IB WJEC Mathematics: Simple Harmonic Motion Key Points | IB WJEC 数学：简谐运动考点精讲

📚 IB WJEC Mathematics: Simple Harmonic Motion Key Points | IB WJEC 数学：简谐运动考点精讲

Simple harmonic motion (SHM) is a fundamental topic in the IB and WJEC Mathematics curricula, often appearing in the context of differential equations and kinematics. Mastery of SHM involves understanding the second-order differential equation, its general solution, and the relationships between displacement, velocity, acceleration, and energy. This article provides a comprehensive review of the key concepts, formulas, and exam techniques to help students excel.

简谐运动是IB和WJEC数学课程中的基本主题，常出现在微分方程和运动学内容中。掌握SHM需要理解二阶微分方程、其通解以及位移、速度、加速度和能量之间的关系。本文全面回顾核心概念、公式和考试技巧，助力学生取得优异成绩。

1. Introduction to Simple Harmonic Motion | 简谐运动简介

Simple harmonic motion is a type of periodic motion where the acceleration of an object is directly proportional to its displacement from a fixed point and is always directed towards that point. In mathematical terms, the restoring force (and hence acceleration) is proportional to the negative of the displacement.

简谐运动是一种周期运动，其加速度与物体相对于某固定点的位移成正比，且方向始终指向该固定点。用数学语言描述，即回复力（进而加速度）与位移的负值成正比。

The motion is called ‘simple’ because the restoring force is linear with respect to displacement, leading to a straightforward sinusoidal solution. Many physical systems, such as a mass on a spring or a simple pendulum (for small angles), can be modelled as SHM.

之所以称为“简谐”，是因为回复力与位移成线性关系，从而导出简洁的正弦函数解。许多物理系统，例如弹簧振子或单摆（小角度时），都可建模为简谐运动。

2. Defining Equation and Differential Equation | 定义方程与微分方程

The defining property of SHM can be expressed as: the acceleration a is given by a = -ω²x, where x is the displacement from the equilibrium position and ω is a positive constant called the angular frequency. Since acceleration is the second derivative of displacement with respect to time, this gives the fundamental differential equation:

简谐运动的定义性质可表述为：加速度a由a = -ω²x给出，其中x是相对于平衡位置的位移，ω是正的常数，称为角频率。因为加速度是位移对时间的二阶导数，由此得到基本微分方程：

d²x/dt² = -ω²x

This is a second-order linear homogeneous differential equation with constant coefficients. The term ω² must be positive to ensure the motion is oscillatory. In many exam questions, you will either be given this equation directly or be required to derive it from Newton’s second law and a linear restoring force.

这是一个常系数二阶线性齐次微分方程。ω²必须为正，以确保运动具有振荡特性。在许多考题中，要么直接给出此方程，要么需要根据牛顿第二定律和线性回复力自行推导。

The positive constant ω can be linked to the physical properties of the system, such as the spring constant k and mass m in a mass-spring system (ω² = k/m) or the length of a pendulum (ω² = g/l).

正常数ω可与系统的物理属性相关联，例如弹簧振子中的劲度系数k和质量m（ω² = k/m），或单摆的摆长（ω² = g/l）。

3. General Solution of the Differential Equation | 微分方程的通解

To solve d²x/dt² + ω²x = 0, we assume a trial solution of the form x = e^(λt). Substituting gives the characteristic equation λ² + ω² = 0, so λ = ±iω. The general solution can then be written in either of two equivalent forms:

要求解d²x/dt² + ω²x = 0，我们假设试解形如x = e^(λt)。代入后得到特征方程λ² + ω² = 0，因此λ = ±iω。通解可写为以下两种等价形式之一：

x = A cos(ωt) + B sin(ωt)

x = C sin(ωt + φ) or x = C cos(ωt + φ)

where A, B, C, and φ are constants to be determined by initial conditions. The second form explicitly shows the amplitude C and the phase angle φ. Both forms are acceptable in solutions; the choice depends on which is more convenient for the given problem.

其中A、B、C和φ均为待定常数，由初始条件确定。第二种形式明确展示了振幅C和相位角φ。两种形式在解答中均可接受；选择取决于哪种形式对给定问题更方便。

The sine and cosine forms are interchangeable due to trigonometric identities, e.g., sin(ωt + π/2) = cos(ωt). It is crucial to recognise that any expression of the form p cos(ωt) + q sin(ωt) can be combined into R sin(ωt + α) with R = √(p²+q²) and tan α = p/q.

由于三角恒等式，正弦和余弦形式可相互转换，例如 sin(ωt + π/2) = cos(ωt)。关键是要认识到，任何形如p cos(ωt) + q sin(ωt) 的表达式均可合并为 R sin(ωt + α)，其中R = √(p²+q²)，tan α = p/q。

4. Displacement, Velocity and Acceleration Functions | 位移、速度与加速度函数

If the displacement is given by x = A sin(ωt + φ), then differentiating with respect to time gives the velocity and acceleration:

若位移由x = A sin(ωt + φ)给出，则对时间求导可得速度和加速度：

v = dx/dt = Aω cos(ωt + φ)

a = dv/dt = d²x/dt² = -Aω² sin(ωt + φ) = -ω²x

These relationships show that the velocity leads the displacement by π/2 radians (quarter of a cycle), and the acceleration is always opposite in sign to the displacement. The constant ω governs how rapidly the functions oscillate.

这些关系表明，速度超前位移π/2弧度（四分之一周期），而加速度始终与位移符号相反。常数ω控制函数振荡的快慢。

If the alternative form x = A cos(ωt + φ) is used, the derivatives become v = -Aω sin(ωt + φ) and a = -Aω² cos(ωt + φ) = -ω²x, yielding the same core property that acceleration is proportional to negative displacement.

若采用另一种形式x = A cos(ωt + φ)，则导数变为v = -Aω sin(ωt + φ)，a = -Aω² cos(ωt + φ) = -ω²x，同样体现出加速度与位移负值成正比这一核心性质。

5. Amplitude, Period, Frequency and Phase | 振幅、周期、频率和相位

The amplitude A is the maximum displacement from equilibrium, always taken as a positive quantity. The angular frequency ω (in rad/s) is related to the period T (the time for one full oscillation) and the frequency f (the number of oscillations per second) by:

振幅A是相对于平衡位置的最大位移，总是取正值。角频率ω（单位 rad/s）与周期T（一次全振动所需时间）和频率f（每秒振动次数）有如下关系：

T = 2π/ω, f = 1/T = ω/(2π)

The phase angle φ (sometimes written as ε or α) determines the initial state of the motion at t = 0. For example, if φ = 0, the particle starts at x = 0 and moves in the positive direction when using x = A sin(ωt). If φ = π/2, then x = A sin(ωt + π/2) = A cos(ωt), so the particle starts at maximum displacement.

相位角φ（有时写作ε或α）决定了t = 0时的初始运动状态。例如，若φ = 0，当使用x = A sin(ωt)时，粒子从x = 0出发并沿正方向运动。若φ = π/2，则x = A sin(ωt + π/2) = A cos(ωt)，粒子从最大位移处开始运动。

These four parameters fully describe any SHM and are frequently used in exam problems to translate between a given physical situation and its mathematical model.

这四个参数可完整描述任何简谐运动，并常在考试题中用于在给定物理情境与其数学模型之间进行转换。

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

From the velocity expression v = Aω cos(ωt + φ), the maximum speed occurs when cos(ωt + φ) = ±1, i.e., when the particle passes through the equilibrium position. Thus:

由速度表达式v = Aω cos(ωt + φ)可知，当cos(ωt + φ) = ±1时，即粒子通过平衡位置时，速率达到最大。因此：

v_max = ωA

The maximum acceleration occurs when sin(ωt + φ) = ±1, i.e., at the extreme points of the motion where displacement is ±A. Since a = -ω²x, the magnitude of maximum acceleration is:

当sin(ωt + φ) = ±1，即在位移为±A的端点时，加速度达到最大。因为a = -ω²x，最大加速度的大小为：

a_max = ω²A

These two results are extremely useful for checking calculations and for solving problems where either the maximum speed or maximum acceleration is known, allowing the determination of ω and A.

这两个结果对检验计算十分有用，也在已知最大速度或最大加速度的情况下，用来求解ω和A。

7. Using Initial Conditions to Determine Constants | 利用初始条件确定常数

When a specific SHM problem is presented, initial conditions such as the displacement and velocity at t = 0 are given. These are used to find the constants in the general solution. Suppose x = A sin(ωt + φ). Then at t = 0: x₀ = A sin φ and v₀ = Aω cos φ. Dividing gives tan φ = ω x₀ / v₀, while squaring and adding gives A = √(x₀² + (v₀/ω)²).

当遇到具体的SHM问题时，通常会给定t = 0时的位移和速度等初始条件，用于确定通解中的常数。假设x = A sin(ωt + φ)，则在t = 0时：x₀ = A sin φ，v₀ = Aω cos φ。两式相除得tan φ = ω x₀ / v₀，同时通过平方相加可得A = √(x₀² + (v₀/ω)²)。

Alternatively, using the form x = P cos(ωt) + Q sin(ωt), at t = 0 we have x₀ = P and v₀ = Qω, making the constants explicit without further algebra. This form is often more efficient for solving initial-value problems.

另一种方式，采用形式x = P cos(ωt) + Q sin(ωt)，在t = 0时有x₀ = P，v₀ = Qω，使得常数一目了然，无需额外代数运算。这种形式在求解初值问题时通常更高效。

In examinations, always check the given conditions and choose the solution form that minimises the number of steps. Remember that A must be positive; if you obtain a negative A, adjust the phase by π.

考试中，应查看给定条件，选择步骤最少的解形式。请牢记A必须为正；若得到负的A，需将相位调整π。

8. Energy in Simple Harmonic Motion | 简谐运动的能量

For a particle of mass m executing SHM with spring constant k (where ω² = k/m), the total mechanical energy remains constant and is the sum of kinetic and potential energies:

对于质量为m、在劲度系数为k的弹簧（ω² = k/m）作用下做简谐运动的物体，总机械能守恒，为动能与势能之和：

E_total = ½ m v² + ½ k x² = ½ m ω² A²

At the equilibrium position, all energy is kinetic (½ m v_max² = ½ m ω² A²), while at the extremes all energy is stored as potential (½ k A² = ½ m ω² A²). The energy conservation equation can be rearranged to express velocity as a function of displacement:

在平衡位置，所有能量均为动能（½ m v_max² = ½ m ω² A²）；而在端点处，所有能量以势能形式储存（½ k A² = ½ m ω² A²）。能量守恒方程可重排为速度随位移变化的函数：

v = ± ω √(A² – x²)

This formula is particularly useful for finding the speed at any given displacement without needing the time variable. The two signs correspond to the direction of motion.

该公式在无需时间变量的情况下，特别适用于求解任意给定位移处的速率。正负号对应运动方向。

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

There is a deep geometrical link between SHM and uniform circular motion. Consider a point moving with constant angular speed ω around a circle of radius A. The projection of this point onto a diameter executes simple harmonic motion. If the projection is onto the vertical axis, the displacement is y = A sin(ωt + φ), matching the SHM equation exactly.

简谐运动与匀速圆周运动之间存在深刻的几何联系。考虑一点以恒定角速度ω沿半径为A的圆运动，该点在直径上的投影即在做简谐运动。若投影到垂直轴上，位移为y = A sin(ωt + φ)，与SHM方程完全一致。

This insight gives an alternative derivation of velocity and acceleration: the velocity of the projected point is the component of the circular velocity v_circ = ωA along the diameter; the acceleration is the centripetal acceleration a_circ = ω²A projected back onto the line, giving a = -ω²x.

这一洞见提供了速度和加速度的另一种推导：投影点的速度是圆周速度v_circ = ωA沿直径的分量；加速度则是向心加速度a_circ = ω²A投影回直线上的分量，得到a = -ω²x。

10. Example: Mass-Spring System | 实例：质量-弹簧系统

A classic example of SHM is a mass m attached to a light spring of stiffness k on a smooth horizontal surface. Using Hooke’s law F = -kx and Newton’s second law, we obtain ma = -kx, which leads to the differential equation d²x/dt² = -(k/m)x. Comparing with the standard SHM equation gives ω² = k/m, and consequently:

简谐运动的一个经典实例是，在光滑水平面上，质量为m的物体连接在一个劲度系数为k的轻质弹簧上。利用胡克定律F = -kx和牛顿第二定律，可得ma = -kx，进而导出微分方程d²x/dt² = -(k/m)x。与标准SHM方程对比，得到ω² = k/m，进而有：

T = 2π/ω = 2π √(m/k)

This analysis shows that the period depends only on the mass and the spring constant, not on the amplitude, which is a hallmark of linear SHM. Exam questions often ask for the derivation of the equation and the calculation of T or ω.

分析表明，周期仅取决于质量和弹簧劲度系数，而与振幅无关，这是线性简谐运动的标志。考试题型常要求推导方程并计算T或ω。

11. Example: Simple Pendulum | 实例：单摆

For a simple pendulum of length L and mass m swinging with small angular amplitude (so that sin θ ≈ θ), the restoring force along the arc is -mg sin θ. Using tangential acceleration a_t = L d²θ/dt², we get L d²θ/dt² = -g θ, or:

对于摆长为L、质量为m的单摆，当其以小角度振幅摆动时（满足sin θ ≈ θ），沿弧线的回复力为 -mg sin θ。利用切向加速度a_t = L d²θ/dt²，可得L d²θ/dt² = -g θ，也即：

d²θ/dt² = -(g/L) θ

This matches the SHM differential equation with angular displacement θ instead of x and ω² = g/L. Hence the period is:

这与SHM微分方程一致，只不过位移由角位移θ替代，且ω² = g/L。因此周期为：

T = 2π √(L/g)

Remember that this formula is only valid for small oscillations (typically θ < 10°). The approximation sin θ ≈ θ is essential to linearise the equation; without it the motion is not simple harmonic.

请记住此公式仅对小幅度摆动有效（一般θ < 10°）。近似sin θ ≈ θ是线性化方程的关键；没有这一近似，运动将不再是简谐运动。

12. Common Exam Questions and Tips | 常见考试题型与技巧

Typical SHM exam questions require you to: (1) show that a given physical setup leads to the SHM differential equation; (2) write down ω² from the coefficients; (3) state the period; (4) use initial conditions to find the full displacement equation; (5) find velocity or acceleration at a specific time or position.

典型SHM考题要求考生能够：(1) 证明给定的物理装置可导出SHM微分方程；(2) 由系数写出ω²；(3) 给出周期；(4) 利用初始条件求出完整的位移方程；(5) 求特定时刻或位置的速度或加速度。

A crucial step is often to recognise SHM by rewriting acceleration in the form a = – (constant) × x, and then identifying ω² as that constant. When using the velocity formula v = ±ω √(A² – x²), take care to select the correct sign based on the direction of motion described in the problem.

关键一步通常是将加速度重写为a = – (常数) × x的形式，从而识别出SHM，并将该常数定为ω²。使用速度公式v = ±ω √(A² – x²)时，务必根据题目所述运动方向正确选择正负号。

Time management is essential; practice rewriting expressions like R sin(ωt + α) and differentiating efficiently. Check that your solution satisfies the initial conditions by substitution. Finally, whenever possible, verify your maximum values using v_max = ωA or a_max = ω²A as a quick sanity check.

时间管理至关重要；练习将R sin(ωt + α)类表达式高效地进行变换与求导。通过代入初始条件检查解是否满足。最后，只要可能，用v_max = ωA或a_max = ω²A快速验证最大值，作为合理性检验。

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IB WJEC Mathematics: Simple Harmonic Motion Key Points | IB WJEC 数学：简谐运动 考点精讲