Simple Harmonic Motion Key Points | 简谐运动考点精讲

📚 Simple Harmonic Motion Key Points | 简谐运动考点精讲

Simple Harmonic Motion (SHM) is a fundamental topic in Edexcel A Level Mathematics, specifically within the Mechanics component. Mastery of SHM requires a deep understanding of its defining equation, the sinusoidal solutions for displacement and velocity, and the ability to apply energy methods and periodic motion concepts to real-world systems such as springs and pendulums. This article provides a concise yet thorough revision of all key points you need to succeed in the exam.

简谐运动 (SHM) 是爱德思 A Level 数学力学部分的核心课题。掌握 SHM 需要深入理解其定义方程、位移与速度的正弦型解，并能将能量法和周期运动概念应用于弹簧、摆等实际系统。本文对考试重点进行精炼而全面的梳理，助你备考成功。

1. Understanding Simple Harmonic Motion | 理解简谐运动

Simple harmonic motion is a type of periodic motion where the restoring force is directly proportional to the displacement from a fixed equilibrium position and always acts towards that position. In mathematical terms, the acceleration a of a particle undergoing SHM is given by a = −ω²x, where ω is the angular frequency and x is the displacement.

简谐运动是一种周期运动，其中回复力与物体偏离固定平衡位置的位移成正比，且始终指向平衡位置。数学上，物体作 SHM 时的加速度 a 满足 a = −ω²x，其中 ω 是角频率，x 为位移。

The negative sign indicates that acceleration is always opposite in direction to displacement. This linear restoring force leads to sinusoidal functions for displacement, velocity, and acceleration over time.

负号表示加速度的方向始终与位移相反。这种线性回复力导致位移、速度和加速度随时间呈正弦函数变化。

2. The Fundamental Equation: a = −ω²x | 基本方程 a = −ω²x

At the heart of SHM is the equation a = −ω²x. This second-order differential equation arises from Newton’s second law when the net force is −kx, with k being the spring constant or equivalent. Comparing F = ma = −kx gives a = −(k/m)x, hence ω = √(k/m). For any SHM system, once you identify the effective spring constant and mass, you can find ω.

SHM 的核心是方程 a = −ω²x。该二阶微分方程源自牛顿第二定律，当合力为 −kx 时，k 为弹性系数或等效常数。由 F = ma = −kx 得 a = −(k/m)x，由此 ω = √(k/m)。对于任何 SHM 系统，找出等效弹性系数和质量即可确定 ω。

a = −ω²x

This condition implies that the motion is isochronous—the period does not depend on the amplitude of oscillation, provided the motion remains within the linear elastic limit.

该条件意味着运动具有等时性——只要在弹性限度内，周期与振幅无关。

3. General Solution for Displacement | 位移的通解

Solving the differential equation d²x/dt² = −ω²x yields the general solution x = A sin(ωt) + B cos(ωt), which can be written in more convenient forms such as x = R sin(ωt + φ) or x = R cos(ωt + φ), where R is the amplitude and φ is the phase angle. The choice of form depends on initial conditions.

解微分方程 d²x/dt² = −ω²x 可得通解 x = A sin(ωt) + B cos(ωt)，也可写成更便捷的形式：x = R sin(ωt + φ) 或 x = R cos(ωt + φ)，其中 R 为振幅，φ 为初相。形式选择取决于初始条件。

If at t = 0, the particle is at its maximum positive displacement (x = A), we use x = A cos(ωt). If it passes through the equilibrium position moving in the positive direction at t = 0, we use x = A sin(ωt). Edexcel questions often expect you to select the correct function based on the scenario.

若 t = 0 时粒子位于最大正向位移处 (x = A)，则用 x = A cos(ωt)；若 t = 0 时通过平衡位置并正向运动，则用 x = A sin(ωt)。爱德思考题常要求根据情景选择正确的函数形式。

x = A cos(ωt) or x = A sin(ωt)

4. Velocity in SHM | 简谐运动中的速度

The velocity v is the derivative of displacement with respect to time. For x = A sin(ωt), v = ωA cos(ωt) = ω√(A² − x²). A very useful form for problem-solving is v² = ω²(A² − x²). This expression shows that speed is maximum at the equilibrium position and zero at the amplitudes.

速度 v 是位移对时间的导数。若 x = A sin(ωt)，则 v = ωA cos(ωt) = ω√(A² − x²)。解题时非常实用的形式是 v² = ω²(A² − x²)。该式表明平衡位置处速度最大，振幅处速度为零。

The maximum speed v_max = ωA occurs when x = 0. The sign of v indicates the direction of motion, which is determined by the quadrant of the phase angle.

最大速度 v_max = ωA 出现在 x = 0 时。速度的符号表示运动方向，由相角的象限决定。

v = ±ω√(A² − x²)

v_max = ωA

5. Period and Frequency | 周期与频率

The period T is the time taken to complete one full oscillation. It is related to angular frequency by T = 2π/ω. Frequency f = 1/T = ω/(2π). The unit of angular frequency is rad s⁻¹. These relationships are independent of amplitude, which is a key characteristic of SHM.

周期 T 是完成一次全振动所需的时间，与角频率的关系为 T = 2π/ω。频率 f = 1/T = ω/(2π)。角频率的单位是 rad s⁻¹。这些关系与振幅无关，这是 SHM 的关键特征。

From x = A cos(ωt), the cosine function repeats every 2π, so the time interval for a complete cycle is T = 2π/ω. In mechanics problems, always ensure ω is in rad/s.

由 x = A cos(ωt)，余弦函数每 2π 重复一次，因此全周期的时间间隔为 T = 2π/ω。在力学题目中，务必确保 ω 单位为弧度/秒。

T = 2π/ω, f = ω/(2π)

6. The Mass-Spring System | 质量-弹簧系统

A classic example of SHM is a mass m attached to a light spring of force constant k oscillating horizontally on a smooth surface. The net force is F = −kx, leading to ω = √(k/m). Therefore, the period is T = 2π√(m/k). For a vertical spring with mass, the equilibrium extension changes, but the period formula remains the same because gravity adds a constant force that does not affect ω.

典型的 SHM 模型是质量为 m 的物体连接在劲度系数为 k 的轻弹簧上，在光滑水平面上振动。合力 F = −kx，由此 ω = √(k/m)，周期为 T = 2π√(m/k)。对于竖直悬挂的弹簧，虽然平衡伸长量改变，但周期公式不变，因为重力提供恒力，不影响 ω。

T = 2π√(m/k)

In exam questions, you may need to combine two springs in series or parallel, or consider a block oscillating between two springs. The key is always to determine the effective spring constant k_eff and mass m.

在试题中，你可能需要处理弹簧串联或并联，或物块在两弹簧间振动的情况。关键是确定等效弹簧系数 k_eff 与质量 m。

7. The Simple Pendulum | 单摆

A simple pendulum consists of a point mass m suspended by a light inextensible string of length L. When displaced by a small angle θ (in radians), the restoring force is −mg sin θ. For small angles, sin θ ≈ θ, giving a = −(g/L)x, where x ≈ Lθ is the arc length. Thus ω = √(g/L) and the period is T = 2π√(L/g).

单摆由通过长为 L 的轻质不可伸长的细线悬挂的质点构成。当偏移小角度 θ (弧度) 时，回复力为 −mg sin θ。对于小角度，sin θ ≈ θ，加速度 a = −(g/L)x，其中 x ≈ Lθ 为弧长。因此 ω = √(g/L)，周期为 T = 2π√(L/g)。

The approximation is valid for amplitudes less than about 10° (≈0.17 rad). The period is independent of the mass and the amplitude (for small oscillations), which is why pendulums were historically used in clocks.

这一近似适用于振幅小于约 10° (≈0.17 rad) 的情形。周期与质量及（小振动的）振幅无关，因此摆曾用于钟表。

T = 2π√(L/g)

8. Energy Considerations in SHM | 能量分析

In SHM, the total mechanical energy is conserved (assuming no damping). It can be expressed as the sum of kinetic energy (KE) and potential energy (PE). For a mass-spring system, PE = ½kx², and KE = ½mv². Using v² = ω²(A² − x²) and ω² = k/m, we find Total Energy = ½kA² = ½mω²A². This is constant, directly proportional to the square of the amplitude.

在 SHM 中，如果无阻尼，总机械能

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Simple Harmonic Motion Key Points | 简谐运动 考点精讲