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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

2026-07-14 23:39 · 本文部分内容由AI辅助创作

📚 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' sponding 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

h. 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 / mT / sT² / s²
0.451.351.82
0.601.562.43
0.801.803.24
1.002.014.04
1.202.204.84

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

m 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' ts 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

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。


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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)

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。

cept 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)

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