📚 A-Level Physics: Experimental Investigation in Unit 4 (June 2022) | A-Level 物理：Unit 4 实验探究 (2022年6月)
The AQA Unit 4 June 2022 question paper features a classic experimental investigation: determining the acceleration due to gravity, g, using a simple pendulum. This experiment not only tests practical skills but also challenges students to process data, linearise relationships, interpret gradients, and evaluate uncertainties. Mastery of these techniques is essential for A2 success and forms the backbone of many examination questions.
AQA Unit 4 2022 年 6 月试卷中包含一个经典实验探究：利用单摆测定重力加速度 g。这个实验不仅考查实践技能，还要求学生处理数据、线性化关系、解释斜率并评估不确定度。熟练掌握这些技巧对 A2 阶段的成功至关重要，也是许多考题的核心。
1. The Simple Pendulum and Its Theory | 单摆与理论
A simple pendulum consists of a small, dense bob suspended from a fixed point by a light, inextensible string. For small angular displacements (θ ≤ 10°), the motion approximates simple harmonic motion (SHM). The period T is given by T = 2π√(L/g), where L is the length from the pivot to the centre of mass of the bob and g is the gravitational field strength.
单摆由一个固定点悬挂的小而密的摆球和一根轻质、不可伸长的细线组成。在小角度摆动（θ ≤ 10°）时，其运动近似为简谐运动。周期 T 由公式 T = 2π√(L/g)给出，其中 L 是从悬点到摆球质心的长度，g 是重力场强度。
This relationship can be rearranged to give T² = (4π²/g)L. Plotting T² against L yields a straight line through the origin with gradient = 4π²/g. Hence, g can be found from g = 4π²/gradient.
该关系可改写为 T² = (4π²/g)L。绘制 T² 对 L 的图线将得到一条过原点的直线，其斜率为 4π²/g。因此，可通过 g = 4π²/斜率 求得 g。
In the June 2022 paper, students were required to deduce this linear form, identify variables for the axes, and calculate g from a provided or self‑obtained graph.
在 2022 年 6 月试卷中，学生需要推导这种线性形式，确定坐标轴变量，并根据给出的或自行获得的图线计算 g。
2. Experimental Setup and Procedure | 实验装置与步骤
The apparatus includes a clamp stand, a split cork or clamp to hold the string, a metre ruler, a stopwatch, a protractor, and a small metal bob (e.g., a 100 g mass with a hook). The effective length L is measured from the point of suspension to the centre of the bob. The bob’s radius must be added to the length of the string when determining L.
实验装置包括铁架台、固定细线的开口软木塞或夹子、米尺、秒表、量角器和小金属摆球（例如带钩的 100 g 砝码）。有效长度 L 应从悬点量至摆球中心。确定 L 时，必须将摆球半径加到绳长上。
The accepted procedure is: set L to a chosen value (e.g., 1.000 m), displace the bob by less than 10° from the vertical, release it, and time a number of complete oscillations (say 20) to reduce the human reaction‑time error. Repeat the timing several times and calculate the mean period T = total time / number of oscillations.
认可的操作步骤是：将 L 设置为选定值（如 1.000 m），使摆球偏离竖直方向小于 10°，释放后用秒表记录多次完整摆动（如 20 次）的时间，以减小人为反应时间误差。重复计时多次，并计算平均周期 T = 总时间 / 摆动次数。
Change L over a wide range (e.g., 0.800 m to 1.200 m in 0.100 m steps) and record the corresponding T for each length. Keep the amplitude small and constant to satisfy the SHM condition.
在较宽的范围内改变 L（例如从 0.800 m 到 1.200 m，步长 0.100 m），记录每个长度对应的 T。保持摆幅小且恒定以满足简谐运动条件。
3. Data Collection and Tabulation | 数据收集与表格
In the exam, a table of results may be provided or you may need to design one. A typical table includes columns for L/m, time t₁ for 20 oscillations, time t₂, mean time t_mean, period T/s, and T²/s². All measurements include uncertainties (e.g., ±0.001 m for length, ±0.01 s for the stopwatch).
考试中可能会提供数据表，或者需要你自行设计。典型表格包含以下列：L/m、20 次摆动的时间 t₁、时间 t₂、平均时间 t_mean、周期 T/s 和 T²/s²。所有测量均带有不确定度（例如长度 ±0.001 m，秒表 ±0.01 s）。
Below is a sample data table similar to the one analysed in the June 2022 paper:
下面的样本数据表类似于 2022 年 6 月试卷中分析的表：
| L / m (±0.001 m) | t₁ for 20T / s | t₂ for 20T / s | Mean t / s | T / s | T² / s² |
|---|---|---|---|---|---|
| 0.800 | 35.80 | 35.92 | 35.86 | 1.793 | 3.215 |
| 0.900 | 38.00 | 38.10 | 38.05 | 1.903 | 3.621 |
| 1.000 | 40.10 | 40.22 | 40.16 | 2.008 | 4.032 |
| 1.100 | 42.10 | 42.18 | 42.14 | 2.107 | 4.439 |
| 1.200 | 44.00 | 44.06 | 44.03 | 2.202 | 4.849 |
Notice that the period is calculated to four significant figures initially, matching the precision of the raw data. In exam answers, the final values should reflect the appropriate level of precision.
请注意，周期最初计算至四位有效数字，与原始数据的精度匹配。在考试答案中，最终数值应反映出适当的精度水平。
4. Data Processing and Linearisation | 数据处理与线性化
The direct relationship T = 2π√(L/g) is not linear. The exam question specifically asks candidates to plot T² on the y‑axis against L on the x‑axis. This linearisation yields the equation T² = (4π²/g)L, which is of the form y = mx + c with c = 0 theoretically.
直接关系 T = 2π√(L/g) 不是线性的。考题明确要求考生在 y 轴上绘制 T² 对 x 轴上的 L。这种线性化得到方程 T² = (4π²/g)L，其形式为 y = mx + c，理论上 c = 0。
Calculate T² for each row and record it to an appropriate number of significant figures. In the sample table above, T² varies from 3.22 to 4.85 s², so three significant figures are usually sufficient.
计算每一行的 T²，并以合适的有效数字记录。在上面的样本表中，T² 范围从 3.22 到 4.85 s²，因此通常三位有效数字就足够了。
When handling uncertainties in derived quantities, the absolute uncertainty in T² can be found using the percentage uncertainty method: %U(T²) = 2 × (%U(T)), where %U(T) = (ΔT/T) × 100%.
在处理导出量的不确定度时，T² 的绝对不确定度可以用百分比不确定度法求得：%U(T²) = 2 × (%U(T))，其中 %U(T) = (ΔT/T) × 100%。
5. Graphical Analysis: Plotting T² against L | 图形分析：绘制 T² 对 L 图
Plot the data points on a graph with sensible scales, label axes with quantities and units, and draw a line of best fit. The line should pass through the origin if the model is correct, but small systematic errors may produce a non‑zero intercept. In the 2022 paper, candidates were asked to determine the gradient and comment on the intercept.
在坐标纸上以合适的刻度绘制数据点，标注坐标轴物理量和单位，并画出最佳拟合线。如果模型正确，该直线应通过原点，但微小的系统误差可能产生非零截距。2022 年试卷要求考生确定斜率并评论截距。
Using the table values, a sample gradient calculation:
gradient = (ΔT²) / (ΔL) = (4.849 – 3.215) / (1.200 – 0.800) = 1.634 / 0.400 = 4.085 s² m⁻¹
This gradient has units of s² m⁻¹. More accurate tests would use the full set of data and a least‑squares method, but the exam typically expects a large triangle method with clear working.
该斜率的单位是 s² m⁻¹。更精确的分析会使用全部数据和最小二乘法，但考试通常期望用大三角形法清晰求解。
6. Determining g from the Gradient | 从斜率求 g
Since gradient = 4π²/g, we rearrange to g = 4π²/gradient. Using the above gradient:
由于斜率 = 4π²/g，重新整理得 g = 4π²/斜率。代入上述斜率：
g = 4π² / 4.085 ≈ 39.48 / 4.085 ≈ 9.67 m s⁻²
This value is close to the accepted value of 9.81 m s⁻², indicating good experimental technique. In the June 2022 question, students had to compare their calculated g with the standard value and suggest reasons for any discrepancy.
该值接近公认值 9.81 m s⁻²，表明实验技术良好。在 2022 年 6 月考题中，学生需要将计算出的 g 与标准值进行比较，并对任何差异提出可能的原因。
7. Uncertainty Analysis in the Experiment | 实验中的不确定度分析
Uncertainties arise from measurements of length and time. For L, the ruler has a precision of ±1 mm, but the actual uncertainty may be larger because of judging the centre of mass. For time, the stopwatch has a precision of ±0.01 s, but the dominant uncertainty is human reaction time (typically ±0.2 s per reading). By timing 20 oscillations, the uncertainty per period is reduced to about ±0.01 s.
不确定度来源于长度和时间的测量。对于 L，尺子的精度为 ±1 mm，但由于需要判断质心位置，实际不确定度可能更大。对于时间，秒表精度为 ±0.01 s，但主要不确定度是人的反应时间（通常每次读数 ±0.2 s）。通过计时 20 次摆动，每个周期的不确定度降低至约 ±0.01 s。
The percentage uncertainty in T² is twice that in T. The overall percentage uncertainty in g is approximately the sum of the percentage uncertainty in gradient, which depends on the spread of points, and any systematic uncertainty in L. Candidates in the 2022 paper were expected to estimate this and quote g with an appropriate uncertainty range.
T² 的百分比不确定度是 T 的两倍。g 的总百分比不确定度大约等于斜率的百分比不确定度加上 L 的任何系统不确定度之和。2022 年试卷期望考生估算这些不确定度并给出带有适当不确定范围的 g 值。
8. Sources of Error and Possible Improvements | 误差来源与改进方法
The main sources of inaccuracy include: (i) measuring L from the pivot to the centre of the bob; using a pointer or vernier calipers to locate the centre reduces this error. (ii) Reaction time when using a stopwatch; a light gate connected to a data logger can eliminate human reaction. (iii) Wide angular amplitude causing the SHM approximation to break down; use a protractor to keep the angle < 10°. (iv) Air resistance and damping; a heavier bob with a streamlined shape minimises this effect.
不准度的主要来源包括：(i) 从悬点到摆球中心的长度测量；使用指针或游标卡尺定位中心可减小此误差。(ii) 使用秒表时的反应时间；连接到数据记录器的光闸可以消除人为反应。(iii) 大角度摆动导致简谐运动近似失效；使用量角器将角度保持在 10° 以下。(iv) 空气阻力和阻尼；更重且流线型的摆球可将此影响降至最低。
An exam question might ask: ‘Suggest one improvement to the procedure and explain how it increases accuracy.’ A model answer would be: ‘Use a fiducial marker (e.g., a split cork and pin) at the equilibrium position, and start the stopwatch when the bob passes this marker, because timing over many oscillations from the same point reduces reaction‑time error.’
考题可能会问：“建议一项程序改进并解释它如何提高准确度。”标准答案可以是：“在平衡位置使用参考标记（如对开的软木塞和一根针），当摆球经过此标记时启动秒表，因为从同一点计时多次摆动可减少反应时间误差。”
9. Interpreting the Graph: Intercept and Anomalous Points | 图形解读：截距与异常点
In the ideal T²–L graph, the line passes through the origin. A small positive intercept on the y‑axis could indicate that the measured L is systematically shorter than the true effective length (e.g., the radius of the bob was not added correctly). The June 2022 paper featured a question requiring students to explain a non‑zero intercept and calculate the systematic error from it.
在理想的 T²–L 图中，直线通过原点。y 轴上的小正截距可能表明测得的 L 系统性地短于真实有效长度（例如摆球半径未正确加上）。2022 年 6 月试卷中包含一道要求学生解释非零截距并据此计算系统误差的题目。
Anomalies in the graph may come from miscounting oscillations or large amplitude swings. Students should be prepared to identify these points, explain their cause, and state how they would treat them (e.g., repeat the measurement or exclude them from the line of best fit).
图中的异常点可能来自摆动次数计数错误或摆幅过大。学生应准备好识别这些点，解释它们的原因，并说明如何处理（例如重复测量或将其从最佳拟合线中排除）。
10. Exam‑Style Questions and Commentary | 考试题型与点评
Typical marks in the Unit 4 paper are allocated for: (a) Describing the procedure and explaining why certain steps are taken (3 marks). (b) Completing a table and calculating uncertainties (4 marks). (c) Plotting T² against L, drawing the best‑fit line, and determining the gradient (5 marks). (d) Calculating g from the gradient and commenting on accuracy (4 marks). (e) Identifying sources of error and suggesting improvements (4 marks).
Unit 4 试卷中典型分值分配为：(a) 描述操作步骤并解释为何采取某些步骤（3 分）。(b) 完成表格并计算不确定度（4 分）。(c) 绘制 T²–L 图线，画最佳拟合线并求斜率（5 分）。(d) 由斜率计算 g 并评论准确度（4 分）。(e) 识别误差来源并提出改进建议（4 分）。
When tackling these questions, always use clear subheadings, show all working, and state any assumptions (e.g., small angle approximation). Units must be consistent. For the June 2022 paper, many students lost marks by not converting centimetres to metres or by forgetting to square the period when plotting.
在解答这些题目时，始终使用清晰的子标题，呈现所有计算步骤，并说明任何假设（如小角近似）。单位必须一致。对于 2022 年 6 月试卷，许多学生因未将厘米换算为米或在绘图时忘记将周期平方而失分。
11. Conclusion | 结论
The simple pendulum experiment is a superb vehicle for assessing a wide range of practical and analytical skills. The June 2022 Unit 4 paper skilfully weaves together measurement, data processing, graphical interpretation, and error analysis. By mastering the linearisation strategy T² = (4π²/g)L, students can confidently determine g and handle the associated uncertainties. Consistent practice with marking schemes and past papers will ensure top performance in this section.
单摆实验是评估多种实践与分析技能的绝佳载体。2022 年 6 月 Unit 4 试卷巧妙地将测量、数据处理、图形解读和误差分析融合在一起。通过掌握线性化策略 T² = (4π²/g)L，学生可以自信地测定 g 并处理相关不确定度。结合评分方案和历年真题的持续练习，将确保在这一部分取得优异成绩。
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