum | 单摆

A simple pendulum of length l, for small angular displacements (typically θ < 10°), approximately follows SHM. The restoring tangential acceleration is g sin θ ≈ g θ, and using the arc length x = lθ, we obtain:

长为 l 的单摆，在小角度位移下（通常 θ < 10°），近似做简谐运动。回复切向加速度为 g sin θ ≈ g θ，利用弧长关系 x = lθ，得到：

ω = √(g/l)   T = 2π√(l/g)

The period is independent of amplitude for small swings, a property called isochronism. OCR questions often ask you to derive ω or T by resolving forces and using the small-angle approximation.

在小幅摆动下周期与振幅无关，这一性质称为等时性。OCR 考题常要求通过受力分析并利用小角度近似，自行推导 ω 或 T。

---

10. Recognising SHM from an Equation | 从方程中识别 SHM

Sometimes the equation of motion is given in a form that needs rearranging, e.g. d²x/dt² + 4x = 0. Comparing with d²x/dt² = -ω²x gives ω² = 4, so ω = 2, and the motion is SHM. The key is that acceleration is expressed as a negative constant times displacement.

有时运动方程需要稍作变形，例如 d²x/dt² + 4x = 0。与标准形式 d²x/dt² = -ω²x 对比，得 ω² = 4，ω = 2，该运动即为简谐运动。关键点在于加速度可表示为位移乘上一个负常数。

---

11. Worked Example | 典型例题精讲

Example: A particle of mass 0.5 kg is attached to a spring of stiffness 32 N m⁻¹ on a smooth horizontal surface. It is displaced 0.2 m from equilibrium and released from rest. Find (a) the angular frequency and period, (b) the maximum speed, and (c) the speed when the displacement is 0.1 m.

例题：质量 0.5 kg 的物体系于劲度系数 32 N m⁻¹ 的弹簧，置于光滑水平面上。将物体从平衡位置拉开 0.2 m 后静止释放。求 (a) 角频率和周期, (b) 最大速率, (c) 位移为 0.1 m 时的速率。

Solution (a): ω = √(k/m) = √(32/0.5) = √64 = 8 rad s⁻¹. T = 2π/ω = 2π/8 = π/4 s.
解 (a)：ω = √(k/m) = √(32/0.5) = √64 = 8 rad s⁻¹. T = 2π/ω = 2π/8 = π/4 s.

(b): Amplitude R = 0.2 m. v_max = ωR = 8 × 0.2 = 1.6 m s⁻¹.
(b)：振幅 R = 0.2 m。v_max = ωR = 8 × 0.2 = 1.6 m s⁻¹。

(c): Using v² = ω²(R² – x²) = 64(0.2² – 0.1²) = 64(0.04 – 0.01) = 64 × 0.03 = 1.92. Thus v = √1.92 ≈ 1.386 m s⁻¹.
(c)：利用 v² = ω²(R² – x²) = 64(0.2² – 0.1²) = 64(0.04 – 0.01) = 64 × 0.03 = 1.92。因此 v = √1.92 ≈ 1.386 m s⁻¹。

Always pay attention to units and ensure that displacement x is measured from the equilibrium position.

始终注意单位一致，并确保位移 x 是从平衡位置量起。

---

12. Key Exam Tips | 考试要点提醒

• Memorise the two fundamental formulas: a = -ω²x and v² = ω²(R² – x²). They appear in nearly every SHM problem.
• Be able to derive the period of a mass-spring system and a simple pendulum from first principles.
• Check that the phase angle φ is consistent with initial velocity direction.
• When using energy methods, clearly state whether the system is undamped.
• For vertical springs, always adjust the equilibrium position to include the static extension.

• 牢记两条基本公式：a = -ω²x 和 v² = ω²(R² – x²)。几乎每道 SHM 题都会用到它们。
• 能够从基本原理出发推导弹簧振子和单摆的周期公式。
• 验证初相 φ 与初始速度方向是否一致。
• 运用能量方法时，要明确说明系统是否无阻尼。
• 对于竖直弹簧，务必计入静态伸长量来确定平衡位置。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）