📚 Mastering Simple Harmonic Motion for A-Level Maths | 简谐运动考点精讲
Simple harmonic motion (SHM) is a fundamental topic in A-Level Mathematics, bridging mechanics and trigonometry. It describes systems where the restoring force is proportional to the displacement and always acts towards an equilibrium position. Mastering SHM requires understanding its differential equation, the sinusoidal forms of displacement, velocity and acceleration, and the physical interpretation of amplitude, period and energy. This article breaks down the key concepts, formulas and exam techniques you need, with each point clearly explained in both English and Chinese to support bilingual learners.
简谐运动是 A-Level 数学中将力学与三角学联系起来的基础课题。它描述恢复力与位移成正比且始终指向平衡位置的运动系统。要掌握简谐运动，需要理解其微分方程，位移、速度和加速度的正弦形式，以及振幅、周期和能量的物理意义。本文拆解了你需要的核心概念、公式和应试技巧，每个要点都用中英双语清晰解释，以帮助双语学生更好地掌握。
1. Definition of SHM | 简谐运动的定义
A particle undergoes simple harmonic motion if its acceleration a is directly proportional to its displacement x from a fixed equilibrium position and is always directed towards that point. This relationship is written as a ∝ −x. Introducing a positive constant ω², we obtain the fundamental equation a = −ω²x. The negative sign indicates that acceleration and displacement are always opposite in direction.
如果一个质点的加速度 a 与其离固定平衡位置的位移 x 成正比，且始终指向该点，则该质点做简谐运动。这个关系可写为 a ∝ −x。引入一个正常数 ω²，我们得到基本方程 a = −ω²x。负号表示加速度和位移的方向总是相反。
a = −ω²x
The constant ω (omega) is called the angular frequency and determines how quickly the motion oscillates. The larger ω, the greater the acceleration for a given displacement, and the more rapid the oscillation.
常数 ω 称为角频率，它决定振动的快慢。ω 越大，给定位移下的加速度就越大，振动也就越迅速。
2. The Differential Equation of SHM | 简谐运动的微分方程
Since acceleration is the second derivative of displacement with respect to time, we can write d²x/dt² = −ω²x. This is the standard second-order linear differential equation for SHM. Its general solution involves sine and cosine functions. Depending on the initial conditions, the displacement can be expressed as:
由于加速度是位移对时间的二阶导数，我们可以写出 d²x/dt² = −ω²x。这是简谐运动的标准二阶线性微分方程。其通解包含正弦和余弦函数。根据初始条件的不同，位移可以表示为：
x = A sin(ωt) or x = A cos(ωt) or x = A sin(ωt + φ)
If the motion starts from the equilibrium position (x=0) at t=0, the solution x = A sin(ωt) is usually preferred. If it starts at maximum displacement (x=A), then x = A cos(ωt) is more natural. The constant φ (phi) is the initial phase angle and allows any starting point to be described. At A-Level, you are often expected to form and solve the differential equation and then apply boundary conditions to find A and φ.
如果运动在 t=0 时从平衡位置（x=0）开始，通常采用 x = A sin(ωt)；如果从最大位移（x=A）开始，则用 x = A cos(ωt) 更自然。常数 φ 是初相位，它允许描述任意起始位置。在 A-Level 中，往往要求你建立并求解该微分方程，然后应用边界条件求出 A 和 φ。
3. Displacement, Velocity and Acceleration | 位移、速度和加速度
Taking x = A sin(ωt) as an example, we can find velocity v and acceleration a by differentiating. Velocity is the first derivative: v = dx/dt = Aω cos(ωt). Acceleration is the second derivative: a = d²x/dt² = −Aω² sin(ωt) = −ω²x. Notice that v and x are π/2 out of phase: when x = 0, v = ±Aω (maximum); when x = ±A, v = 0.
以 x = A sin(ωt) 为例，我们可以通过微分求出速度 v 和加速度 a。速度是一阶导数：v = dx/dt = Aω cos(ωt)。加速度是二阶导数：a = d²x/dt² = −Aω² sin(ωt) = −ω²x。注意 v 与 x 相位差 π/2：当 x = 0 时，v = ±Aω（最大）；当 x = ±A 时，v = 0。
If you prefer the cosine form x = A cos(ωt), then v = −Aω sin(ωt) and a = −Aω² cos(ωt) = −ω²x. The relationship a = −ω²x remains the hallmark of SHM in all cases, independent of the chosen form.
如果你选用余弦形式 x = A cos(ωt)，则 v = −Aω sin(ωt)，a = −Aω² cos(ωt) = −ω²x。无论哪种形式，a = −ω²x 都是简谐运动的标志性关系。
The three quantities—displacement, velocity, acceleration—can also be linked without explicit time: v² = ω²(A² − x²). This is a very useful energy-style relationship that avoids dealing with trigonometric functions.
位移、速度和加速度也可以通过不含时间的公式联系起来：v² = ω²(A² − x²)。这是一个极为实用的、类似于能量的关系式，可以避免处理三角函数。
4. Amplitude, Period and Frequency | 振幅、周期和频率
Amplitude A is the maximum displacement from equilibrium. It is always positive and depends on the initial energy of the system. The period T is the time taken for one complete oscillation, and the frequency f is the number of oscillations per second. They are related to the angular frequency ω by:
振幅 A 是离开平衡位置的最大位移。它总是正值，且取决于系统的初始能量。周期 T 是完成一次全振动所需的时间，频率 f 是每秒振动的次数。它们与角频率 ω 的关系为：
T = 2π/ω and f = 1/T = ω/(2π)
Therefore, ω = 2πf. In many A-Level problems, you will be given T or f and must find ω before writing the displacement equation. The SI unit of ω is rad/s.
因此，ω = 2πf。在许多 A-Level 题目中，你会已知 T 或 f，必须先求出 ω 才能写出位移方程。角频率 ω 的国际单位是 rad/s。
For a given SHM system, the period T is independent of the amplitude—a property called isochronism—provided the motion remains simple harmonic. This is true for the mass-spring system and the simple pendulum at small angles.
对于给定的简谐运动系统，只要运动保持简谐，周期 T 与振幅无关，这一性质被称为等时性。对于弹簧振子和小角度单摆，这都成立。
5. Phase and Phase Difference | 相位与相位差
The phase of SHM at time t is the argument of the sine or cosine function, e.g., (ωt + φ). The constant φ is the initial phase (at t=0). Phase difference describes how much one oscillation leads or lags another. For example, in x = A sin(ωt), the velocity v = Aω cos(ωt) can be rewritten as v = Aω sin(ωt + π/2), showing that velocity leads displacement by π/2. Acceleration leads velocity by π/2, but is exactly out of phase (π) with displacement.
在时刻 t 的相位是正弦或余弦函数的自变量，例如 (ωt + φ)。常数 φ 是初相位。相位差描述一个振动超前或滞后于另一个振动的程度。例如，在 x = A sin(ωt) 中，速度 v = Aω cos(ωt) 可以改写为 v = Aω sin(ωt + π/2)，这表明速度超前位移 π/2。加速度超前速度 π/2，但与位移相位差恰好为 π（反相）。
Understanding phase relationships helps in interpreting graphs and in comparing the motion of two particles. If two particles have the same ω but different initial phases, their motions are shifted in time. A phase difference of 2π corresponds to one full period.
理解相位关系有助于解读图像和比较两个质点的运动。如果两个质点具有相同的 ω 但初相位不同，它们的运动在时间上就会错开。相位差 2π 对应一个完整周期。
6. Maximum Values and Turning Points | 最大值与转折点
The following table summarises the extreme values of velocity and acceleration and where they occur.
下表总结了速度和加速度的极值及其发生位置。
| Quantity | Expression | Occurs at |
|---|---|---|
| Maximum speed vmax | ωA | Equilibrium (x = 0) |
| Minimum speed | 0 | x = ±A (turning points) |
| Maximum acceleration amax | ω²A | x = ±A |
| Minimum acceleration | 0 | Equilibrium (x = 0) |
At the equilibrium position, the restoring force is zero, so acceleration is zero but speed is greatest. At the extremes, the particle momentarily stops (turning points), so speed is zero and the acceleration reaches its maximum magnitude, directed back towards equilibrium.
在平衡位置，回复力为零，加速度为零但速率最大。在端点处，质点瞬间停止（转折点），速率为零，加速度的大小达到最大，方向指向平衡位置。
Many exam questions ask you to find vmax and amax directly from given A and ω, or to deduce ω from the maximum acceleration if the amplitude is known. Always check whether the particle is at the centre or the extreme before using these formulas.
许多考题会要求你从已知的 A 和 ω 直接求出 vmax 和 amax，或者在已知振幅的情况下由最大加速度推出 ω。使用这些公式前，一定要先判断质点处于中心还是端点。
7. Energy in SHM | 简谐运动中的能量
For an undamped SHM system, the total mechanical energy E is constant and can be expressed in terms of either the maximum kinetic energy at the centre or the maximum potential energy at the extremes. For a horizontal mass-spring system, the elastic potential energy is U = ½ kx² and kinetic energy is K = ½ mv². Using k = mω² (since ω² = k/m), we obtain:
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