Formula Derivations in A-Level Physics Unit 3 (June 2022) | A-Level物理Unit 3 (2022年6月)公式推导

📚 Formula Derivations in A-Level Physics Unit 3 (June 2022) | A-Level物理Unit 3 (2022年6月)公式推导

In Edexcel International A-Level Physics, Unit 3 (Practical Skills in Physics I) assesses your ability to design experiments, process data, and evaluate uncertainties. The June 2022 paper (WPH13/01) featured several contexts where deriving the correct formula from experimental measurements was essential. This article revisits the key formula derivations that frequently appear in Unit 3, using examples inspired by the June 2022 examination. Understanding these derivations not only helps you answer calculation questions but also deepens your grasp of experimental physics.

在爱德思国际A-Level物理中，Unit 3（物理实验技能I）考查实验设计、数据处理和不确定度评估能力。2022年6月的试卷（WPH13/01）包含多个需要根据实验测量值推导公式的情境。本文以2022年6月试题为背景，回顾Unit 3中常见的核心公式推导。理解这些推导不仅能帮助你解答计算题，还能加深对实验物理的理解。

1. Free Fall with Time Delay – Deriving g via Linearisation | 含时间延迟的自由落体 – 线性化推导g

In many free fall experiments, an electromagnet holds a steel ball, and a timer starts when the circuit is broken. However, there is often a small time delay δt before the ball actually begins to fall because residual magnetism holds it momentarily. The actual fall time is therefore t + δt, where t is the measured time. The distance fallen from rest is s = ½ g (t + δt)².

在许多自由落体实验中，电磁铁吸住钢球，电路断开时计时器启动。然而，由于剩磁暂时吸住小球，在球实际开始下落前通常存在一个微小的时间延迟 δt。因此实际下落时间为 t + δt（t 为测量时间）。从静止下落的距离为 s = ½ g (t + δt)²。

To eliminate the unknown δt and find g, we linearise the equation. Taking square roots gives √(2s/g) = t + δt, which can be rearranged as t = √(2/g) √s – δt. This is of the form y = mx + c, where y = t, x = √s, slope m = √(2/g), and intercept c = –δt.

为了消去未知的 δt 并求出 g，我们将方程线性化。开平方得 √(2s/g) = t + δt，可整理为 t = √(2/g) √s – δt。这具有 y = mx + c 的形式，其中 y = t，x = √s，斜率 m = √(2/g)，截距 c = –δt。

By plotting a graph of t against √s for several values of s and t, you obtain a straight line. From the slope m, the acceleration of free fall is derived as g = 2 / m². The intercept gives –δt, allowing you to quantify the systematic delay.

通过针对不同 s 和 t 值作 t–√s 图，可得到一条直线。由斜率 m 可推导出重力加速度 g = 2 / m²。截距给出 –δt，从而可以量化系统延迟的大小。

t = √(2/g) √s – δt → g = 2 / (slope)²

2. Simple Pendulum – Deriving g = 4π²L/T² | 单摆 – 重力加速度 g 的推导

For a simple pendulum of length L swinging through a small angle, the restoring force along the arc is F = –mg sinθ ≈ –mgθ. With θ = x/L, this becomes F = –(mg/L)x, which is proportional to displacement and directed towards equilibrium. Hence the motion is simple harmonic with spring constant equivalent k = mg/L.

对于摆长为 L、以小角度摆动的单摆，沿弧线的回复力为 F = –mg sinθ ≈ –mgθ。代入 θ = x/L，得 F = –(mg/L)x，力与位移成正比且指向平衡位置。因此这一运动是简谐运动，等效劲度系数 k = mg/L。

The angular frequency is ω = √(k/m) = √(g/L). Since ω = 2π/T, the period is T = 2π/ω = 2π√(L/g). Squaring both sides yields T² = 4π²L/g, which when rearranged gives the well‑known formula for g.

角频率为 ω = √(k/m) = √(g/L)。因 ω = 2π/T，周期为 T = 2π/ω = 2π√(L/g)。两边平方得 T² = 4π²L/g，移项后得到著名的 g 公式。

T = 2π√(L/g) → g = 4π²L / T²

In Unit 3, you are often required to explain how a graph of T² against L can be used to find g. The slope of such a graph is 4π²/g, so g = 4π² / slope. This method reduces the effect of timing errors.

在Unit 3中，常要求说明如何利用 T²–L 图求出 g。该

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