AQA Mathematics: Simple Harmonic Motion – Exam Essentials | AQA 数学：简谐运动 考点精讲

📚 AQA Mathematics: Simple Harmonic Motion – Exam Essentials | AQA 数学：简谐运动 考点精讲

Simple harmonic motion (SHM) is a core topic in AQA A-level Mathematics (Mechanics) and appears regularly in exams. It models oscillatory systems where the acceleration is directly proportional to the displacement from equilibrium and acts in the opposite direction. Understanding the governing equations, deriving parameters for spring-mass systems and pendulums, and applying energy principles are essential for success. This guide covers all key points you need to master.

简谐运动（SHM）是AQA数学A-level（力学）的核心主题，经常在考试中出现。它描述了加速度与偏离平衡位置的位移成正比、且方向相反的振动系统。理解基本方程、推导弹簧振子和单摆的参数，以及应用能量原理是取得高分的关键。本指南涵盖了你需要掌握的所有考点。

1. Definition and Core Equation | 定义与核心方程

SHM is defined by a linear restoring law: the acceleration a of a particle is directly proportional to its displacement x from the equilibrium position and is always directed towards that position. This is expressed as:

a = −ω²x

where ω is the angular frequency (rad s⁻¹) of the motion. The negative sign indicates that acceleration and displacement are in opposite directions.

简谐运动的定义基于线性回复规律：质点的加速度 a 与其相对于平衡位置的位移 x 成正比，且总是指向平衡位置。表示为：

a = −ω²x

其中 ω 是运动的角频率（单位 rad s⁻¹）。负号表示加速度与位移方向相反。

2. Solutions to the SHM Differential Equation | 简谐运动微分方程的解

The equation a = −ω²x can be written as d²x/dt² = −ω²x. The general solution is a combination of sine and cosine functions. Two standard forms commonly used in AQA exams are:

x = A cos(ωt)   or   x = A sin(ωt)

where A is the amplitude (maximum displacement). The choice between sine and cosine depends on the initial conditions. A more general form including a phase constant φ is x = A sin(ωt + φ) or x = A cos(ωt + φ).

方程 a = −ω²x 可写作 d²x/dt² = −ω²x。它的通解是正弦和余弦函数的组合。AQA 考试中常用的两种标准形式为：

x = A cos(ωt)   或   x = A sin(ωt)

其中 A 是振幅（最大位移）。选择正弦还是余弦取决于初始条件。包含相位常数 φ 的更一般形式为 x = A sin(ωt + φ) 或 x = A cos(ωt + φ)。

3. Period and Frequency | 周期与频率

The angular frequency ω is related to the period T and frequency f of the oscillation by:

T = 2π / ω   and   f = 1 / T = ω / 2π

The period T is the time for one complete oscillation (in seconds), and the frequency f is the number of oscillations per second (in hertz, Hz). Note that ω has units of rad s⁻¹ and is not the same as f.

角频率 ω 与振荡的周期 T 和频率 f 之间的关系为：

T = 2π / ω   和   f = 1 / T = ω / 2π

周期 T 是一次完整振动所需的时间（单位为秒），频率 f 是每秒振动的次数（单位为赫兹 Hz）。注意 ω 的单位是 rad s⁻¹，与 f 不同。

4. Velocity in SHM | 简谐运动中的速度

Differentiating the displacement with respect to time gives the velocity v. For x = A cos(ωt), we obtain:

v = dx/dt = −Aω sin(ωt)

A more useful form that does not involve time is derived using sin²θ + cos²θ = 1:

v = ± ω √(A² − x²)

The maximum speed occurs at the equilibrium position (x = 0) and is v_max = ωA. The speed is zero at the extreme displacements (x = ±A).

将位移对时间求导即得速度 v。对于 x = A cos(ωt)，有：

v = dx/dt = −Aω sin(ωt)

利用 sin²θ + cos²θ = 1 可重写为一个不显含时间的实用形式：

v = ± ω √(A² − x²)

最大速率出现在平衡位置（x = 0），为 v_max = ωA。在端点（x = ±A）速率为零。

5. Acceleration in SHM | 简谐运动中的加速度

Acceleration is obtained by differentiating velocity or directly from the defining equation. For x = A cos(ωt):

a = dv/dt = −Aω² cos(ωt) = −ω²x

The maximum acceleration occurs where displacement is greatest (x = ±A) and equals a_max = ω²A. At the equilibrium position, the acceleration is zero.

加速度可通过速度求导或直接从定义方程得到。对于 x = A cos(ωt)：

a = dv/dt = −Aω² cos(ωt) = −ω²x

最大加速度出现在位移最大处（x = ±A），等于 a_max = ω²A。在平衡位置，加速度为零。

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

A classic example is a mass m attached to a light spring of stiffness k on a smooth horizontal surface. The restoring force is given by Hooke’s law F = −kx. Using Newton’s second law:

m a = −kx   ⇒   a = −(k/m)x

Comparing with a = −ω²x gives ω² = k/m. Hence:

ω = √(k/m),   T = 2π √(m/k)

The amplitude is determined by how far the mass is pulled initially, not by ω.

一个典型例子是置于光滑水平面上的轻弹簧（劲度系数 k）和质量为 m 的物体。回复力由胡克定律给出 F = −kx。应用牛顿第二定律：

m a = −kx   ⇒   a = −(k/m)x

与 a = −ω²x 对比可得 ω² = k/m。因此：

ω = √(k/m),   T = 2π √(m/k)

振幅由初始拉伸量决定，与 ω 无关。

7. The Simple Pendulum | 单摆

For a simple pendulum of length L, the restoring force for small angular displacements θ (less than about 10°) is −mg sinθ ≈ −mgθ. The tangential displacement is x = Lθ, so the tangential acceleration is a = − (g/L) x. This matches SHM with ω² = g/L, giving:

ω = √(g/L),   T = 2π √(L/g)

Notice that the period depends only on the pendulum length and gravitational acceleration, not on mass or amplitude (for small angles).

对于长度为 L 的单摆，当角位移 θ 很小（约小于 10°）时，回复力为 −mg sinθ ≈ −mgθ。切向位移为 x = Lθ，因此切向加速度为 a = −(g/L)x。这符合 SHM 且 ω² = g/L，得：

ω = √(g/L),   T = 2π √(L/g)

注意周期仅取决于摆长和重力加速度，与质量和振幅（小角度时）无关。

8. Vertical Spring-Mass System | 竖直弹簧振子系统

When a mass hangs on a spring, gravity causes a static extension e given by mg = ke. If we measure displacement y downwards from this new equilibrium position, the net force is mg − k(e + y) = −ky. Newton’s second law yields m a = −ky, so a = −(k/m)y. Therefore the system still undergoes SHM with the same angular frequency ω = √(k/m). The equilibrium is simply shifted.

当重物悬挂在弹簧下时，重力会使弹簧产生一个静态伸长量 e，满足 mg = ke。若从此新平衡位置向下测量位移 y，则合力为 mg − k(e + y) = −ky。牛顿第二定律给出 m a = −ky，故 a = −(k/m)y。因此系统仍作角频率 ω = √(k/m) 的简谐运动，只是平衡位置发生了平移。

9. Energy in SHM | 简谐运动中的能量

For a horizontal spring system, the elastic potential energy is U = ½ kx² = ½ mω²x², and kinetic energy K = ½ mv². Using v² = ω²(A² − x²), we find:

K = ½ m ω² (A² − x²)

The total mechanical energy E remains constant and is

E = K + U = ½ m ω² A² = ½ k A²

Energy continuously interchanges between potential and kinetic forms. At the equilibrium position, energy is all kinetic; at extreme displacements, it is all potential. For a simple pendulum, analogous energy expressions hold with effective potential energy ½ m ω²x² (for small angles).

对于水平弹簧系统，弹性势能为 U = ½ kx² = ½ mω²x²，动能为 K = ½ mv²。利用 v² = ω²(A² − x²)，可得：

K = ½ m ω² (A² − x²)

总机械能 E 守恒，且满足

E = K + U = ½ m ω² A² = ½ k A²

能量在势能和动能之间不断转换。在平衡位置，能量全部为动能；在最大位移处，全部为势能。对于单摆（小角度），也有类似能量关系，有效势能为 ½ m ω²x²。

10. Using Initial Conditions | 利用初始条件解题

Exam questions often ask you to determine A and the appropriate form of solution from given starting conditions. Some common cases:

• If the particle starts from rest at displacement x = ±A: use x = A cos(ωt). Then v = −Aω sin(ωt) and at t=0, v=0.

  如果质点从位移 x = ±A 处由静止释放：使用 x = A cos(ωt)。此时 v = −Aω sin(ωt)，t=0 时 v=0。

• If the particle starts at equilibrium x = 0 with velocity u (e.g., given an impulse): use x = (u/ω) sin(ωt), so amplitude A = u/ω.

  如果质点从平衡位置 x = 0 以初速度 u 开始运动（如受到瞬时冲量）：使用 x = (u/ω) sin(ωt)，振幅 A = u/ω。

• For general initial displacement x₀ and velocity v₀, A and phase φ can be found from A² = x₀² + (v₀/ω)² and tan φ = ω x₀ / v₀ for the sine form, but AQA often sets simpler cases.

  对于一般初始位移 x₀ 和初速度 v₀，可通过 A² = x₀² + (v₀/ω)² 和 tan φ = ω x₀ / v₀（正弦形式）求出振幅和相位，但 AQA 通常出题会简单一些。

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11. SHM and Circular Motion | 简谐运动与圆周运动

SHM can be modelled as the projection of uniform circular motion onto a diameter. Consider a point moving in a circle of radius A with constant angular speed ω. Its displacement along one axis is x = A cos(ωt) or x = A sin(ωt), depending on the starting angular position. The velocity and acceleration components match those derived above. This link can help visualise phase and maxima.

简谐运动可以看作匀速圆周运动在直径上的投影。考虑一个半径为 A、以恒定角速度 ω 做圆周运动的点，它在某一轴上的位移为 x = A cos(ωt) 或 x = A sin(ωt)，具体取决于起始角度。速度和加速度分量与之前推导的一致。这一联系有助于理解相位和最大值。

12. Exam Tips and Common Pitfalls | 考试技巧与常见误区

Always use radians: In calculations involving ωt or differentiation, ensure your calculator is in radian mode, especially when finding time from displacement.

务必使用弧度制：在涉及 ωt 的运算或求导时，务必将计算器设为弧度模式，尤其是根据位移求时间时。

Don’t confuse f and ω: Remember ω = 2πf. A common error is to use frequency f in place of ω in the equation a = −ω²x. Check the units.

不要混淆 f 与 ω：记住 ω = 2πf。常见错误是在方程 a = −ω²x 中将频率 f 当作 ω 用。要检查单位。

Choose the right form for x: If the particle is released from rest at a positive maximum, use cosine. If it passes through equilibrium with maximum speed at t=0, use sine.

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