📚 Exploring AS Physics Unit 2 June 2022 Experiment: Measuring Young’s Modulus | 探究 AS 物理单元 2 2022 年 6 月实验：测量杨氏模量
The June 2022 AQA AS Physics Unit 2 paper featured a practical investigation into the elastic properties of a copper wire, specifically the determination of its Young’s modulus. This experiment is a classic example of linking mechanical measurements with material science concepts. In this article, we will deconstruct the experimental setup, the step‑by‑step procedure, data handling, sources of uncertainty, and how to extract a reliable value for Young’s modulus. Whether you are preparing for exams or reinforcing your lab skills, this breakdown will guide you through every critical stage of the investigation.
2022 年 6 月 AQA AS 物理单元 2 试卷中有一道关于铜线弹性性质的实验探究题，重点在于测定其杨氏模量。该实验是将力学测量与材料科学概念联系起来的经典案例。本文将详细拆解实验装置、分步操作流程、数据处理方法、不确定度来源以及如何提取可靠的杨氏模量值。无论你是在备考还是巩固实验技能，这篇解析都将带你走完探究的每一个关键环节。
1. Experimental Objective and Background | 实验目的与背景
The primary aim of the investigation is to determine the Young' material and is defined as the ratio of tensile stress σ to tensile strain ε: E = σ/ε. Stress is the force applied per unit cross‑ tional extension (ε = ΔL/L₀), where L₀ is the original length and ΔL the change in length.
本探究的主要目的是通过测量铜线在逐渐增加的拉伸载荷下的伸长量来确定其杨氏模量。杨氏模量 E 表征材料的刚度，定义为拉伸应力 σ 与拉伸应变 ε 之比：E = σ/ε。应力是单位横截面积所受的力（σ = F/A），应变则是长度变化分数（ε = ΔL/L₀），其中 L₀ 为原长，ΔL 为长度变化量。
Understanding this relationship allows physicists and engineers to predict how structures will deform under load. In the AS practical, students are required to gather raw data, plot a suitable graph, and use its gradient to compute E. The demand for careful measurement and rigorous uncertainty treatment reflects the real‑world significance of such material tests.
理解这一关系有助于物理学家和工程师预测结构在载荷下的变形情况。在 AS 实验中，学生需要采集原始数据，绘制合适的图像，并利用其斜率计算 E。对精细测量和严谨不确定度处理的要求，正体现了这类材料测试的实际意义。
E = σ/ε = (F/A) / (ΔL/L₀)
2. Apparatus and Setup | 实验仪器与装置
The experiment requires a long, thin copper wire, typically 2 m in length and approximately 0.3 mm in diameter. One end of the wire is firmly clamped to a rigid support, often a sturdy laboratory stand, while the other end is attached to a weight holder. A metre rule or a travelling microscope is aligned parallel to the wire to measure extension accurately. A small marker, such as a piece of tape, is fixed to the wire to serve as a reference point for extension readings.
实验需要一根细长的铜线，通常长约 2 m，直径约 0.3 mm。铜线一端牢固地夹在刚性支架（通常是稳固的实验室铁架台）上，另一端连接砝码盘。需要在铜线旁平行放置一把米尺或一个读数显微镜，以准确测量伸长量。此外，在铜线上固定一个小标记（如胶带），作为读取伸长量的参考点。
the wire at several points along its length, enabling the calculation of the average cross‑sectional area A = π(d/2)². Weights of known mass (e.g. 100 g increments) provide the tensile force F = mg, where g = 9.81 m s⁻². Safety screens are recommended in case the wire snaps under high tension.
使用螺旋测微器在铜线多处测量其直径，以计算平均横截面积 A = π(d/2)²。已知质量的砝码（例如以 100 g 递增）提供拉伸力 F = mg，其中 g = 9.81 m s⁻²。建议配备防护屏，以防铜线在高张力下突然断裂。
3. Step‑by‑Step Experimental Procedure | 实验步骤详解
Step 1 – rk using a metre rule. Repeat and record the mean value to minimise reading errors. Step 2 – o ensure any ovality is captured. Calculate the mean diameter and then the radius r = d/2.
步骤 1 – 用米尺测量从夹持点到参考标记之间的铜线原长 L₀。重复测量并记录平均值，以减小读数误差。步骤 2 – 用螺旋测微器在五个不同位置测量直径 d，转动铜线以检查是否存在不圆度。计算平均直径，进而得到半径 r = d/2。
Step 3 – Attach the weight holder and add a small initial mass to straighten the wire without causing significant extension. Record this starting setup. Step 4 – Increase the load in equal increments, each time recording the new position of the marker and hence the extension ΔL. Crucially, readings must be taken both while loading and while unloading to check for hysteresis and elastic limit.
步骤 3 – 挂上砝码盘并加一个小的初始质量以拉直铜线，但不产生显著伸长。记录此起始状态。步骤 4 – 以等增量增加负荷，每次记录标记的新位置，从而得到伸长量 ΔL。关键是，必须在加载和卸载过程中分别读数，以检查是否存在迟滞现象及是否超出弹性极限。
Step 5 – Continue adding masses until the wire shows signs of plastic deformation or until the maximum safe load is reached. As a safety measure, never exceed the manufacturer’s recommended maximum tension for the wire.
步骤 5 – 持续增加砝码，直到铜线开始出现塑性变形迹象或达到最大安全负荷。作为安全措施，切勿超过铜线制造商推荐的最大张力。
4. Raw Data Recording | 原始数据记录
de their absolute uncertainties. For example, the metre rule may have an uncertainty of ±1 mm, the micrometer ±0.01 mm, and the set of masses ±1 g. These values propagate into the final result.
系统化的数据采集至关重要。下表展示了典型的记录方式。所有读数必须包含绝对不确定度。例如，米尺的不确定度可能是 ±1 mm，螺旋测微器为 ±0.01 mm，砝码组为 ±1 g。这些值会传递到最终结果中。
| Mass m / kg | Force F / N | Marker position / mm | Extension ΔL / mm |
| 0.100 | 0.981 | 2.0 | 0.0 |
| 0.200 | 1.962 | 3.4 | 1.4 |
| 0.300 | 2.943 | 4.7 | 2.7 |
| 0.400 | 3.924 | 6.1 | 4.1 |
| 0.500 | 4.905 | 7.5 | 5.5 |
Note: the first row with zero extension indicates the straightened condition after the initial small load. From here, the true extension under each additional load is measured.
注意：第一行伸长量为零表示施加初始小负荷后的拉直状态。此后，每次额外加载所产生的真实伸长量均可据此测量。
5. Calculating Stress and Strain | 计算应力和应变
d, the tensile stress is σ = F/A, where A = πr². With a measured average diameter d = 0.31 mm, the radius is 0.155 mm = 1.55 × 10⁻⁴ m. Therefore, A = π × (1.55 × 10⁻⁴)² = 7.55 × 10⁻⁸ m². This small area explains why a moderate force produces a large stress.
对于每个负荷，拉伸应力 σ = F/A，其中 A = πr²。若测得平均直径 d = 0.31 mm，则半径 r = 0.155 mm = 1.55 × 10⁻⁴ m。因此 A = π × (1.55 × 10⁻⁴)² = 7.55 × 10⁻⁸ m²。这一微小的横截面积解释了为何中等大小的力就能产生较大的应力。
Tensile strain is ε = ΔL/L₀. With L₀ = 1.800 m (chosen for demonstration), convert the extension values into metres: e.g. for a 0.200 kg load, ΔL = 1.4 mm = 1.4 × 10⁻³ m, so ε = 1.4 × 10⁻³ / 1.800 = 7.78 × 10⁻⁴. These calculations are repeated for all data points.
拉伸应变 ε = ΔL/L₀。设原长 L₀ = 1.800 m（示例取值），将伸长量换算为米：例如 0.200 kg 负荷时，ΔL = 1.4 mm = 1.4 × 10⁻³ m，则 ε = 1.4 × 10⁻³ / 1.800 = 7.78 × 10⁻⁴。对所有数据点重复此计算。
Using the table above, the derived stress and strain pairs are shown in the next section. The key is to maintain consistent SI units to avoid exponent errors.
根据上表，求得的应力-应变数据对将在下一节展示。关键在于保持统一的国际单位制，以避免指数运算错误。
6. Graph Plotting and Analysis | 绘图与图像分析
Following standard practice, a graph of stress σ (y‑axis) against strain ε (x‑axis) should be plotted. Since both quantities are derived from measurements, each has an associated uncertainty, which can be represented as error bars. In the linear elastic region, the data points should follow a straight line passing through the origin. The gradient of this line gives the Young’s modulus E.
按照常规做法，应以应力 σ（y 轴）对应变 ε（x 轴）作图。由于两者均由测量数据导出，各自都有相应不确定度，可用误差棒表示。在线弹性区域内，数据点应分布在一条通过原点的直线上。该直线的斜率即为杨氏模量 E。
It is essential to use only the points that lie in the Hooke’s law region. If a point deviates significantly at higher stress, the elastic limit has been exceeded and that point must be excluded from the gradient calculation. The best‑fit line is drawn either by eye or using a computer; the gradient is then computed from a large triangle, avoiding the use of plotted points.
必须仅使用处于胡克定律区域内的点。如果某点在较大应力下明显偏离，说明已超出弹性极限，该点应从斜率计算中剔除。最佳拟合线可通过目测或计算机绘制；随后从一个较大的三角形计算斜率，避免直接使用数据点本身。
For example, from the graph, gradient Δσ/Δε may be 1.15 × 10¹¹ Pa, leading to an experimental E value of 1.15 × 10¹¹ Pa, close to the textbook value for copper (≈ 1.2 × 10¹¹ Pa).
例如，从图像中得到的斜率 Δσ/Δε 可能为 1.15 × 10¹¹ Pa，由此得出的实验 E 值为 1.15 × 10¹¹ Pa，接近铜的教科书值（约 1.2 × 10¹¹ Pa）。
7. Determination of Young’s Modulus via the Gradient | 通过斜率确定杨氏模量
The relationship E = σ/ε is mathematically equivalent to the gradient of the stress‑ n, then E = (σ₂ − σ₁)/(ε₂ − ε₁) = Δσ/Δε. By taking two widely separated points on the best‑fit line, the percentage uncertainty in the gradient can be minimised.
关系式 E = σ/ε 在数学上等同于弹性范围内应力‑应变图像的斜率。若直线通过原点，则有 E = (σ₂ − σ₁)/(ε₂ − ε₁) = Δσ/Δε。通过在最佳拟合线上选取相距较远的两点，可以最大限度地减小斜率的百分不确定度。
Suppose the chosen points are (0,0) and (8.0 × 10⁻⁴, 9.6 × 10⁷ Pa). Then gradient = (9.6 × 10⁷ − 0)/(8.0 × 10⁻⁴ − 0) = 1.20 × 10¹¹ result can be compared with the known value to comment on the accuracy of the experiment.
假设所取两点为 (0,0) 和 (8.0 × 10⁻⁴, 9.6 × 10⁷ Pa)，则斜率 = (9.6 × 10⁷ − 0)/(8.0 × 10⁻⁴ − 0) = 1.20 × 10¹¹ Pa。可将该结果与已知值进行比较，以评价实验的准确性。
Many June 2022 candidates found that their experimental value was slightly lower than the standard value. This discrepancy usually arises from systematic errors such as the wire not being perfectly straight initially, or from slip at the clamps.
许多 2022 年 6 月考生发现自己的实验值略低于标准值。这种差异通常源于系统误差，例如铜线初始未完全拉直，或夹持处存在滑移。
8. Uncertainty Analysis | 不确定度分析
Every measurement contributes uncertainty. For the metre rule (±1 mm), the fractional uncertainty in L₀ = 1 mm / 1800 mm = 5.6 × 10⁻⁴ (0.056%). The micrometer (±0.01 mm) over a 0.31 mm diameter gives a fractional uncertainty of 0.01/0.31 = 3.2% for d. Since A depends on d², the fractional uncertainty in area is 2 × 3.2% = 6.4%.
每一项测量都会引入不确定度。对于米尺（±1 mm），L₀ 的相对不确定度为 1 mm / 1800 mm = 5.6 × 10⁻⁴（0.056%）。对于 0.31 mm 的直径，螺旋测微器（±0.01 mm）带来的相对不确定度为 0.01/0.31 = 3.2%。由于 A 取决于 d²，面积的相对不确定度为 2 × 3.2% = 6.4%。
a simple ruler is used. A travelling microscope can reduce the reading uncertainty to ±0.1 mm. In the example, a typical ΔL of 1.4 mm with ±0.1 mm uncertainty has a percentage uncertainty of (0.1/1.4) × 100% = 7.1% ractional uncertainties in stress and strain gives the total uncertainty in E of around 10–15%.
伸长量 ΔL 的测量通常具有最大的不确定度，尤其在使用普通直尺时。读数显微镜可将读数不确定度降至 ±0.1 mm。在本例中，一个典型的 1.4 mm 伸长量伴随 ±0.1 mm 不确定度，其百分比不确定度为 (0.1/1.4) × 100% = 7.1%。将应力和应变的相对不确定度相加，可得出杨氏模量 E 的总不确定度约为 10%–15%。
Candidates in the June 2022 exam were expected to combine these uncertainties correctly, either by adding percentage uncertainties or by constructing worst‑case calculation lines on the graph.
2022 年 6 月的考生应能正确合成这些不确定度，可采用百分比不确定度相加法，或在图像上构建最差情况计算线。
9. Common Sources of Error and Improvements | 常见误差来源与改进
A frequent error is using a wire with kinks that does not hang truly vertically. This causes the measured extension to be less than the actual strain, underestimating E. To reduce this, a pre‑tensioning small weight and a plumb line can ensure perfectly vertical alignment. Another systematic error arises from slippage at the clamp – the wire can slowly pull through when loaded, giving a larger apparent extension. Using a split‑clamp or wrapping the wire around a peg improves grip.
常见错误之一是使用了带有扭结的铜线，致使悬垂不竖直。这会导致测得的伸长量小于实际应变，从而低估了 E。为减少此类误差，可施加预张紧的小砝码并使用铅垂线，以确保完全垂直对齐。另一个系统误差来源于夹持处的滑移——加载时铜线可能缓慢地被拉出，使表观伸长量偏大。采用分瓣夹头或将铜线缠绕在销钉上，可以改善夹持效果。
Zero error in the micrometer and parallax error when reading the marker against the ruler also compromise precision. Candidates should check the micrometer for zero error and align their eye perpendicularly to the scale. Furthermore, the temperature of the wire should remain constant, as thermal expansion can mimic extension.
螺旋测微器的零点误差，以及从直尺读取标记位置时的视差，也会影响精度。考生应检查测微器零点，并使视线与刻度垂直。此外，铜线的温度应保持恒定，因为热膨胀也会被误判为伸长。
For higher accuracy, multiple loading‑unloading cycles can be performed and averaged, and an optical lever or a strain gauge can replace the simple ruler. These enhancements are often discussed in the exam’s follow‑up questions.
为获得更高准确度，可进行多次加载‑卸载循环并取平均值，也可用光杠杆或应变片替代简单的直尺。考试中的延伸问题常涉及这些改进措施。
10. Safety Considerations in the Lab | 实验室安全注意事项
When a wire under high tension snaps, it can recoil with considerable energy, posing a risk of injury to the eyes and face. Therefore, safety spectacles must be worn throughout the experiment, and a transparent safety screen should be placed between the apparatus and the experimenter. A sand tray or a soft landing placed below the masses prevents damage to the floor and reduces the shock if weights fall.
当处于高张力下的铜线突然断裂时，会以相当大的能量回弹，有伤及眼睛和面部的风险。因此，实验全程必须佩戴安全护目镜，并在仪器与实验者之间放置透明安全防护屏。在砝码下方放置沙盘或软垫，可防止损坏地面，并减轻重物坠落的冲击。
The load should never exceed the elastic limit of the wire as determined by preliminary tests; overloading can cause the wire to snap suddenly. Students should also be careful when handling masses to avoid dropping them on their feet.
所加载荷切不可超过经预实验确定的铜线弹性极限；超载会导致铜线突然断裂。学生在操作砝码时也需小心，避免砸到脚部。
During the June 2022 assessment, examiners noted that many candidates mentioned safety goggles but failed to specify protection against snapping wires. A precise answer earns full marks, so being specific about risks and remedies is crucial.
在 2022 年 6 月的考评中，考官注意到许多考生提到了护目镜，但未具体说明是防止铜线断裂弹射。精准的答案才能获得满分，因此明确表述风险与防护措施至关重要。
11. Connecting to the June 2022 Exam Questions | 联系 2022 年 6 月真题
The AS Paper 2 included a series of questions surrounding this practical. One typical question asked students to describe how they would measure the diameter of the wire to obtain an accurate average. This required not just naming the instrument (micrometer) but also detailing taking measurements at multiple orientations to account for non‑uniformity, and calculating the mean with proper significant figures.
AS 试卷 2 中包含了一系列围绕该实验的问题。一个典型问题是：请描述如何测量铜线直径以获得准确的平均值。这不仅要求说出仪器名称（螺旋测微器），还须详述在不同方向多次测量以考虑不均匀性，并用恰当的有效数字计算平均值。
Another question presented a graph of load versus extension and asked for the calculation of Young’s modulus, requiring students to convert load to stress using the given diameter. Candidates were also prompted to identify a major source of error and suggest a practical improvement, tying seamlessly into the safety and procedural discussions above.
另一道题给出了一幅载荷‑伸长量关系图，要求计算杨氏模量，这就需要考生利用已知直径将载荷转换为应力。试卷还让考生指出一项主要误差来源并提出一项实际改进措施，与上文的安全与流程讨论无缝衔接。
Thus, mastering the experimental detail and underlying physics is essential not only for practical endorsements but also for the written paper. Students who practised the full experiment were able to answer these questions more confidently and accurately.
因此，掌握实验细节和底层物理原理，不仅对实验资格认证至关重要，对笔试也同样关键。那些完整操作过实验的学生，回答这些问题时会更加自信和准确。
12. Summary and Exam Preparation Tips | 总结与备考建议
To sum up, determining Young’s modulus of copper involves careful measurement of length, diameter, and extension under known forces. A stress‑strain graph yields the modulus as its gradient. Success depends on minimising uncertainties, recognising systematic errors, and maintaining safe laboratory practice. The June 2022 paper tested these competencies thoroughly.
总而言之，测定铜的杨氏模量需要仔细测量长度、直径以及已知作用力下的伸长量，并通过应力‑应变图像的斜率求得模量。成功的关键在于尽可能降低不确定度、识别系统误差并遵守安全操作规程。2022 年 6 月的试卷对这些能力进行了全面考查。
When revising, reconstruct the data table, practise plotting the graph with error bars, and compute the modulus along with its uncertainty. Anticipate follow‑up questions on alternative measurement techniques, such as using a Searle’s apparatus, and think about how they reduce error. This holistic preparation will serve you well in both practical assessments and written exams.
在复习中，不妨重新构建数据表，练习绘制带误差棒的图像，并计算模量及其不确定度。预想关于替代测量技术的延伸问题，例如使用西尔勒仪（Searle's apparatus），并思考这些技术如何减小误差。这种全面的准备将使你在实验考核与笔试中都能游刃有余。
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