📚 Mastering the A-Level Physics Insert (Unit 1, Jan 2021): Young Modulus Experiment | 掌握A-Level物理配套材料（2021年1月单元1）：杨氏模量实验
The A-Level Physics Paper 1 Insert for January 2021 provides a set of experimental measurements taken during a classic investigation to determine the Young modulus of a metal wire. This article will guide you through the skills needed to analyse such data, plot graphs, calculate uncertainties, and evaluate the procedure. Mastering these skills is essential for both your practical endorsement and written examination.
2021年1月的A-Level物理试卷1配套资料提供了一组用于测定金属丝杨氏模量的经典实验测量数据。本文将引导你掌握分析这类数据、绘制图像、计算不确定度以及评估实验步骤所需的技能。熟练掌握这些技能对实践考核和笔试都至关重要。
1. Introduction to the Setup | 实验装置简介
The experiment aims to find the Young modulus of a metal wire by measuring how much it stretches under known loads. A long wire is clamped at one end, passes over a pulley, and hangs vertically with a weight hanger at the free end. A marker on the wire is observed against a fixed metre rule to record its position. A micrometer screw gauge measures the wire’s diameter at several points along its length.
该实验的目的是通过测量金属丝在已知载荷下的伸长量来求其杨氏模量。一根长线一端被固定夹紧，绕过滑轮垂直悬挂，自由端带有砝码盘。线上的标记相对于固定米尺的位置被记录下来。用千分尺在线的不同位置多次测量其直径。
2. Understanding the Insert Data | 理解配套资料中的数据
The January 2021 insert contains a table of mass and corresponding scale readings. A typical set of data might appear as follows. Notice the original length L₀ of the wire is 2.500 m, and the no-load scale reading s₀ is 163.7 mm. The diameter d is measured with the micrometer, and three repeated readings are given: 0.46 mm, 0.48 mm, 0.47 mm.
2021年1月的配套资料包含一个质量与对应标尺读数的表格。一组典型数据如下所示。注意金属丝的原长 L₀ = 2.500 m，空载标尺读数 s₀ = 163.7 mm。直径 d 用千分尺测量，给出了三次重复读数：0.46 mm、0.48 mm、0.47 mm。
| Mass m / g | Scale reading s / mm |
|---|---|
| 0 | 163.7 |
| 100 | 155.2 |
| 200 | 146.9 |
| 300 | 138.8 |
| 400 | 130.9 |
| 500 | 123.2 |
3. Calculating Extension and Force | 计算伸长量与力
The extension ΔL for each mass is found by subtracting the current scale reading from the no-load reading: ΔL = s₀ − s. For example, with m = 100 g, ΔL = 163.7 mm − 155.2 mm = 8.5 mm = 8.5 × 10⁻³ m. The force F applied by the weight is F = mg, where g = 9.81 N kg⁻¹. Working in SI units is essential: m = 0.100 kg gives F = 0.100 × 9.81 = 0.981 N.
每个质量下的伸长量 ΔL 由空载读数减去当前标尺读数得到：ΔL = s₀ − s。例如，m = 100 g 时，ΔL = 163.7 mm − 155.2 mm = 8.5 mm = 8.5 × 10⁻³ m。重物施加的力为 F = mg，此处 g = 9.81 N kg⁻¹。使用国际单位至关重要：m = 0.100 kg 则 F = 0.100 × 9.81 = 0.981 N。
To aid further analysis, it is useful to compile a new table, converting all mass values to forces in newtons and all extensions to metres. This step significantly reduces unit errors when determining the gradient and the final Young modulus.
为进一步分析，建议编制一个新表格，将所有质量值换算为以牛顿为单位的力，所有伸长量换算为米。该步骤可大幅减少在计算斜率和最终杨氏模量时出现的单位错误。
4. Linearising the Relationship | 线性化关系
The Young modulus E is defined by stress/strain = (F/A) / (ΔL/L₀). Rearranging gives F = (EA/L₀) ΔL. This shows that a graph of force F on the y-axis against extension ΔL on the x-axis should yield a straight line through the origin, with gradient = EA/L₀, provided the elastic limit is not exceeded.
杨氏模量 E 定义为应力/应变 = (F/A) / (ΔL/L₀)。整理后得 F = (EA/L₀) ΔL。这表明，只要未超过弹性极限，以力 F 为 y 轴、伸长量 ΔL 为 x 轴绘制的图像应是一条通过原点的直线，其斜率 = EA/L₀。
The linear relationship allows us to use the gradient to find E, without needing to know the absolute value of any individual force–extension pair after plotting the best-fit line. Plotting F against ΔL therefore makes the experiment’s analysis both straightforward and robust.
利用这种线性关系，我们可以在绘制最佳拟合线后，通过斜率求出 E，而不必依赖任何单独的力-伸长量数据对。因此，作 F-ΔL 图像使实验分析既简便又可靠。
5. Plotting an Accurate Graph | 绘制精确图像
Use graph paper or software to plot force F (N) on the vertical axis and extension ΔL (m) on the horizontal axis. Each data point should be marked with a small cross (×), and a clear best-fit straight line must be drawn. Do not force the line through the origin unless the data strongly support it; in this experiment a small intercept may appear due to initial slack in the wire.
使用坐标纸或软件，将力 F (N) 置于纵轴，伸长量 ΔL (m) 置于横轴。每个数据点用小十字 (×) 标出，并绘制一条清晰的最佳拟合直线。除非数据强烈支持，否则不要强制直线经过原点；本实验中由于线材初始松弛，可能出现微小截距。
Axes must be labelled with the quantity and unit, and the scales chosen so the plotted points occupy more than half the graph area. An incorrectly scaled graph is a common source of lost marks in A-Level physics exams.
坐标轴必须标注物理量和单位，刻度选择应使所绘点占据图面一半以上。坐标轴比例不当是A-Level物理考试中常见的失分原因。
6. Determining the Gradient and Young Modulus | 计算斜率与杨氏模量
Select two widely spaced points on the best-fit line (not data points) to calculate the gradient. Using the displayed data and a plausible best-fit line, suppose the line passes through (ΔL = 2.0 × 10⁻³ m, F = 0.24 N) and (ΔL = 12.0 × 10⁻³ m, F = 1.42 N). The gradient k = ΔF/ΔL = (1.42 − 0.24) / (12.0 × 10⁻³ − 2.0 × 10⁻³) = 1.18 / 10.0 × 10⁻³ = 118 N m⁻¹.
在最佳拟合直线上选取两个相距较远的点（非数据点）计算斜率。利用所给数据及一条合理的最佳拟合直线，假设直线经过 (ΔL = 2.0 × 10⁻³ m, F = 0.24 N) 和 (ΔL = 12.0 × 10⁻³ m, F = 1.42 N)。斜率 k = ΔF/ΔL = (1.42 − 0.24) / (12.0 × 10⁻³ − 2.0 × 10⁻³) = 1.18 / 10.0 × 10⁻³ = 118 N m⁻¹。
Next, calculate the cross-sectional area A of the wire. The mean diameter dₘ = (0.46 + 0.48 + 0.47) / 3 = 0.470 mm = 4.70 × 10⁻⁴ m. Then A = π (dₘ/2)² = π dₘ²/4 = π (4.70 × 10⁻⁴)² / 4 = 1.735 × 10⁻⁷ m².
The Young modulus E = (gradient × L₀) / A = (118 × 2.500) / 1.735 × 10⁻⁷ = 295 / 1.735 × 10⁻⁷ ≈ 1.70 × 10¹¹ Pa.
接着计算金属丝的横截面积 A。平均直径 dₘ = (0.46 + 0.48 + 0.47) / 3 = 0.470 mm = 4.70 × 10⁻⁴ m。则 A = π (dₘ/2)² = π dₘ²/4 = π (4.70 × 10⁻⁴)² / 4 = 1.735 × 10⁻⁷ m²。
杨氏模量 E = (斜率 × L₀) / A = (118 × 2.500) / 1.735 × 10⁻⁷ = 295 / 1.735 × 10⁻⁷ ≈ 1.70 × 10¹¹ Pa。
7. Measurement Uncertainties | 测量不确定度
The main uncertainties arise from the diameter, the original length, and the scale readings. The micrometer typically has a resolution of 0.01 mm, giving an absolute uncertainty of ±0.01 mm. The metre rule for L₀ might have an uncertainty of ±2 mm. The scale used for extension readings has a resolution of 1 mm, but repeated readings suggest an uncertainty of about ±0.5 mm in extension.
主要的不确定度来源于直径、原长和标尺读数。千分尺的分辨力通常为 0.01 mm，绝对不确定度为 ±0.01 mm。测量 L₀ 的米尺不确定度约为 ±2 mm。用于读伸长量的标尺分辨力为 1 mm，但多次读数显示伸长量的不确定度约为 ±0.5 mm。
Percentage uncertainty in diameter is (0.01 / 0.470) × 100% ≈ 2.13%. Since area depends on d², its percentage uncertainty is 2 × 2.13% = 4.26%. The percentage uncertainty in L₀ is (0.002 / 2.500) × 100% = 0.08%. For extension, a typical value of 10 mm gives (0.5 / 10) × 100% = 5%. The force measurement (using calibrated masses) has a negligible uncertainty (<0.1%).
直径的百分不确定度为 (0.01 / 0.470) × 100% ≈ 2.13%。由于面积取决于 d²，其百分不确定度为 2 × 2.13% = 4.26%。L₀ 的百分不确定度为 (0.002 / 2.500) × 100% = 0.08%。对伸长量，取典型值 10 mm 得 (0.5 / 10) × 100% = 5%。力测量（使用校准砝码）的不确定度可忽略 (<0.1%)。
8. Combining Uncertainties and Final Result | 合成不确定度与最终结果
The total percentage uncertainty in the Young modulus is obtained by adding the percentage uncertainties of the quantities that are multiplied or divided. Here, %U(E) ≈ %U(ΔL) + %U(L₀) + 2 × %U(d) = 5% + 0.08% + 4.26% ≈ 9.34%. The contributions from the numerical constant π and the gradient are assumed small.
杨氏模量的总百分不确定度由相乘或相除的各量的百分不确定度相加得到。此处，%U(E) ≈ %U(ΔL) + %U(L₀) + 2 × %U(d) = 5% + 0.08% + 4.26% ≈ 9.34%。常数 π 和斜率带来的不确定度假定为较小。
Absolute uncertainty in E = 9.34% × 1.70 × 10¹¹ Pa ≈ 0.16 × 10¹¹ Pa. Therefore, the Young modulus can be quoted as E = (1.70 ± 0.16) × 10¹¹ Pa. This value is close to the accepted figure for copper (about 1.1–1.3 × 10¹¹ Pa) or steel (~2.0 × 10¹¹ Pa), suggesting the wire is likely a steel alloy.
E 的绝对不确定度 = 9.34% × 1.70 × 10¹¹ Pa ≈ 0.16 × 10¹¹ Pa。因此杨氏模量可表示为 E = (1.70 ± 0.16) × 10¹¹ Pa。该值接近钢材的公认值（约 2.0 × 10¹¹ Pa），表明该金属丝可能是一种钢合金。
9. Evaluating Sources of Error | 评估误差来源
Systematic errors may arise if the wire is not perfectly straight at the start or if the pulley introduces friction. The no-load reading s₀ might change if the wire permanently stretches after heavy loading. Parallax errors when reading the marker against the ruler can also skew the data.
如果线材在初始时并非完全笔直，或者滑轮引入摩擦，则会产生系统误差。如果线材在大载荷下发生永久伸长，空载读数 s₀ 可能改变。标记对标尺读数时的视差也会使数据偏移。
Random errors are reduced by taking repeated diameter readings at different positions along the wire. The extension readings are particularly sensitive; a fine pointer or Vernier scale can improve precision. The original length L₀ should be measured with the wire under a small tension to remove kinks.
通过在线材不同位置重复测量直径，可以减少随机误差。伸长量读数尤其敏感；使用细指针或游标尺可提高精度。测量原长 L₀ 时，应使线材承受微小张力，消除卷曲。
10. Exam Tips and Common Mistakes | 考试提示与常见错误
Always calculate extensions using s₀ minus s, not s minus s₀. Convert all lengths to metres before substituting into equations. When drawing the graph, use a sharp pencil and ensure the line of best fit has a balanced distribution of points on either side. Do not use data points to calculate the gradient—use points taken from the line.
始终用 s₀ − s 计算伸长量，而非 s − s₀。代入方程前将所有长度换算为米。绘图时使用尖细的铅笔，确保最佳拟合线两侧点的分布平衡。不要用数据点计算斜率——应使用取自直线的点。
In uncertainty calculations, remember that the diameter uncertainty is squared because the cross-sectional area depends on d². Finally, when comparing your result to an accepted value, express the difference as a percentage: % difference = |E_experiment − E_accepted| / E_accepted × 100%, and relate it to the total percentage uncertainty to judge reliability.
在不确定度计算中，记住由于横截面积取决于 d²，直径的不确定度需加倍。最后，将你的结果与公认值比较时，用百分差表示：% 差异 = |E_实验 − E_公认| / E_公认 × 100%，并结合总百分不确定度判断结果是否可靠。
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