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) is a key topic in the CCEA GCSE Further Mathematics Mechanics unit. It describes a special type of periodic motion where the acceleration of a particle is directly proportional to its displacement from a fixed point and always directed towards that point. Mastering SHM involves understanding displacement–time, velocity–time and acceleration–time relationships, as well as applying standard formulas to calculate period, maximum velocity and maximum acceleration.

简谐运动是 CCEA GCSE 进阶数学力学单元的核心考点。它描述了一种特殊的周期性运动：物体的加速度与其离开平衡位置的位移成正比，并且方向始终指向该平衡点。掌握简谐运动需要理解位移–时间、速度–时间和加速度–时间的关系，并能运用标准公式计算周期、最大速度和最大加速度。

1. Definition and Restoring Condition | 定义与回复条件

In SHM, the resultant force (and hence acceleration) acting on a particle must obey a = −ω²x, where x is displacement from the equilibrium position and ω is a positive constant called the angular frequency. The negative sign indicates that acceleration always opposes displacement.

在简谐运动中，作用在质点上的合力（从而加速度）必须满足 a = −ω²x，其中 x 是离开平衡位置的位移，ω 是正常数，称为角频率。负号表示加速度始终与位移反向。

The condition can also be written as d²x/dt² = −ω²x. This second-order differential equation is the defining equation of SHM.

该条件也可写作 d²x/dt² = −ω²x。这个二阶微分方程是简谐运动的定义方程。

Any system that obeys this equation will execute SHM about the equilibrium point x = 0.

任何满足该方程的系统都将围绕平衡位置 x = 0 作简谐运动。

2. Basic Solutions for Displacement | 位移的基本解

The general solution to d²x/dt² = −ω²x can be expressed as x = A cos(ωt) + B sin(ωt), where A and B are constants determined by initial conditions. In exam questions, you will often see x = A cos(ωt) or x = A sin(ωt) depending on where the particle starts.

方程 d²x/dt² = −ω²x 的通解可表示为 x = A cos(ωt) + B sin(ωt)，其中 A 与 B 是由初始条件确定的常数。考试题目中，常会见到 x = A cos(ωt) 或 x = A sin(ωt)，取决于质点的起始位置。

If the motion begins from maximum displacement (A), use x = A cos(ωt). If it starts from the equilibrium position moving in the positive direction, use x = A sin(ωt).

若运动从最大位移 A 处开始，使用 x = A cos(ωt)；若从平衡位置出发并向正方向运动，使用 x = A sin(ωt)。

The amplitude A is the greatest distance from the equilibrium position and is always taken as positive.

振幅 A 是离开平衡位置的最大距离，始终取正值。

3. Velocity in SHM | 简谐运动中的速度

Velocity is obtained by differentiating displacement with respect to time: v = dx/dt. For x = A cos(ωt), v = −Aω sin(ωt). For x = A sin(ωt), v = Aω cos(ωt).

速度由位移对时间求导得到：v = dx/dt。对于 x = A cos(ωt)，v = −Aω sin(ωt)；对于 x = A sin(ωt)，v = Aω cos(ωt)。

A particularly useful formula linking v, x and A is v² = ω²(A² − x²). This allows you to find speed at any displacement without involving time.

一条极其有用的关联 v、x 和 A 的公式是 v² = ω²(A² − x²)。这使你可以直接求取任意位移处的速率，无需涉及时间。

The maximum speed occurs when x = 0: vₘₐₓ = Aω. At the extreme positions (x = ±A), speed is zero.

最大速率发生在 x = 0 时：vₘₐₓ = Aω。在极限位置（x = ±A）处，速率为零。

4. Acceleration in SHM | 简谐运动中的加速度

Acceleration is the derivative of velocity: a = dv/dt = d²x/dt² = −ω²x. The maximum magnitude of acceleration is aₘₐₓ = ω²A, occurring at the endpoints x = ±A.

加速度是速度的导数：a = dv/dt = d²x/dt² = −ω²x。加速度的最大量值为 aₘₐₓ = ω²A，出现在端点 x = ±A 处。

At the equilibrium position, x = 0, acceleration is zero. The direction of acceleration is always towards the centre, opposing the displacement.

在平衡位置 x = 0 时，加速度为零。加速度的方向始终指向中心，与位移反向。

This linear relationship between acceleration and displacement is the hallmark of SHM and is used to determine ω from experimental data or given forces.

这种加速度与位移之间的线性关系是简谐运动的标志，可用于从实验数据或给定力中确定 ω。

5. Period and Frequency | 周期与频率

The angular frequency ω is related to the period T and frequency f by ω = 2π/T = 2πf. Therefore, the period of SHM is T = 2π/ω, independent of amplitude (isochronous).

角频率 ω 与周期 T 和频率 f 的关系为 ω = 2π/T = 2πf。因此，简谐运动的周期 T = 2π/ω，与振幅无关（等时性）。

For a mass–spring system, ω = √(k/m), giving T = 2π√(m/k). For a simple pendulum with small amplitude, ω = √(g/l), giving T = 2π√(l/g). You must know these derivations and be able to apply them.

对于质量–弹簧系统，ω = √(k/m)，故 T = 2π√(m/k)；对于小振幅单摆，ω = √(g/l)，故 T = 2π√(l/g)。你必须掌握这些推导并能加以应用。

Questions may ask you to calculate the period from given values or to find the original length of a pendulum from its motion.

题目可能要求你根据给定数值计算周期，或根据运动情况求单摆的原长。

6. Mass–Spring System | 质量–弹簧系统

A particle of mass m attached to the end of a light elastic spring which obeys Hooke’s law will execute SHM if displaced from equilibrium. The effective force is −kx, where k is the spring constant, leading to acceleration a = −(k/m)x. Hence ω² = k/m.

质量为 m 的质点固定在轻质弹性弹簧末端，弹簧满足胡克定律，若偏离平衡位置，质点将作简谐运动。回复力为 −kx，其中 k 为弹簧劲度系数，导致加速度 a = −(k/m)x，因此 ω² = k/m。

Exam tip: always take the equilibrium position as the new origin for measuring displacement in vertical spring problems. The extension due to weight does not appear in the SHM equation.

应试贴士：在竖直弹簧问题中，始终以平衡位置作为新的原点来量度位移。由重力引起的伸长量不会出现在简谐运动方程中。

You should be able to find the maximum extension and the amplitude by equating energies or using initial conditions.

你应该能够通过能量守恒或利用初始条件求出最大伸长量和振幅。

7. Simple Pendulum | 单摆

For a simple pendulum of length l, the restoring force is mg sin θ. For small angles (θ in radians), sin θ ≈ θ, and the equation of motion becomes d²θ/dt² = −(g/l)θ, which is SHM in the angular displacement θ. Hence ω² = g/l.

对于长度为 l 的单摆，回复力为 mg sin θ。对于小角度（θ 以弧度计），sin θ ≈ θ，运动方程变为 d²θ/dt² = −(g/l)θ，表明角位移 θ 作简谐运动。因此 ω² = g/l。

The period T = 2π√(l/g) is independent of the mass of the bob and the amplitude (provided small angle approximation holds).

周期 T = 2π√(l/g) 与摆锤质量和振幅无关（前提是小角度近似成立）。

Be prepared to solve problems involving seconds pendulums (T = 2 s) or to find g experimentally from measured length and period.

准备好求解涉及秒摆（T = 2 s）的问题，或通过测量摆长和周期实验测定 g 值。

8. Graphical Representation | 图像表示

You must be able to sketch and interpret displacement–time, velocity–time and acceleration–time graphs for SHM. All are sinusoidal waves.

你必须能够绘制和解读简谐运动的位移–时间、速度–时间和加速度–时间图像。所有图像均为正弦波。

For x = A sin(ωt): v = Aω cos(ωt) leads x by π/2 (quarter of a period). a = −Aω² sin(ωt) leads x by π (half a period), meaning it is in antiphase with displacement.

对于 x = A sin(ωt)：v = Aω cos(ωt) 超前 x 相位 π/2（四分之一周期）；a = −Aω² sin(ωt) 超前 x 相位 π（半个周期），即与位移反相。

Key points to label on graphs: amplitude, period, intercepts, and the relationships between the zero, maximum and minimum points of the three quantities.

图中需标注的要点：振幅、周期、截距，以及这三个量的零点、极大点与极小点之间的对应关系。

9. Energy Considerations | 能量分析

The total mechanical energy in an undamped SHM system is conserved and can be expressed as E = ½m v² + ½k x² = ½k A² = ½m ω² A², where k = mω². This is useful for finding velocities or amplitudes without using calculus.

无阻尼简谐运动系统的总机械能守恒，可表示为 E = ½m v² + ½k x² = ½k A² = ½m ω² A²，其中 k = mω²。这在无需微积分求速度或振幅时非常有用。

Kinetic energy is maximum at the equilibrium position and zero at extremes; potential energy (elastic or gravitational) is maximum at extremes and zero at equilibrium.

动能于平衡位置达到最大，于端点处为零；势能（弹性势能或重力势能）于端点处最大，于平衡位置为零。

Energy–displacement graphs are parabolas: KE is ½m ω² (A² − x²), PE is ½m ω² x². The total energy line is horizontal.

能量–位移图呈抛物线状：KE = ½m ω² (A² − x²)，PE = ½m ω² x²，而总能量线为水平线。

10. Initial Conditions and Particular Solutions | 初始条件与特解

When solving SHM problems, you often need to use initial displacement x₀ and initial velocity v₀ to determine the constants A and the phase constant φ if using the form x = A sin(ωt + φ) or x = A cos(ωt + φ).

在求解简谐运动问题时，常需利用初始位移 x₀ 和初始速度 v₀ 确定常数 A 和相位常数 φ，若使用 x = A sin(ωt + φ) 或 x = A cos(ωt + φ) 形式。

Use x(0) = A sin φ (or A cos φ) and v(0) = Aω cos φ (or −Aω sin φ) to solve for A and φ simultaneously. Often A = √(x₀² + (v₀/ω)²).

利用 x(0) = A sin φ（或 A cos φ）及 v(0) = Aω cos φ（或 −Aω sin φ）联立求解 A 和 φ。通常 A = √(x₀² + (v₀/ω)²)。

CCEA questions may give a mix of numerical and algebraic initial data, requiring you to set up the correct form.

CCEA 的考题可能给出数值与代数混合的初始数据，要求你建立正确的表达式形式。

11. Worked Example Type | 典型例题讲解

Example: A particle of mass 0.5 kg moves with SHM of amplitude 0.4 m and period 2 s. Find (a) the maximum speed, (b) speed when displacement is 0.3 m, (c) maximum acceleration.

例题：一质量为 0.5 kg 的质点作振幅 0.4 m、周期 2 s 的简谐运动。求 (a) 最大速率，(b) 当位移为 0.3 m 时的速率，(c) 最大加速度。

Solution: ω = 2π/T = π rad/s. (a) vₘₐₓ = Aω = 0.4 × π = 1.257 m s⁻¹. (b) v = ω √(A² − x²) = π √(0.4² − 0.3²) = π × √0.07 ≈ 0.831 m s⁻¹. (c) aₘₐₓ = ω² A = π² × 0.4 ≈ 3.95 m s⁻².

解答：ω = 2π/T = π rad/s。(a) vₘₐₓ = Aω = 0.4 × π = 1.257 m s⁻¹。(b) v = ω √(A² − x²) = π √(0.4² − 0.3²) = π × √0.07 ≈ 0.831 m s⁻¹。(c) aₘₐₓ = ω² A = π² × 0.4 ≈ 3.95 m s⁻²。

Always check units and give answers to 3 significant figures unless otherwise stated. Show your substitution clearly to gain method marks.

除非另有说明，始终注意单位，答案保留三位有效数字。清晰列出代换过程以获得方法分。

12. Common Mistakes and Exam Tips | 常见错误与应试技巧

Do not confuse amplitude with the total length of swing. Amplitude is measured from equilibrium to one extreme.

不要混淆振幅与总摆动距离。振幅是从平衡位置量到一个极端位置的距离。

Remember to convert degrees to radians when using small-angle approximation for pendulums, and ensure your calculator is in radian mode for ω.

在对单摆使用小角度近似时，记得将度数转换为弧度，并确保计算器处于弧度模式来计算 ω。

When using v² = ω² (A² − x²), the sign of x is irrelevant because it is squared. Speed is always positive.

使用 v² = ω² (A² − x²) 时，x 的符号无关紧要，因为作了平方。速率始终为正数。

Always define the positive direction and show that the acceleration is in the opposite direction for full marks.

始终定义正方向，并表明加速度与之反向方可获得满分。

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Simple Harmonic Motion | 简谐运动 考点精讲