📚 AS Physics: Formula Derivations for Oscillations and Waves (OxfordAQA International) | AS 物理：振动与波公式推导（OxfordAQA 国际版）
Understanding how key formulas in oscillations and waves are derived is essential for mastering AS Physics. This article walks you through the most important derivations required by the OxfordAQA International AS specification, from simple harmonic motion to travelling wave equations. Each step is explained in both English and Chinese, ensuring clarity for bilingual learners.
理解振动与波中关键公式的推导过程，是掌握 AS 物理的重要环节。本文带你逐一推演 OxfordAQA 国际 AS 考纲要求的最重要公式——从简谐运动到行波方程。每一步都提供中英双语解释，让双语学习者都能清晰掌握。
1. Defining Simple Harmonic Motion (SHM) | 简谐运动的定义
Simple harmonic motion is defined by the condition that the acceleration of an object is directly proportional to its displacement from a fixed point and always directed towards that point. Mathematically, this is expressed as:
简谐运动的定义是：物体的加速度与其相对某一固定点的位移成正比，并且始终指向该固定点。数学表达式为：
a = -ω²x
Here, a is acceleration, x is displacement, and ω is the angular frequency. The negative sign indicates that acceleration and displacement are in opposite directions. This equation is the foundation from which all other SHM relationships follow.
其中 a 是加速度，x 是位移，ω 是角频率。负号表示加速度与位移方向相反。这一方程是所有其他简谐运动关系式的推导基础。
2. Deriving the Displacement–Time Equation | 位移-时间方程的推导
From the definition a = -ω²x, and knowing that acceleration is the second derivative of displacement with respect to time, we can write the differential equation: d²x/dt² = -ω²x. A solution to this is a sinusoidal function. Assuming the object starts at maximum displacement A (amplitude) when t = 0, we take the cosine form:
根据定义 a = -ω²x，并且知道加速度是位移对时间的二阶导数，我们可以写出微分方程：d²x/dt² = -ω²x。该方程的解是正弦或余弦函数。假设物体在 t = 0 时位于最大位移 A（振幅）处，我们采用余弦形式：
x = A cos(ωt)
If the motion starts from the equilibrium position (x = 0) at t = 0, a sine function is more appropriate: x = A sin(ωt). Both forms satisfy the differential equation, as demonstrated by differentiating twice.
如果运动在 t = 0 时从平衡位置（x = 0）开始，则正弦形式更合适：x = A sin(ωt)。两种形式都满足微分方程，可通过两次微分加以验证。
3. Deriving Velocity in Terms of Displacement | 速度-位移关系的推导
To find velocity without involving time directly, we can use energy conservation or differentiate the displacement function and apply the trigonometric identity sin²θ + cos²θ = 1. Let’s use differentiation on x = A cos(ωt):
要得到不显含时间的速度表达式，可以利用能量守恒，或对位移函数求导并应用三角恒等式 sin²θ + cos²θ = 1。我们对 x = A cos(ωt) 求导：
v = dx/dt = -Aω sin(ωt)
Since x = A cos(ωt), we have sin(ωt) = √(1 – cos²(ωt)) = √(1 – (x/A)²) for the magnitude. Therefore:
由于 x = A cos(ωt)，因此 sin(ωt) = √(1 – cos²(ωt)) = √(1 – (x/A)²)（取绝对值）。于是：
v = ± ω √(A² – x²)
The plus or minus sign indicates direction – the object moves back and forth. This equation is particularly useful for calculating speed at any given displacement.
正负号表示方向——物体来回运动。该公式在计算任意位移处的速率时特别有用。
4. Energy in Simple Harmonic Motion | 简谐运动中的能量
In SHM, energy continuously transforms between kinetic and potential forms while the total energy remains constant (assuming no damping). Kinetic energy K = ½mv². Substituting v = ω√(A² – x²):
在简谐运动中，能量在动能和势能之间不断转换，而总能量保持恒定（假设无阻尼）。动能 K = ½mv²。代入 v = ω√(A² – x²)：
K = ½ m ω² (A² – x²)
Potential energy U is derived from the work done against the restoring force. Since F = -mω²x, the average force over a displacement from 0 to x is ½ mω²x, giving U = ½ mω²x². Adding K and U yields the total energy:
势能 U 从克服回复力做功导出。因为 F = -mω²x，从 0 到 x 位移过程中平均力为 ½ mω²x，因此 U = ½ mω²x²。将 K 与 U 相加得到总能量：
Etotal = ½ m ω² A²
The total energy depends only on the amplitude and the square of the angular frequency.
总能量仅取决于振幅与角频率的平方。
5. Period of a Mass–Spring System | 弹簧振子的周期推导
For a mass m attached to a spring of spring constant k, Hooke’s law gives restoring force F = -kx. Using Newton’s second law:
对于连接在劲度系数为 k 的弹簧上的质量 m，胡克定律给出回复力 F = -kx。利用牛顿第二定律：
ma = -kx → a = -(k/m)x
Comparing with the SHM definition a = -ω²x, we identify ω² = k/m, so ω = √(k/m). Since the period T = 2π/ω, we obtain:
与简谐运动定义 a = -ω²x 对比，得到 ω² = k/m，所以 ω = √(k/m)。由于周期 T = 2π/ω，得到：
T = 2π √(m/k)
This shows the period is independent of amplitude – a key characteristic of isochronous oscillators.
这表明周期与振幅无关——这是等时振子的关键特征。
6. Deriving the Period of a Simple Pendulum | 单摆周期的推导
Consider a pendulum of length l displaced by a small angle θ (in radians). The restoring force along the arc is -mg sinθ. For small angles, sinθ ≈ θ. The linear displacement x along the arc is x = lθ, and the tangential acceleration is a = lα (where α = d²θ/dt²). Applying F = ma:
考虑摆长为 l 的单摆，偏离小角度 θ（以弧度计）。沿弧线的回复力为 -mg sinθ。小角度下 sinθ ≈ θ。沿弧线的线位移 x = lθ，切向加速度 a = lα（α = d²θ/dt²）。应用 F = ma：
-mg θ = m l (d²θ/dt²) → d²θ/dt² = -(g/l) θ
This has the same form as a = -ω²x, with ω² = g/l. Therefore:
这与 a = -ω²x 具有相同形式，其中 ω² = g/l。因此：
T = 2π √(l/g)
The approximation sinθ ≈ θ is valid only for small amplitudes (typically less than about 10°), making this derivation essential for accurate practical predictions.
近似 sinθ ≈ θ 仅适用于小振幅（通常小于约10°），这使得该推导对准确的实际预测至关重要。
7. Deriving the Travelling Wave Equation | 行波方程的推导
A wave travelling along the positive x-axis transfers a disturbance without moving material. The displacement y of a particle at position x and time t can be modelled by considering a wave shape moving at speed v. If the source vibrates with y = A sin(ωt), then a point at a distance x will experience the same vibration after a time delay x/v. Thus:
沿 x 轴正方向传播的波，只传递扰动而不迁移物质。位置 x 处质点在时刻 t 的位移 y，可以通过考虑以波速 v 移动的波形来建模。如果波源以 y = A sin(ωt) 振动，则距离 x 处的点将经历延迟 x/v 的相同振动。因此：
y = A sin[ω(t – x/v)]
Using the wave number k = 2π/λ and angular frequency ω = 2πf, along with v = fλ, the argument simplifies to (ωt – kx). The general form is:
利用波数 k = 2π/λ 和角频率 ω = 2πf，以及 v = fλ，函数参数简化为 (ωt – kx)。一般形式为：
y = A sin(ωt – kx + φ)
where φ is the phase constant determined by initial conditions. A negative sign before kx indicates the wave travels in the positive x-direction; a plus sign indicates the opposite direction.
其中 φ 是由初始条件决定的相位常数。kx 前取负号表示波沿正 x 方向传播；取正号则表示相反方向。
8. Deriving v = fλ and Its Importance | 波速公式 v = fλ 的推导及其重要性
The wave speed v, frequency f, and wavelength λ are linked by a straightforward relationship. By definition, speed is distance divided by time. In one period T, a wave advances by one wavelength λ. Therefore:
波速 v、频率 f 和波长 λ 之间存在一个直接关系。根据定义，速度等于距离除以时间。在一个周期 T 内，波前进一个波长 λ。因此：
v = λ/T
Since frequency f = 1/T, we obtain the well-known equation v = fλ. This relation holds for all types of waves – mechanical, electromagnetic, and sound. It is crucial for interpreting experimental data, such as measuring the speed of sound using a resonance tube or determining the wavelength of light from a double-slit experiment.
由于频率 f = 1/T，我们得到著名的关系式 v = fλ。该关系对所有类型的波都成立——机械波、电磁波和声波。它对于解释实验数据至关重要，例如用共振管测量声速，或通过双缝实验确定光的波长。
9. Phase Difference and Path Difference | 相位差与路径差
When two points on a wave (or from two sources) are compared, their phase difference Δφ and path difference Δx are directly related. A full cycle corresponds to a phase change of 2π radians over a distance of one wavelength λ. Hence, the phase difference for a path difference Δx is:
当比较波上两点（或两个波源）时，它们的相位差 Δφ 与路径差 Δx 直接相关。一个完整周期对应相位变化 2π 弧度，距离为一个波长 λ。因此，对于路径差 Δx，相位差为：
Δφ = (2π/λ) × Δx
This formula underpins interference phenomena. Constructive interference occurs when Δφ = 2nπ (n = 0, 1, 2, …), meaning Δx = nλ. Destructive interference occurs when Δφ = (2n+1)π, so Δx = (n+½)λ. Mastering this link is essential for solving superposition problems in the AS exam.
该公式是干涉现象的基础。相长干涉发生在 Δφ = 2nπ（n = 0, 1, 2, …）时，即 Δx = nλ。相消干涉发生在 Δφ = (2n+1)π 时，即 Δx = (n+½)λ。掌握这一联系对于解答 AS 考试中的叠加问题至关重要。
10. Summary of Key Derivations | 关键推导总结
In this article, we have derived the core equations for AS-level oscillations and waves, starting from the fundamental SHM definition a = -ω²x. We constructed the displacement–time solution, linked velocity to displacement, and used energy conservation to find total energy. We also showed how the period formulas for the mass–spring system and simple pendulum emerge naturally from Newton’s laws. For waves, we built the travelling wave equation from the concept of time delay and derived the fundamental relation v = fλ, as well as the essential phase-difference formula. Practising these derivations step by step will deepen your understanding and prepare you for any calculation or explanation question in the OxfordAQA International AS Physics examination.
本文从简谐运动的基本定义 a = -ω²x 出发，推导了 AS 级别振动与波的核心方程。我们构建了位移-时间解，将速度与位移联系起来，并通过能量守恒求得总能量。我们还展示了弹簧振子和单摆的周期公式如何从牛顿定律自然导出。在波动部分，利用时间延迟概念构建了行波方程，并推导了基本关系 v = fλ 以及重要的相位差公式。一步步练习这些推导，可以加深理解，为 OxfordAQA 国际 AS 物理考试中的任何计算或解释题做好充分准备。
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