📚 A-Level Physics: Formula Derivation from Unit 3 January 2021 Paper | A-Level 物理：2021年1月单元3试卷公式推导
In the January 2021 Edexcel IAL Physics Unit 3 (WPH13/01) paper, a key question challenged students to derive gravitational field strength g from raw experimental data obtained with a simple pendulum. Mastering this type of formula derivation is central to the practical skills assessed in Unit 3 and builds confidence for tackling data‑analysis problems under exam conditions.
在2021年1月的爱德思国际A-Level物理单元3 (WPH13/01) 试卷中，一道关键题目要求学生从单摆实验获得的原始数据推导重力场强度 g。掌握这类公式推导是单元3所评估的实践技能的核心，也能增强在考试条件下解决数据分析问题的信心。
1. Introduction to the Exam Question | 考题介绍
The question presented a student’s investigation into how the period T of a simple pendulum depends on its length l. A table of measured lengths and corresponding periods was supplied, and candidates were reminded of the standard formula T = 2π√(l/g). Their task was to manipulate this relationship into a linear graph, use the graph’s slope to deduce a value for g, and discuss uncertainties.
题目展示了一位学生对单摆周期 T 与摆长 l 之间关系的研究。试卷给出了摆长与对应周期的测量数据表，并提醒考生标准公式 T = 2π√(l/g)。考生的任务是将这一关系式变形为直线图，利用图形斜率推算出 g 值，并讨论不确定度。
2. Recalling the Pendulum Formula | 回顾单摆公式
The well‑known period of a simple pendulum for small oscillations is given by:
小角度振动下单摆的著名周期公式为：
T = 2π √(l / g)
Here T is the time for one complete oscillation, l is the length from the pivot to the centre of mass of the bob, and g is the gravitational field strength. The formula assumes oscillations with an angular amplitude less than about 10°, where the sinθ ≈ θ approximation holds.
式中 T 为一次完整振动的时间，l 为从支点到摆锤质心的长度，g 为重力场强度。该公式假设振动角振幅小于约10°，此时 sinθ ≈ θ 近似成立。
3. Understanding the Data Table | 理解数据表格
In the exam, students were provided with a table similar to the one below. They were expected to calculate T² for each value of l and then plot a graph of T² against l.
考试中，学生获得了一个类似下表的表格。他们需要计算每个 l 对应的 T²，然后绘制 T² 对 l 的图形。
| l / m | T / s | T² / s² |
|---|---|---|
| 0.45 | 1.35 | 1.82 |
| 0.60 | 1.56 | 2.43 |
| 0.80 | 1.80 | 3.24 |
| 1.00 | 2.01 | 4.04 |
| 1.20 | 2.20 | 4.84 |
Notice that raw T is recorded to three significant figures, and T² is calculated accordingly. Such raw‑data handling is a vital skill in Unit 3.
注意原始周期 T 记录为三位有效数字，并相应地计算 T²。这种原始数据处理是单元3中的一项重要技能。
4. Purpose of the Derivation | 推导的目的
The core aim is not simply to quote g = 9.81 m s⁻² but to show how g can be extracted from a straight‑line graph. Linearising the equation makes it possible to use a graph’s gradient (and optionally its intercept) to determine the constant g, while also providing a visual check on the data’s quality.
核心目标不仅仅是引用 g = 9.81 m s⁻²，而是要展示如何从一条直线图中提取 g。将方程线性化后，就可以利用图形的斜率（也可选截距）来确定常数 g，同时也能直观地检验数据质量。
5. Linearising the Pendulum Equation | 将单摆方程线性化
Begin by squaring both sides of T = 2π√(l/g) to remove the square root:
首先将 T = 2π√(l/g) 两边平方，以消去根号：
T² = (4π² / g) l
This equation now has the form y = m x + c, where y = T², x = l, m = 4π²/g, and c = 0 if the idealised model holds perfectly.
该方程现在具有 y = m x + c 的形式，其中 y = T²，x = l，斜率 m = 4π²/g，并且在理想模型严格成立时截距 c = 0。
This step is the most critical in the derivation—once the equation is linearised, all subsequent calculations become straightforward.
这一步是推导中最关键的一步——一旦方程线性化，所有后续计算都变得简单了。
6. Plotting the Linear Graph | 绘制线性图形
Students were required to plot T² on the vertical axis and l on the horizontal axis. With real data, the points will not all lie exactly on a straight line due to random uncertainties. The best‑fit line is drawn to pass through as many error bars as possible, ignoring any clear outliers.
学生需要将 T² 画在纵轴、l 画在横轴上。由于存在随机不确定度，真实数据点不会全部精确落在一条直线上。最佳拟合线应尽量穿过所有误差棒，并忽略任何明显的异常点。
Even if the line does not pass exactly through the origin, the gradient still provides a reliable value for g. A systematic offset may indicate a small systematic error, such as an incorrect zero for length.
即使直线不完全通过原点，斜率仍能提供可靠的 g 值。系统性的偏移可能表明存在小的系统误差，例如长度零点不正确。
7. Identifying Slope and Intercept | 确定斜率和截距
From the best‑fit line, select two well‑separated points (x₁, y₁) and (x₂, y₂) that lie exactly on the line—not data points—to compute the gradient m:
从最佳拟合线上选取两个相距较远且恰好位于线上的点 (x₁, y₁) 和 (x₂, y₂)（而非数据点），来计算斜率 m：
m = (y₂ – y₁) / (x₂ – x₁) = (T₂² – T₁²) / (l₂ – l₁)
The intercept c can be read directly from the graph where the line crosses the T² axis. In the idealised model, c = 0, but a small positive or negative intercept is common in student experiments.
截距 c 可以直接从图上直线与 T² 轴的交点读出。在理想模型中 c = 0，但在学生实验中常出现小的正截距或负截距。
8. Deriving g from the Slope | 由斜率推导 g
Equate the measured gradient to the theoretical expression:
将测量得到的斜率与理论表达式建立等式：
m = 4π² / g
Rearrange to solve for g:
移项求解 g：
g = 4π² / m
This is the final derived formula. If the slope is measured in s² m⁻¹, then g will be in m s⁻². The numerical value of π is taken as 3.142 or left in symbolic form until the very last step to avoid rounding errors.
这就是最终的推导公式。如果斜率以 s² m⁻¹ 为单位，则 g 的单位是 m s⁻²。π 的数值取 3.142 或保持符号形式到最后一步，以避免舍入误差。
9. Handling a Non‑Zero Intercept | 处理非零截距
Sometimes the graph shows an intercept c that is not zero. In the pendulum equation, the intercept should theoretically be zero. A non‑zero intercept could be caused by a systematic error in measuring l—for instance, if the length was measured to the top of the bob rather than to its centre.
有时图形显示的截距 c 不为零。在单摆方程中，截距理论上应为零。非零截距可能由测量 l 的系统误差引起——例如，测量长度时测到了摆锤顶部而非其中心。
If c is small compared to typical T² values, g is still calculated solely from the slope using g = 4π² / m. The intercept is treated as an indication of systematic uncertainty, not as part of the g calculation.
如果 c 与典型的 T² 值相比很小，g 仍仅由斜率使用 g = 4π² / m 计算。截距只视为系统不确定度的迹象，而不参与 g 的计算。
10. Numerical Example of the Derivation | 推导的数值实例
Suppose from the T² vs l graph the gradient m is found to be 4.05 s² m⁻¹. Substituting into the derived formula gives:
假设从 T² 对 l 的图中求得斜率 m 为 4.05 s² m⁻¹。代入推导公式，得到：
g = 4π² / 4.05 = (4 × 3.1416²) / 4.05 ≈ 9.77 m s⁻²
This value is close to the accepted 9.81 m s⁻², indicating the experiment is reasonably accurate. Slight deviations are expected due to experimental uncertainties.
该值接近公认的 9.81 m s⁻²，表明实验相当准确。由于实验不确定度，微小偏差是意料之中的。
11. Propagating Uncertainty to g | g 的不确定度传播
The exam question often asks for the uncertainty in the derived g. Since g depends only on the gradient m, the percentage uncertainty in g equals the percentage uncertainty in m:
考题常常要求给出推导出的 g 的不确定度。由于 g 仅依赖于斜率 m，g 的百分不确定度等于 m 的百分不确定度：
Δg / g = Δm / m
Therefore, once the absolute uncertainty Δm in the gradient is estimated from the steepest and shallowest acceptable lines, the uncertainty Δg can be found by:
因此，一旦从最陡和最平缓可接受直线估算出斜率的绝对不确定度 Δm，便可通过下式求得 Δg：
Δg = g × (Δm / m)
Presenting the final result as g ± Δg, with an appropriate number of significant figures, is essential for meeting Unit 3 mark scheme demands.
将最终结果表示为 g ± Δg，并带适当的有效数字，对于满足单元3评分方案的要求至关重要。
12. Key Takeaways for the Exam | 考试关键要点
- Always write the basic formula and identify which variable becomes y and which becomes x before plotting.
- 在画图前，始终写出基本公式，并确定哪个变量作为 y、哪个作为 x。
- Square the period equation to obtain a linear relationship: T² = (4π²/g) l.
- 将周期方程平方以得到线性关系：T² = (4π²/g) l。
- Use the gradient m to derive g = 4π²/m; do not use individual data pairs.
- 利用斜率 m 推导 g = 4π²/m；切勿使用单个数据对。
- Quote g to two or three significant figures, and always include the unit m s⁻².
- 用两位或三位有效数字报告 g，且务必包含单位 m s⁻²。
- Handle uncertainty by evaluating the range of possible gradients.
- 通过评估可能斜率的范围来处理不确定度。
- Practice this derivation with different datasets so that the method becomes second nature.
- 用不同数据集练习这一推导，使方法
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