Equations of s.h.m. | 简谐运动方程

📚 Equations of s.h.m. | 简谐运动方程

Simple harmonic motion (s.h.m.) is one of the most important oscillatory models in A-Level Physics. It describes systems in which the restoring force is always directed towards an equilibrium position and is proportional to the displacement from that position. The equations of s.h.m. link displacement, velocity, acceleration, time, energy and the constants of the motion.

简谐运动（s.h.m.）是 A-Level 物理中最重要的振动模型之一。它描述的系统，其回复力始终指向平衡位置，且大小与离开平衡位置的位移成正比。简谐运动方程把位移、速度、加速度、时间、能量以及运动常量联系在一起。

---

1. Defining Simple Harmonic Motion | 简谐运动的定义

In simple harmonic motion, the acceleration a of a particle is directly proportional to its displacement x from a fixed equilibrium position, and is always directed towards that position.

在简谐运动中，质点的加速度 a 与其离开固定平衡位置的位移 x 成正比，并且始终指向平衡位置。

a = −ω²x

The constant ω is called the angular frequency. The negative sign shows that acceleration and displacement are always in opposite directions.

常数 ω 称为角频率。负号表示加速度与位移的方向始终相反。

This defining relationship is the test for s.h.m. If the resultant force on an object gives an acceleration of the form a = −ω²x, the object oscillates with simple harmonic motion.

这一特征关系是判断简谐运动的标准。如果物体所受合力产生的加速度具有 a = −ω²x 的形式，该物体就做简谐运动。

---

2. The Defining Equation a = −ω²x | 特征方程 a = −ω²x

The equation a = −ω²x is fundamental. It tells us that acceleration is proportional to displacement but acts in the opposite direction. This is why the object keeps returning towards equilibrium.

方程 a = −ω²x 是最基本的。它表明加速度与位移成正比，但方向相反。这就是物体不断回到平衡位置的原因。

If a graph of acceleration a against displacement x is a straight line through the origin with a negative slope, the motion is simple harmonic. The gradient of the line is −ω².

如果以加速度 a 对位移 x 作图，得到一条过原点且斜率为负的直线，则该运动为简谐运动。图线的斜率为 −ω²。

Many CIE questions ask you to prove that a given system performs s.h.m. Start from Newton’s second law F = ma, express the resultant force in terms of displacement, and rearrange to obtain a = −(constant) × x.

许多 CIE 考题要求你证明某个系统做简谐运动。可以从牛顿第二定律 F = ma 出发，用位移表示合力，并整理得到 a = −(常数) × x。

---

3. Solutions to the SHM Equation | 简谐运动方程的解

Since acceleration is the second derivative of displacement, the defining equation can be written as d²x/dt² = −ω²x. This is a second-order differential equation.

由于加速度是位移对时间的二阶导数，特征方程可以写成 d²x/dt² = −ω²x。这是一个二阶微分方程。

x = x₀ sin(ωt + φ) or x = x₀ cos(ωt + φ)

The general solution can be expressed in sine or cosine form. The choice depends on the initial position and direction of motion at time t = 0.

通解可以表示为正弦或余弦形式。选择哪一种取决于 t = 0 时的初始位置和运动方向。

If the particle starts at maximum positive displacement, use x = x₀ cos(ωt). If it starts at equilibrium and moves in the positive direction, use x = x₀ sin(ωt).

如果质点从最大正位移处开始运动，用 x = x₀ cos(ωt)。如果从平衡位置开始并向正方向运动，用 x = x₀ sin(ωt)。

---

4. Displacement–Time Equation | 位移–时间方程

For CIE examinations, the displacement equation is usually written as x = x₀ sin ωt or x = x₀ cos ωt when the initial phase is zero. A phase constant φ shifts the curve horizontally.

在 CIE 考试中，当初相为零时，位移方程通常写作 x = x₀ sin ωt 或 x = x₀ cos ωt。初相常量 φ 使曲线水平平移。

x = x₀ sin(ωt + φ)

Here t is time, x₀ is the maximum displacement called the amplitude, and the angle (ωt + φ) is the phase. Angular frequency is related to period T by ω = 2π/T and to frequency f by ω = 2πf.

这里 t 是时间，x₀ 是最大位移，称为振幅，角度 (ωt + φ) 为相位。角频率与周期 T 的关系为 ω = 2π/T，与频率 f 的关系为 ω = 2πf。

Quantity	Symbol	SI unit
Displacement	x	m
Amplitude	x₀	m
Angular frequency	ω	rad s⁻¹
Phase constant	φ	rad
Period	T	s
Frequency	f	Hz

The displacement equation shows that x repeats every time the phase increases by 2π. This gives the period T = 2π/ω.

位移方程表明，每当相位增加 2π，x 就重复一次。由此得到周期 T = 2π/ω。

---

5. Velocity–Time Equation | 速度–时间方程

The velocity in s.h.m. can be found by differentiating displacement with respect to time. For x = x₀ sin(ωt + φ):

简谐运动中的速度可通过位移对时间求导得到。对于 x = x₀ sin(ωt + φ)：

v = dx/dt = ωx₀ cos(ωt + φ)

The maximum speed is v_max = ωx₀, occurring when the particle passes through equilibrium. The speed is zero at maximum displacement because the particle momentarily stops before changing direction.

最大速率为 v_max = ωx₀，发生在质点经过平衡位置时。在最大位移处速率为零，因为质点在改变方向前会瞬间停止。

A very useful form links speed to displacement without using time:

另一个非常有用的形式把速率与位移联系起来，不含时间：

v = ±ω√(x₀² − x²)

The ± sign shows direction depends on whether the particle is moving towards positive or negative x. The magnitude of the velocity is ω√(x₀² − x²).

正负号表示方向取决于质点朝向正 x 还是负 x 运动。速度的大小为 ω√(x₀² − x²)。

This equation is especially useful

Published by TutorHao | A-Level Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）