📚 Year 13 SQA Physics: Case Study in Action | SQA Year 13 物理：案例分析实战演练
In Year 13 SQA Physics, case study questions demand a blend of practical skills, data analysis, and scientific reasoning. This article walks you through a complete worked example, from planning an investigation to evaluating results, so you can tackle any case study with confidence.
在SQA Year 13 物理中，案例分析题需要你综合运用实验技能、数据处理和科学推理。本文将通过一个完整的实例，从设计探究到评价结果，帮助你自信应对任何案例分析。
1. The Nature of SQA Case Studies | SQA案例分析的性质
SQA Advanced Higher Physics case studies are not traditional ‘theory’ questions. They present a real‑world scenario, often with raw experimental data, and ask you to plan, process, evaluate, and draw conclusions. Mark schemes reward logical structure, correct use of uncertainties, and critical comparison with accepted values.
SQA高级物理的案例分析并非传统的理论题。它们给出一个真实情境，常附有原始实验数据，要求你进行计划、处理、评价并得出结论。评分标准看重逻辑结构、正确的不确定度处理以及与公认值的批判性比较。
In such questions, you are effectively acting as a researcher. The examiner wants to see that you can think independently, identify the relevant physics, and communicate your findings clearly. A strong answer always links numerical results back to the underlying principles.
在这类题目中，你实际上扮演着研究者的角色。考官希望看到你能独立思考、识别相关物理原理并清晰地表达结果。优秀的答案总是将数值结果与背后的原理联系起来。
2. Defining the Problem: Determining g Using a Pendulum | 定义问题：利用单摆测定 g
A classic case study asks you to determine the acceleration due to gravity, g, using a simple pendulum. The underlying theory gives the period T = 2π√(L/g). By measuring the period for different lengths L, you can find a value for g and assess its uncertainty.
一个经典的案例分析要求你利用单摆测定重力加速度 g。基本理论给出周期 T = 2π√(L/g)。通过测量不同摆长 L 对应的周期，你可以求出 g 并评估其不确定度。
The aim must be stated precisely: ‘To obtain an experimental value for g by measuring the period of a simple pendulum at various lengths and analysing the linearised relationship T² against L.’ A clear aim guides the entire investigation.
实验目的必须准确表述：“通过测量不同摆长下单摆的周期，并分析 T² 与 L 的线性关系，获得重力加速度 g 的实验值。”清晰的目的能够指引整个探究过程。
3. Experimental Plan: Variables and Procedure | 实验计划：变量与步骤
Independent variable: length of the pendulum (L), measured from the point of suspension to the centre of the bob. Dependent variable: period (T), derived from the time for 20 complete oscillations. Controlled variables: amplitude (keep angle less than 10°), mass and shape of the bob, and point of suspension.
自变量：摆长 L，从悬点到摆球中心的距离。因变量：周期 T，由 20 次全振动时间算出。控制变量：振幅（保持摆角小于 10°）、摆球的质量和形状以及悬点。
A typical procedure: Set up the pendulum with L ≈ 0.500 m. Measure L with a metre rule and a set square to minimise parallax. Displace the bob by a small angle (<10°) and release. Use a stopwatch to time 20 complete swings. Repeat three times for each length. Change L in steps up to 1.200 m.
典型步骤：搭建摆长约 0.500 m 的单摆。用米尺和三角尺测量 L，以减小视差。将摆球拉离平衡位置一个小角度（<10°）后释放。用秒表测量 20 次全摆动的时间。每个摆长重复三次。逐步改变 L 直至 1.200 m。
4. Recording Data: Tables and Significant Figures | 记录数据：表格与有效数字
Data must be presented in a clear table with headings, units, and consistent significant figures. Below is a sample set of measurements collected during the pendulum experiment.
数据必须用表格清晰呈现，包括标题、单位和一致的有效数字。下方是单摆实验中收集的一组示例数据。
| L / m (±0.002 m) | Time for 20T / s (trial 1) | Time for 20T / s (trial 2) | Time for 20T / s (trial 3) | Mean 20T / s | Mean T / s | T² / s² |
|---|---|---|---|---|---|---|
| 0.500 | 28.5 | 28.3 | 28.4 | 28.4 | 1.420 | 2.016 |
| 0.800 | 35.9 | 35.7 | 35.8 | 35.8 | 1.790 | 3.204 |
| 1.000 | 40.2 | 40.0 | 40.1 | 40.1 | 2.005 | 4.020 |
| 1.200 | 44.0 | 43.8 | 43.9 | 43.9 | 2.195 | 4.818 |
Significant figures mirror the precision of the instruments: L is given to 3 decimal places (0.500 m) because the metre rule reads to 1 mm. The mean periods are quoted to 3 decimal places, and T² is kept to 3 decimal places for consistency, though further calculation may use more digits.
有效数字应反映仪器精度：L 给出三位小数（0.500 m），因为米尺可读到 1 mm。平均周期也取三位小数，T² 为保持一致同样取三位小数，但后续计算中可能保留更多位数。
5. Uncertainty Analysis: Instrumental and Random Errors | 不确定度分析：仪器和随机误差
Every measurement carries uncertainty. For L, the instrumental uncertainty is ±1 mm, but an additional parallax error may increase it to ±0.002 m, so we adopt ΔL = ±0.002 m. For the time, the stopwatch reading uncertainty is ±0.01 s, but human reaction time (±0.1 s) dominates. Since we time 20 oscillations, the absolute uncertainty in the mean 20T is taken as the half‑range of the three trials: for L=0.500 m, max 28.5 s, min 28.3 s, so Δ(20T) = (28.5−28.3)/2 = ±0.1 s.
每次测量都带有不确定度。长度 L 的仪器不确定度为 ±1 mm，但视差可能使之增大到 ±0.002 m，因此采用 ΔL = ±0.002 m。时间方面，秒表读数不确定度为 ±0.01 s，但人的反应时间（±0.1 s）起主导作用。由于计时 20 次摆动，平均 20T 的绝对不确定度取三次测量半极差：L=0.500 m 时，最大值 28.5 s，最小值 28.3 s，故 Δ(20T) = (28.5−28.3)/2 = ±0.1 s。
The period T is obtained by dividing 20T by 20, so ΔT = Δ(20T)/20 = ±0.005 s. This random uncertainty is larger than the instrumental uncertainty and is therefore used in subsequent calculations. The relative uncertainty in T² is then 2ΔT/T.
周期 T 由 20T 除以 20 得到，因此 ΔT = Δ(20T)/20 = ±0.005 s。该随机不确定度大于仪器不确定度，因此用于后续计算。T² 的相对不确定度则为 2ΔT/T。
6. Transforming Data: Linearising the Relationship | 数据转换：使关系线性化
The relationship T = 2π√(L/g) is not linear. Squaring both sides gives T² = (4π²/g) L. Thus, plotting T² on the y‑axis against L on the x‑axis should yield a straight line through the origin, with gradient m = 4π²/g. This linearisation allows a simple graphical determination of g.
T = 2π√(L/g) 并非线性关系。两边平方后得到 T² = (4π²/g) L。因此，以 T² 为 y 轴、L 为 x 轴绘图，应得到一条过原点的直线，其斜率 m = 4π²/g。这一线性化处理使得用图像求 g 变得简便。
In our case, the data are already tabulated. We can now assign uncertainties to the plotted quantities: for L, horizontal error bars are ±0.002 m; for T², vertical error bars are calculated using Δ(T²) = 2T × ΔT. For L=0.500 m, T=1.420 s, ΔT=0.005 s, so Δ(T²) = 2 × 1.420 × 0.005 = 0.014 s².
在我们的案例中，数据已经列表。现在可以为绘图量分配不确定度：L 的水平误差棒为 ±0.002 m；T² 的垂直误差棒用 Δ(T²) = 2T × ΔT 计算。以 L=0.500 m 为例，T=1.420 s，ΔT=0.005 s，因此 Δ(T²) = 2 × 1.420 × 0.005 = 0.014 s²。
7. Plotting the Graph: T² Against L | 绘制图像：T² 对 L
Plot the four data points on graph paper or using software. Mark axes clearly: x‑axis ‘Length L / m’, y‑axis ‘T² / s²’. Use sensible scales, do not force the origin to be a data point, and draw error bars for both axes. A best‑fit straight line should pass as close as possible to all points, respecting the error bars.
在坐标纸或软件上绘制四个数据点。坐标轴标记清楚：x 轴“Length L / m”，y 轴“T² / s²”。使用合理的分度，不必将原点作为数据点，并画出两轴的误差棒。一条最佳拟合直线应尽可能靠近所有点，并顾及误差棒的范围。
From the plot, the gradient is determined by taking two well‑separated points on the line, not necessarily the data points. Suppose the line passes through (0.200, 0.800) and (1.200, 4.800). Then gradient m = (4.800 − 0.800) / (1.200 − 0.200) = 4.000 / 1.000 = 4.00 s² m⁻¹.
根据图像，在直线上选取两个距离较远的点（不一定是原始数据点）计算斜率。假设直线通过 (0.200, 0.800) 和 (1.200, 4.800)，则斜率 m = (4.800 − 0.800) / (1.200 − 0.200) = 4.000 / 1.000 = 4.00 s² m⁻¹。
8. Calculating g from Graph Gradient | 从图像斜率计算 g
Since m = 4π²/g, rearranging gives g = 4π²/m. Using m = 4.00 s² m⁻¹, we obtain g = 4 × (3.1416)² / 4.00 ≈ 39.478 / 4.00 = 9.87 m s⁻². This value is slightly higher than the accepted 9.81 m s⁻² but lies within a reasonable range.
因为 m = 4π²/g，整理得 g = 4π²/m。代入 m = 4.00 s² m⁻¹，得到 g = 4 × (3.1416)² / 4.00 ≈ 39.478 / 4.00 = 9.87 m s⁻²。该值略高于公认值 9.81 m s⁻²，但仍处于合理范围内。
Compare with the theoretical value using percentage difference: |9.87−9.81|/9.81 × 100% ≈ 0.6%. This agreement is excellent for a school laboratory. However, the true test is whether the uncertainty range covers the accepted value.
与理论值比较的百分比差为 |9.87−9.81|/9.81 × 100% ≈ 0.6%。对学校实验室而言，这一符合程度相当出色。但真正的检验在于不确定度范围是否包含了公认值。
9. Uncertainty in g Using Error Bars | 使用误差棒求 g 的不确定度
To find the uncertainty in the gradient, draw the ‘steepest’ and ‘shallowest’ reasonable straight lines that still touch all error bars. The steepest line might pass through (0.200, 0.820) and (1.200, 4.820), giving mmax = (4.820−0.820)/1.000 = 4.000 s² m⁻¹. The shallowest line could go through (0.200, 0.780) and (1.200, 4.780), giving mmin = 4.000 s² m⁻¹. In a more realistic scenario with scatter, these would differ. Here assume mmax = 4.10 and mmin = 3.90, giving Δm = (4.10−3.90)/2 = ±0.10 s² m⁻¹.
欲求斜率的不确定度，可画出与所有误差棒依然相切的最陡和最缓的合理直线。最陡线可能通过 (0.200, 0.820) 和 (1.200, 4.820)，得 mmax = 4.000 s² m⁻¹，最缓线通过 (0.200, 0.780) 和 (1.200, 4.780)，得 mmin = 4.000 s² m⁻¹。在数据有离散的实际情况下，这两个斜率会不同。此处假定 mmax = 4.10，mmin = 3.90，故 Δm = (4.10−3.90)/2 = ±0.10 s² m⁻¹。
Using the relationship Δg/g = Δm/m (since g ∝ 1/m), we get Δg = g × (Δm/m) = 9.87 × (0.10/4.00) = 9.87 × 0.025 = ±0.25 m s⁻². Thus g = 9.87 ± 0.25 m s⁻². The accepted value 9.81 lies within this range, confirming consistency.
利用关系式 Δg/g = Δm/m（因为 g ∝ 1/m），得到 Δg = g × (Δm/m) = 9.87 × (0.10/4.00) = 9.87 × 0.025 = ±0.25 m s⁻²。因此 g = 9.87 ± 0.25 m s⁻²。公认值 9.81 在此范围内，说明结果具有一致性。
10. Evaluation and Improvements | 评价与改进
A sound evaluation identifies the main sources of error and suggests specific improvements. In this experiment, reaction time in starting and stopping the stopwatch is the dominant random error. Using a light gate and electronic timer would eliminate this. The measurement of L could be improved by using a clamp‑style ruler with a zero‑error correction.
一个好的评价应指出主要的误差来源并提出具体的改进措施。本实验中，启动和停止秒表的反应时间是主要的随机误差。使用光门和电子计时器可以消除它。长度 L 的测量可采用带零误差修正的卡式直尺来改进。
Systematic errors may include the approximation that the pendulum behaves as a simple pendulum only for small angles. Keeping the angle below 10° minimises this, but a slight angle dependence may still contribute. Additionally, the mass of the string and air resistance slightly increase the effective length. Using a thinner, lighter string and a denser bob would help.
系统误差可能包括仅在摆角很小时单摆才近似成立这一前提。将摆角保持在 10° 以下可减小这一影响，但微小角度依赖依然存在。此外，绳的质量和空气阻力会略微增加等效摆长。使用更细、更轻的绳和密度更大的摆球会有所帮助。
11. Conclusion and Exam Strategy | 结论与考试策略
In summary, the experiment yielded g = 9.87 ± 0.25 m s⁻², agreeing with the standard value within experimental uncertainty. The case study method mirrors the process of an authentic scientific investigation: define the problem, take careful measurements, process data with uncertainty, linearise, analyse graphically, and evaluate critically.
总之，实验测得 g = 9.87 ± 0.25 m s⁻²，在实验不确定度范围内与标准值一致。案例分析的方法模拟了真实科学研究的过程：定义问题、细致测量、处理数据并考虑不确定度、线性化、图形分析以及批判性评价。
For the SQA exam, always show your working step by step. Use the correct number of significant figures, label graphs fully, and discuss uncertainties in a structured way. A common pitfall is forgetting to convert raw times to periods or to plot the correct derived quantity. Practice with varied scenarios—such as using a mass‑spring system or a diffraction grating—to build flexibility.
参加 SQA 考试时，务必分步展示推导过程。使用正确的有效数字位数，完整标记图表，并有条理地讨论不确定度。一个常见误区是忘记将原始时间转换为周期或绘制错误的导出量。通过多样化的情境练习——如弹簧振子系统或衍射光栅——来培养灵活应变的能力。
Published by TutorHao | Physics Revision Series | aleveler.com
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