Topic 1 测量与不确定度 / Measurements & Uncertainties
SI 单位 · 科学记数法 · 有效数字 · 误差分析 · 不确定度传播 · 图像法
IB SL / HLTopic 1测量是物理的基础
物理学的核心是对自然现象进行定量测量。没有测量就没有物理定律。任何测量都包含数值和单位两部分，且不可避免地带有不确定度。本章学习正确处理测量数据——这是所有实验工作的基础。
Physics is fundamentally about quantitative measurement of natural phenomena. Every measurement consists of a numerical value and a unit, and inevitably carries uncertainty. This chapter teaches proper handling of measurement data — the foundation of all experimental work.
1.1 SI 单位与科学记数法 / SI Units & Scientific Notation
IB 1.1核心基础
SI 基本单位 / SI Base Units
$$ ext{七个 SI 基本单位：kg, m, s, A, K, mol, cd}$$
国际单位制（SI）有七个基本单位，所有其他单位都是导出单位。在 IB 物理中，最常用的基本单位是 kg（千克，质量）、m（米，长度）和 s（秒，时间）。
The International System (SI) has seven base units; all other units are derived. In IB Physics, the most frequently used base units are kg (mass), m (length), and s (time).
| 基本单位 / Base Unit | 符号 / Symbol | 物理量 / Quantity |
|---|---|---|
| 千克 / kilogram | kg | 质量 / Mass |
| 米 / metre | m | 长度 / Length |
| 秒 / second | s | 时间 / Time |
| 安培 / ampere | A | 电流 / Electric current |
| 开尔文 / kelvin | K | 热力学温度 / Thermodynamic temperature |
| 摩尔 / mole | mol | 物质的量 / Amount of substance |
| 坎德拉 / candela | cd | 发光强度 / Luminous intensity |
SI 前缀 / SI Prefixes
$$10^{12} ext{ (T)} o 10^{9} ext{ (G)} o 10^{6} ext{ (M)} o 10^{3} ext{ (k)} o 10^{-2} ext{ (c)} o 10^{-3} ext{ (m)} o 10^{-6}\ (\mu ext{)} o 10^{-9} ext{ (n)} o 10^{-12} ext{ (p)}$$
SI 前缀用于简化极大或极小的数值表示。考前必须熟记：T（太拉）、G（吉咖）、M（兆）、k（千）、c（厘）、m（毫）、$\mu$（微）、n（纳）、p（皮）。
SI prefixes simplify very large or very small numbers. Essential for exams: T (tera), G (giga), M (mega), k (kilo), c (centi), m (milli), $\mu$ (micro), n (nano), p (pico).
| 前缀 / Prefix | 符号 / Symbol | 倍数 / Factor | 示例 / Example |
|---|---|---|---|
| 太 / tera | T | $10^{12}$ | 1 TW = $10^{12}$ W |
| 吉 / giga | G | $10^{9}$ | 1 GHz = $10^{9}$ Hz |
| 兆 / mega | M | $10^{6}$ | 1 MJ = $10^{6}$ J |
| 千 / kilo | k | $10^{3}$ | 1 km = $10^{3}$ m |
| 厘 / centi | c | $10^{-2}$ | 1 cm = $10^{-2}$ m |
| 毫 / milli | m | $10^{-3}$ | 1 mA = $10^{-3}$ A |
| 微 / micro | $\mu$ | $10^{-6}$ | 1 $\mu$m = $10^{-6}$ m |
| 纳 / nano | n | $10^{-9}$ | 1 nF = $10^{-9}$ F |
| 皮 / pico | p | $10^{-12}$ | 1 pF = $10^{-12}$ F |
科学记数法 / Scientific Notation
$$a imes 10^n \quad ext{其中 } 1 \le a < 10,\ n \in \mathbb{Z}$$
科学记数法将数值表示为 1 到 10 之间的系数乘以 10 的整数次幂。IB 考试中，选择题和数据分析题均要求使用科学记数法表达结果。
Scientific notation expresses numbers as a coefficient between 1 and 10 multiplied by a power of 10. IB exams require results in scientific notation for both multiple-choice and data analysis questions.
📐 详细说明 / Detailed Explanation
科学记数法规则：
① 将小数点移到第一个非零数字之后得到 $a$
② $n$ 等于小数点移动的位数
③ 小数点左移 → $n>0$（大数）；小数点右移 → $n<0$（小数）
示例：
$3450000 = 3.45 imes 10^6$（小数点左移 6 位）
$0.0000782 = 7.82 imes 10^{-5}$（小数点右移 5 位）
工程记数法（IB 中常用）：指数是 3 的倍数（如 $10^3, 10^6, 10^{-3}, 10^{-6}$），便于与 SI 前缀对应。
计算器技巧：使用 $ imes 10^x$ 或 $ ext{EXP}$ 键输入科学记数法。$3.45 imes 10^6$ 输入为 3.45 EXP 6 或 3.45 $ imes 10^x$ 6。
① 将小数点移到第一个非零数字之后得到 $a$
② $n$ 等于小数点移动的位数
③ 小数点左移 → $n>0$（大数）；小数点右移 → $n<0$（小数）
示例：
$3450000 = 3.45 imes 10^6$（小数点左移 6 位）
$0.0000782 = 7.82 imes 10^{-5}$（小数点右移 5 位）
工程记数法（IB 中常用）：指数是 3 的倍数（如 $10^3, 10^6, 10^{-3}, 10^{-6}$），便于与 SI 前缀对应。
计算器技巧：使用 $ imes 10^x$ 或 $ ext{EXP}$ 键输入科学记数法。$3.45 imes 10^6$ 输入为 3.45 EXP 6 或 3.45 $ imes 10^x$ 6。
Scientific notation rules:
① Move decimal to after the first non-zero digit → coefficient $a$
② $n$ = number of places the decimal moved
③ Left → $n>0$ (big numbers); Right → $n<0$ (small numbers)
Examples:
$3450000 = 3.45 imes 10^6$ (decimal moves 6 places left)
$0.0000782 = 7.82 imes 10^{-5}$ (decimal moves 5 places right)
Engineering notation (common in IB): exponents are multiples of 3 ($10^3, 10^6, 10^{-3}, 10^{-6}$), aligning with SI prefixes.
Calculator tip: Use the $ imes 10^x$ or EXP key for scientific notation. Enter $3.45 imes 10^6$ as 3.45 EXP 6.
① Move decimal to after the first non-zero digit → coefficient $a$
② $n$ = number of places the decimal moved
③ Left → $n>0$ (big numbers); Right → $n<0$ (small numbers)
Examples:
$3450000 = 3.45 imes 10^6$ (decimal moves 6 places left)
$0.0000782 = 7.82 imes 10^{-5}$ (decimal moves 5 places right)
Engineering notation (common in IB): exponents are multiples of 3 ($10^3, 10^6, 10^{-3}, 10^{-6}$), aligning with SI prefixes.
Calculator tip: Use the $ imes 10^x$ or EXP key for scientific notation. Enter $3.45 imes 10^6$ as 3.45 EXP 6.
🎯 典型应用 / Applications：① 光速 $c = 3.00 imes 10^8$ m/s——用科学记数法避免写出 300,000,000；② 电子质量 $m_e = 9.11 imes 10^{-31}$ kg——极小的数值用负指数简洁表示；③ 细胞大小 10-100 $\mu$m——$10^{-5}$ m 级别；④ 天文距离——地日距离 1 AU = $1.496 imes 10^{11}$ m。
🎯 Typical Applications: ① Speed of light $c = 3.00 imes 10^8$ m/s — avoids writing 300,000,000; ② Electron mass $m_e = 9.11 imes 10^{-31}$ kg — concise for tiny values; ③ Cell size 10-100 $\mu$m — $10^{-5}$ m scale; ④ Astronomical distances — 1 AU = $1.496 imes 10^{11}$ m.
例 1.1 / Example 1.1
将以下数值转换为科学记数法：(a) 1560000 (b) 0.0000429 (c) $5.2 imes 10^4$ km 转换为 m
Convert to scientific notation: (a) 1560000 (b) 0.0000429 (c) $5.2 imes 10^4$ km to m
解题过程 / Solution
(a) $1.56 imes 10^6$ (b) $4.29 imes 10^{-5}$ (c) $5.2 imes 10^4$ km = $5.2 imes 10^4 imes 10^3$ m = $5.2 imes 10^7$ m
(a) $1.56 imes 10^6$ (b) $4.29 imes 10^{-5}$ (c) $5.2 imes 10^4$ km = $5.2 imes 10^4 imes 10^3$ m = $5.2 imes 10^7$ m
例 1.2 / Example 1.2
地球质量 $5.97 imes 10^{24}$ kg，半径为 $6.37 imes 10^6$ m。求地球的平均密度。
Earth mass $5.97 imes 10^{24}$ kg, radius $6.37 imes 10^6$ m. Find Earth's average density.
解题过程 / Solution
$V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (6.37 \times 10^6)^3 = 1.083 \times 10^{21}$ m$^3$
$\rho = m/V = 5.97 \times 10^{24} / 1.083 \times 10^{21} = 5.51 \times 10^3$ kg/m$^3$ ≈ 5.51 g/cm$^3$
$\rho = m/V = 5.97 \times 10^{24} / 1.083 \times 10^{21} = 5.51 \times 10^3$ kg/m$^3$ ≈ 5.51 g/cm$^3$
$V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (6.37 \times 10^6)^3 = 1.083 \times 10^{21}$ m$^3$
$\rho = m/V = 5.97 \times 10^{24} / 1.083 \times 10^{21} = 5.51 \times 10^3$ kg/m$^3$ ≈ 5.51 g/cm$^3$
$\rho = m/V = 5.97 \times 10^{24} / 1.083 \times 10^{21} = 5.51 \times 10^3$ kg/m$^3$ ≈ 5.51 g/cm$^3$
1.2 有效数字 / Significant Figures
IB 1.1考试高频扣分点
有效数字规则 / Rules for Significant Figures
$$ ext{结果的有效数字位数 = 测量值中精度最低的位数}$$
有效数字（sig figs）表示测量结果的可靠程度。IB 考试中，给出错误位数的有效数字会扣分。
Significant figures indicate the reliability of a measurement. In IB exams, incorrect sig figs lead to mark deductions.
📐 详细规则 / Detailed Rules
有效数字的四条核心规则：
① 非零数字总是有效：1234 → 4 位，56.7 → 3 位
② 非零数字之间的零有效：1005 → 4 位，7.03 → 3 位
③ 前导零无效：0.00034 → 2 位（前导零只是占位）
④ 尾随零：
• 带小数点的尾随零 有效：45.00 → 4 位，3.0 → 2 位
• 无小数点的尾随零 可能无效：1500 → 2-4 位（用科学记数法避免歧义）
运算规则：
• 加减法：以小数位数最少的数为准
$12.5 + 3.45 = 15.95 ightarrow 16.0$（一位小数）
• 乘除法：以有效数字最少的数为准
$3.22 imes 2.0 = 6.44 ightarrow 6.4$（两位有效数字）
• 对数/指数：结果的有效数字与原始值相同
① 非零数字总是有效：1234 → 4 位，56.7 → 3 位
② 非零数字之间的零有效：1005 → 4 位，7.03 → 3 位
③ 前导零无效：0.00034 → 2 位（前导零只是占位）
④ 尾随零：
• 带小数点的尾随零 有效：45.00 → 4 位，3.0 → 2 位
• 无小数点的尾随零 可能无效：1500 → 2-4 位（用科学记数法避免歧义）
运算规则：
• 加减法：以小数位数最少的数为准
$12.5 + 3.45 = 15.95 ightarrow 16.0$（一位小数）
• 乘除法：以有效数字最少的数为准
$3.22 imes 2.0 = 6.44 ightarrow 6.4$（两位有效数字）
• 对数/指数：结果的有效数字与原始值相同
Four core rules for sig figs:
① Non-zero digits are always significant: 1234 → 4 sf, 56.7 → 3 sf
② Zeros between non-zero digits are significant: 1005 → 4 sf, 7.03 → 3 sf
③ Leading zeros are NOT significant: 0.00034 → 2 sf (they're placeholders)
④ Trailing zeros:
• With decimal point: significant — 45.00 → 4 sf, 3.0 → 2 sf
• Without decimal point: ambiguous — 1500 → 2-4 sf (use scientific notation)
Operation rules:
• Addition/Subtraction: round to the fewest decimal places
$12.5 + 3.45 = 15.95 ightarrow 16.0$ (one decimal place)
• Multiplication/Division: round to the fewest sig figs
$3.22 imes 2.0 = 6.44 ightarrow 6.4$ (two sig figs)
• Log/Exp: result has same number of sig figs as original
① Non-zero digits are always significant: 1234 → 4 sf, 56.7 → 3 sf
② Zeros between non-zero digits are significant: 1005 → 4 sf, 7.03 → 3 sf
③ Leading zeros are NOT significant: 0.00034 → 2 sf (they're placeholders)
④ Trailing zeros:
• With decimal point: significant — 45.00 → 4 sf, 3.0 → 2 sf
• Without decimal point: ambiguous — 1500 → 2-4 sf (use scientific notation)
Operation rules:
• Addition/Subtraction: round to the fewest decimal places
$12.5 + 3.45 = 15.95 ightarrow 16.0$ (one decimal place)
• Multiplication/Division: round to the fewest sig figs
$3.22 imes 2.0 = 6.44 ightarrow 6.4$ (two sig figs)
• Log/Exp: result has same number of sig figs as original
🎯 IB 考试要点 / Exam Tips：① 最终答案一般用 2-4 位有效数字；② 中间过程保留更多位数，只在最后一步四舍五入；③ 不确定度通常保留 1-2 位有效数字；④ 常数（如 $\pi$、$g$）不限制有效数字位数；⑤ IB 数据题中，图点大小≈0.5 mm，读数的有效数字由刻度决定。
🎯 IB Exam Tips: ① Final answers: 2-4 sf typically; ② Keep extra digits during intermediate steps, only round at the end; ③ Uncertainties usually have 1-2 sf; ④ Constants ($\pi$, $g$) don't limit sig figs; ⑤ In IB data questions, plotting precision ≈0.5 mm, reading sf determined by scale divisions.
例 1.3 / Example 1.3
计算 (a) $3.56 imes 4.0$ (b) $12.45 + 3.1 + 0.678$，结果保留正确有效数字。
Compute (a) $3.56 imes 4.0$ (b) $12.45 + 3.1 + 0.678$ with correct sig figs.
解题过程 / Solution
(a) $3.56 imes 4.0 = 14.24
ightarrow 14$（2 sf，取决于 4.0 的 2 位）
(b) $12.45 + 3.1 + 0.678 = 16.228 ightarrow 16.2$（1 位小数，取决于 3.1）
(b) $12.45 + 3.1 + 0.678 = 16.228 ightarrow 16.2$（1 位小数，取决于 3.1）
(a) $3.56 imes 4.0 = 14.24
ightarrow 14$ (2 sf, limited by 4.0)
(b) $12.45 + 3.1 + 0.678 = 16.228 ightarrow 16.2$ (1 decimal place, limited by 3.1)
(b) $12.45 + 3.1 + 0.678 = 16.228 ightarrow 16.2$ (1 decimal place, limited by 3.1)
例 1.4 / Example 1.4
长度测量：$L = 12.345$ cm（游标卡尺）。请写出 3 位有效数字、4 位有效数字的结果。
Length measurement: $L = 12.345$ cm (vernier caliper). Write result with 3 sf and 4 sf.
解题过程 / Solution
3 sf: $L = 12.3$ cm 4 sf: $L = 12.35$ cm
注意四舍五入：12.345 → 3 sf 时看第 4 位是 4，舍去 → 12.3；→ 4 sf 时看第 5 位是 5，进位 → 12.35
注意四舍五入：12.345 → 3 sf 时看第 4 位是 4，舍去 → 12.3；→ 4 sf 时看第 5 位是 5，进位 → 12.35
3 sf: $L = 12.3$ cm 4 sf: $L = 12.35$ cm
Rounding: 12.345 → 3 sf, check 4th digit is 4, round down → 12.3; → 4 sf, check 5th digit is 5, round up → 12.35
Rounding: 12.345 → 3 sf, check 4th digit is 4, round down → 12.3; → 4 sf, check 5th digit is 5, round up → 12.35
1.3 不确定度与误差 / Uncertainties & Errors
IB 1.2实验报告核心
误差类型 / Types of Errors
$$ ext{随机误差：不可预测的涨落}\quad ext{系统误差：始终偏向一侧}$$
误差分为两大类：随机误差（random errors）和系统误差（systematic errors）。理解它们的区别是分析实验数据的关键。
Errors fall into two categories: random errors (unpredictable fluctuations) and systematic errors (consistent bias in one direction). Understanding the difference is key to experimental analysis.
📐 详细说明 / Detailed Explanation
随机误差：由不可控因素引起，测量值围绕真值上下波动。
• 来源：环境振动、读数估计、温度漂移
• 特点：正负偏差概率相等，平均值接近真值
• 处理方法：多次测量取平均，增加测量次数可减小随机误差
系统误差：测量结果始终偏大或偏小，不因多次测量而减小。
• 来源：仪器未校准（零误差）、实验方法缺陷、环境条件偏离
• 特点：单向偏差，平均值仍有偏
• 处理方法：校准仪器、改进实验方法、修正公式
• 来源：环境振动、读数估计、温度漂移
• 特点：正负偏差概率相等，平均值接近真值
• 处理方法：多次测量取平均，增加测量次数可减小随机误差
系统误差：测量结果始终偏大或偏小，不因多次测量而减小。
• 来源：仪器未校准（零误差）、实验方法缺陷、环境条件偏离
• 特点：单向偏差，平均值仍有偏
• 处理方法：校准仪器、改进实验方法、修正公式
Random errors: Caused by uncontrollable factors; measurements fluctuate around the true value.
• Sources: vibration, reading estimation, temperature drift
• Feature: equal probability of ± deviation; average approaches true value
• Treatment: repeat measurements and take the mean; more trials reduce random error
Systematic errors: Measurements consistently overestimate or underestimate.
• Sources: uncalibrated instruments (zero error), flawed method, environment
• Feature: one-sided bias; averaging doesn't help
• Treatment: calibrate instruments, improve method, apply correction
• Sources: vibration, reading estimation, temperature drift
• Feature: equal probability of ± deviation; average approaches true value
• Treatment: repeat measurements and take the mean; more trials reduce random error
Systematic errors: Measurements consistently overestimate or underestimate.
• Sources: uncalibrated instruments (zero error), flawed method, environment
• Feature: one-sided bias; averaging doesn't help
• Treatment: calibrate instruments, improve method, apply correction
不确定度表示 / Expressing Uncertainty
$$ ext{测量值} = ext{最佳估计值} \pm ext{不确定度}\quad (x \pm \Delta x)$$
不确定度表示测量值的可信范围。IB 中常用绝对不确定度和相对不确定度。
Uncertainty expresses the range within which the true value lies. IB uses absolute uncertainty and fractional/percentage uncertainty.
$$ ext{绝对：}\Delta x\quad ext{相对：} rac{\Delta x}{x}\quad ext{百分比：} rac{\Delta x}{x} imes 100\%$$
📐 不确定度估算 / Estimation Methods
不确定度的估算方法：
① 模拟仪器读数（如尺子、温度计）：
• 不确定度通常取最小刻度的一半
• 例：米尺最小刻度 1 mm → $\Delta L = \pm 0.5$ mm
② 数字仪器读数（如电子天平、数字秒表）：
• 不确定度通常取末位数字的 ±1
• 例：电子天平读数 12.34 g → $\Delta m = \pm 0.01$ g
③ 多次测量统计法：
• 平均值 $ar{x} = \frac{\sum x_i}{n}$
• 不确定度用范围的一半或标准差表示
• $\Delta x = \frac{x_{\max} - x_{\min}}{2}$（IB 简化法）
① 模拟仪器读数（如尺子、温度计）：
• 不确定度通常取最小刻度的一半
• 例：米尺最小刻度 1 mm → $\Delta L = \pm 0.5$ mm
② 数字仪器读数（如电子天平、数字秒表）：
• 不确定度通常取末位数字的 ±1
• 例：电子天平读数 12.34 g → $\Delta m = \pm 0.01$ g
③ 多次测量统计法：
• 平均值 $ar{x} = \frac{\sum x_i}{n}$
• 不确定度用范围的一半或标准差表示
• $\Delta x = \frac{x_{\max} - x_{\min}}{2}$（IB 简化法）
Uncertainty estimation methods:
① Analog instruments (ruler, thermometer):
• Uncertainty ≈ half the smallest division
• e.g., ruler with 1 mm divisions → $\Delta L = \pm 0.5$ mm
② Digital instruments (electronic balance, digital timer):
• Uncertainty ≈ ±1 of the last displayed digit
• e.g., balance reading 12.34 g → $\Delta m = \pm 0.01$ g
③ Statistical method (multiple trials):
• Mean $ar{x} = \frac{\sum x_i}{n}$
• Uncertainty = half the range or standard deviation
• $\Delta x = \frac{x_{\max} - x_{\min}}{2}$ (IB simplified method)
① Analog instruments (ruler, thermometer):
• Uncertainty ≈ half the smallest division
• e.g., ruler with 1 mm divisions → $\Delta L = \pm 0.5$ mm
② Digital instruments (electronic balance, digital timer):
• Uncertainty ≈ ±1 of the last displayed digit
• e.g., balance reading 12.34 g → $\Delta m = \pm 0.01$ g
③ Statistical method (multiple trials):
• Mean $ar{x} = \frac{\sum x_i}{n}$
• Uncertainty = half the range or standard deviation
• $\Delta x = \frac{x_{\max} - x_{\min}}{2}$ (IB simplified method)
🎯 典型应用 / Applications：① 长度测量——用米尺测桌子长度：$L = 120.5 \pm 0.5$ cm，相对不确定度 0.4%；② 时间测量——用手机计时器测单摆周期 10 次，$T = 1.85 \pm 0.02$ s；③ 质量测量——电子天平读数 $m = 25.37 \pm 0.01$ g；④ 温度测量——温度计最小刻度 1°C，$\Delta T = \pm 0.5$°C。
🎯 Typical Applications: ① Length — table measured with ruler: $L = 120.5 \pm 0.5$ cm, 0.4% relative uncertainty; ② Time — 10 pendulum periods with phone timer: $T = 1.85 \pm 0.02$ s; ③ Mass — electronic balance: $m = 25.37 \pm 0.01$ g; ④ Temperature — thermometer with 1°C divisions: $\Delta T = \pm 0.5$°C.
例 1.5 / Example 1.5
学生测量某物体长度五次：12.3, 12.5, 12.4, 12.2, 12.6 mm。求平均值、绝对不确定度和相对不确定度。
A student measures length five times: 12.3, 12.5, 12.4, 12.2, 12.6 mm. Find mean, absolute uncertainty, and relative uncertainty.
解题过程 / Solution
$ar{x} = (12.3 + 12.5 + 12.4 + 12.2 + 12.6)/5 = 12.4$ mm
$\Delta x = (12.6 - 12.2)/2 = 0.2$ mm
结果：$L = 12.4 \pm 0.2$ mm
相对不确定度：$0.2/12.4 = 0.016 = 1.6\%$
$\Delta x = (12.6 - 12.2)/2 = 0.2$ mm
结果：$L = 12.4 \pm 0.2$ mm
相对不确定度：$0.2/12.4 = 0.016 = 1.6\%$
$ar{x} = (12.3 + 12.5 + 12.4 + 12.2 + 12.6)/5 = 12.4$ mm
$\Delta x = (12.6 - 12.2)/2 = 0.2$ mm
Result: $L = 12.4 \pm 0.2$ mm
Relative uncertainty: $0.2/12.4 = 0.016 = 1.6\%$
$\Delta x = (12.6 - 12.2)/2 = 0.2$ mm
Result: $L = 12.4 \pm 0.2$ mm
Relative uncertainty: $0.2/12.4 = 0.016 = 1.6\%$
例 1.6 / Example 1.6
学生用秒表（最小刻度 0.01 s）测单摆周期。单摆摆动 10 次，总时间 18.54 s。请计算周期及其不确定度。
A student uses a stopwatch (0.01 s resolution) to measure pendulum period. 10 oscillations in 18.54 s. Find period and its uncertainty.
解题过程 / Solution
$T = 18.54/10 = 1.854$ s
秒表不确定度 $\pm 0.01$ s，总时间 $\Delta t = \pm 0.01$ s
周期不确定度：$\Delta T = \Delta t/10 = 0.001$ s
结果：$T = 1.854 \pm 0.001$ s
注意：测量多个周期可减小计时不确定度，这是常用的实验技巧。
秒表不确定度 $\pm 0.01$ s，总时间 $\Delta t = \pm 0.01$ s
周期不确定度：$\Delta T = \Delta t/10 = 0.001$ s
结果：$T = 1.854 \pm 0.001$ s
注意：测量多个周期可减小计时不确定度，这是常用的实验技巧。
$T = 18.54/10 = 1.854$ s
Stopwatch uncertainty $\pm 0.01$ s, total time $\Delta t = \pm 0.01$ s
Period uncertainty: $\Delta T = \Delta t/10 = 0.001$ s
Result: $T = 1.854 \pm 0.001$ s
Note: Measuring multiple cycles reduces timing uncertainty — a common experimental technique.
Stopwatch uncertainty $\pm 0.01$ s, total time $\Delta t = \pm 0.01$ s
Period uncertainty: $\Delta T = \Delta t/10 = 0.001$ s
Result: $T = 1.854 \pm 0.001$ s
Note: Measuring multiple cycles reduces timing uncertainty — a common experimental technique.
1.4 不确定度传播 / Uncertainty Propagation
IB 1.2HL & SL 核心难点
传播规则 / Propagation Rules
当测量值通过数学运算得到新量时，不确定度会随之传播。以下是 IB 中必须掌握的基本规则。
When measured values are combined through mathematical operations, uncertainties propagate. The following rules are essential for IB.
$$ ext{加减法：}\quad \Delta Z = \Delta A + \Delta B$$
加法或减法中，绝对不确定度相加。例如：$(A \pm \Delta A) + (B \pm \Delta B) = (A+B) \pm (\Delta A + \Delta B)$。
For addition/subtraction, absolute uncertainties add. E.g., $(A \pm \Delta A) + (B \pm \Delta B) = (A+B) \pm (\Delta A + \Delta B)$.
📐 推导说明 / Derivation Note
加减法不确定度传播推导：
$Z = A \pm B$，最坏情况下 $A$ 和 $B$ 的误差同向叠加。
$Z_{\max} = (A + \Delta A) \pm (B + \Delta B)$
$Z_{\min} = (A - \Delta A) \pm (B - \Delta B)$
$Z_{\max} - Z_{\min} = 2(\Delta A + \Delta B)$
$\Delta Z = (Z_{\max} - Z_{\min})/2 = \Delta A + \Delta B$
注意：如果多个误差独立且随机，更精确的方法是平方和开方（$\sqrt{\Delta A^2 + \Delta B^2}$），但 IB 大纲要求使用线性相加。IB 考试中直接用 $\Delta Z = \Delta A + \Delta B$。
$Z = A \pm B$，最坏情况下 $A$ 和 $B$ 的误差同向叠加。
$Z_{\max} = (A + \Delta A) \pm (B + \Delta B)$
$Z_{\min} = (A - \Delta A) \pm (B - \Delta B)$
$Z_{\max} - Z_{\min} = 2(\Delta A + \Delta B)$
$\Delta Z = (Z_{\max} - Z_{\min})/2 = \Delta A + \Delta B$
注意：如果多个误差独立且随机，更精确的方法是平方和开方（$\sqrt{\Delta A^2 + \Delta B^2}$），但 IB 大纲要求使用线性相加。IB 考试中直接用 $\Delta Z = \Delta A + \Delta B$。
Derivation for addition/subtraction:
$Z = A \pm B$; in the worst case, errors add in the same direction.
$Z_{\max} = (A + \Delta A) \pm (B + \Delta B)$
$Z_{\min} = (A - \Delta A) \pm (B - \Delta B)$
$Z_{\max} - Z_{\min} = 2(\Delta A + \Delta B)$
$\Delta Z = (Z_{\max} - Z_{\min})/2 = \Delta A + \Delta B$
Note: For independent random errors, quadrature sum ($\sqrt{\Delta A^2 + \Delta B^2}$) is more accurate, but IB syllabus requires linear addition.
$Z = A \pm B$; in the worst case, errors add in the same direction.
$Z_{\max} = (A + \Delta A) \pm (B + \Delta B)$
$Z_{\min} = (A - \Delta A) \pm (B - \Delta B)$
$Z_{\max} - Z_{\min} = 2(\Delta A + \Delta B)$
$\Delta Z = (Z_{\max} - Z_{\min})/2 = \Delta A + \Delta B$
Note: For independent random errors, quadrature sum ($\sqrt{\Delta A^2 + \Delta B^2}$) is more accurate, but IB syllabus requires linear addition.
$$ ext{乘除法：}\quad rac{\Delta Z}{Z} = rac{\Delta A}{A} + rac{\Delta B}{B}$$
乘法或除法中，相对不确定度相加。即 $Z = A imes B$ 或 $Z = A/B$ 时，$\Delta Z/Z = \Delta A/A + \Delta B/B$。
For multiplication/division, fractional uncertainties add. When $Z = A imes B$ or $Z = A/B$, $\Delta Z/Z = \Delta A/A + \Delta B/B$.
📐 推导说明 / Derivation Note
乘法不确定度传播推导：
$Z = A \cdot B$
$\ln Z = \ln A + \ln B$
微分：$\displaystyle \frac{dZ}{Z} = \frac{dA}{A} + \frac{dB}{B}$
取绝对值相加：$\displaystyle \frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$
幂次规则：$Z = A^n$ → $\displaystyle \frac{\Delta Z}{Z} = n \cdot \frac{\Delta A}{A}$
推导：$\ln Z = n \ln A$ → $\displaystyle \frac{dZ}{Z} = n \frac{dA}{A}$
例题：$Z = A^2 / B$，则 $\displaystyle \frac{\Delta Z}{Z} = 2 \frac{\Delta A}{A} + \frac{\Delta B}{B}$
$Z = A \cdot B$
$\ln Z = \ln A + \ln B$
微分：$\displaystyle \frac{dZ}{Z} = \frac{dA}{A} + \frac{dB}{B}$
取绝对值相加：$\displaystyle \frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$
幂次规则：$Z = A^n$ → $\displaystyle \frac{\Delta Z}{Z} = n \cdot \frac{\Delta A}{A}$
推导：$\ln Z = n \ln A$ → $\displaystyle \frac{dZ}{Z} = n \frac{dA}{A}$
例题：$Z = A^2 / B$，则 $\displaystyle \frac{\Delta Z}{Z} = 2 \frac{\Delta A}{A} + \frac{\Delta B}{B}$
Derivation for multiplication:
$Z = A \cdot B$
$\ln Z = \ln A + \ln B$
Differentiate: $\displaystyle \frac{dZ}{Z} = \frac{dA}{A} + \frac{dB}{B}$
Add absolute values: $\displaystyle \frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$
Power rule: $Z = A^n$ → $\displaystyle \frac{\Delta Z}{Z} = n \cdot \frac{\Delta A}{A}$
Derivation: $\ln Z = n \ln A$ → $\displaystyle \frac{dZ}{Z} = n \frac{dA}{A}$
Example: $Z = A^2 / B$, then $\displaystyle \frac{\Delta Z}{Z} = 2 \frac{\Delta A}{A} + \frac{\Delta B}{B}$
$Z = A \cdot B$
$\ln Z = \ln A + \ln B$
Differentiate: $\displaystyle \frac{dZ}{Z} = \frac{dA}{A} + \frac{dB}{B}$
Add absolute values: $\displaystyle \frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$
Power rule: $Z = A^n$ → $\displaystyle \frac{\Delta Z}{Z} = n \cdot \frac{\Delta A}{A}$
Derivation: $\ln Z = n \ln A$ → $\displaystyle \frac{dZ}{Z} = n \frac{dA}{A}$
Example: $Z = A^2 / B$, then $\displaystyle \frac{\Delta Z}{Z} = 2 \frac{\Delta A}{A} + \frac{\Delta B}{B}$
$$ ext{幂次：}\quad Z = A^n \;\Rightarrow\; rac{\Delta Z}{Z} = n \cdot rac{\Delta A}{A}$$
幂次规则是乘除法规则的特例：$A^n$ 相当于 $A$ 自乘 $n$ 次，相对不确定度乘以 $n$。当 $n=1/2$（平方根）时，$\Delta Z/Z = \frac12 \cdot \Delta A/A$。
The power rule is a special case of multiplication: $A^n$ means $A$ multiplied by itself $n$ times, so fractional uncertainty is multiplied by $n$. For $n=1/2$ (square root), $\Delta Z/Z = \frac12 \cdot \Delta A/A$.
| 运算 / Operation | 公式 / Formula | 不确定度规则 / Rule |
|---|---|---|
| 加减 / $\pm$ | $Z = A \pm B$ | $\Delta Z = \Delta A + \Delta B$（绝对） |
| 乘除 / $ imes \div$ | $Z = A \cdot B$ 或 $A/B$ | $\Delta Z/Z = \Delta A/A + \Delta B/B$（相对） |
| 幂次 / Power | $Z = A^n$ | $\Delta Z/Z = n \cdot \Delta A/A$（相对） |
| 常数乘 / Constant | $Z = kA$ | $\Delta Z = k \cdot \Delta A$（绝对） |
🎯 典型应用 / Applications：① 面积计算——$L = 5.0 \pm 0.1$ cm，$W = 3.0 \pm 0.1$ cm，面积 $A = 15.0$ cm²，$\Delta A/A = 0.1/5.0 + 0.1/3.0 = 0.053$，$A = 15.0 \pm 0.8$ cm²；② 密度测量——质量 $m \pm \Delta m$，体积 $V \pm \Delta V$，$
ho = m/V$；③ 速度计算——$v = d/t$；④ 电阻功率——$P = V^2/R$，$\Delta P/P = 2\Delta V/V + \Delta R/R$。
🎯 Typical Applications: ① Area — $L = 5.0 \pm 0.1$ cm, $W = 3.0 \pm 0.1$ cm, $A = 15.0$ cm², $\Delta A/A = 0.1/5.0 + 0.1/3.0 = 0.053$, $A = 15.0 \pm 0.8$ cm²; ② Density — mass $m \pm \Delta m$, volume $V \pm \Delta V$, $
ho = m/V$; ③ Speed — $v = d/t$; ④ Electrical power — $P = V^2/R$, $\Delta P/P = 2\Delta V/V + \Delta R/R$.
例 1.7 / Example 1.7
$A = 5.0 \pm 0.2$，$B = 2.0 \pm 0.1$。求 (a) $Z = A + B$ (b) $Z = A - B$ (c) $Z = A imes B$ (d) $Z = A/B$。
$A = 5.0 \pm 0.2$, $B = 2.0 \pm 0.1$. Find (a) $Z = A + B$ (b) $Z = A - B$ (c) $Z = A imes B$ (d) $Z = A/B$.
解题过程 / Solution
(a) $Z = 5.0 + 2.0 = 7.0$，$\Delta Z = 0.2 + 0.1 = 0.3$ → $Z = 7.0 \pm 0.3$
(b) $Z = 5.0 - 2.0 = 3.0$，$\Delta Z = 0.2 + 0.1 = 0.3$ → $Z = 3.0 \pm 0.3$
(c) $Z = 5.0 imes 2.0 = 10.0$，$\Delta Z/Z = 0.2/5.0 + 0.1/2.0 = 0.04 + 0.05 = 0.09$，$\Delta Z = 10.0 imes 0.09 = 0.9$ → $Z = 10.0 \pm 0.9$
(d) $Z = 5.0/2.0 = 2.5$，$\Delta Z/Z = 0.2/5.0 + 0.1/2.0 = 0.09$，$\Delta Z = 2.5 imes 0.09 = 0.225$ → $Z = 2.5 \pm 0.2$（2 sf）
(b) $Z = 5.0 - 2.0 = 3.0$，$\Delta Z = 0.2 + 0.1 = 0.3$ → $Z = 3.0 \pm 0.3$
(c) $Z = 5.0 imes 2.0 = 10.0$，$\Delta Z/Z = 0.2/5.0 + 0.1/2.0 = 0.04 + 0.05 = 0.09$，$\Delta Z = 10.0 imes 0.09 = 0.9$ → $Z = 10.0 \pm 0.9$
(d) $Z = 5.0/2.0 = 2.5$，$\Delta Z/Z = 0.2/5.0 + 0.1/2.0 = 0.09$，$\Delta Z = 2.5 imes 0.09 = 0.225$ → $Z = 2.5 \pm 0.2$（2 sf）
(a) $Z = 5.0 + 2.0 = 7.0$, $\Delta Z = 0.2 + 0.1 = 0.3$ → $Z = 7.0 \pm 0.3$
(b) $Z = 5.0 - 2.0 = 3.0$, $\Delta Z = 0.2 + 0.1 = 0.3$ → $Z = 3.0 \pm 0.3$
(c) $Z = 5.0 imes 2.0 = 10.0$, $\Delta Z/Z = 0.2/5.0 + 0.1/2.0 = 0.04 + 0.05 = 0.09$, $\Delta Z = 10.0 imes 0.09 = 0.9$ → $Z = 10.0 \pm 0.9$
(d) $Z = 5.0/2.0 = 2.5$, $\Delta Z/Z = 0.09$, $\Delta Z = 2.5 imes 0.09 = 0.225$ → $Z = 2.5 \pm 0.2$ (2 sf)
(b) $Z = 5.0 - 2.0 = 3.0$, $\Delta Z = 0.2 + 0.1 = 0.3$ → $Z = 3.0 \pm 0.3$
(c) $Z = 5.0 imes 2.0 = 10.0$, $\Delta Z/Z = 0.2/5.0 + 0.1/2.0 = 0.04 + 0.05 = 0.09$, $\Delta Z = 10.0 imes 0.09 = 0.9$ → $Z = 10.0 \pm 0.9$
(d) $Z = 5.0/2.0 = 2.5$, $\Delta Z/Z = 0.09$, $\Delta Z = 2.5 imes 0.09 = 0.225$ → $Z = 2.5 \pm 0.2$ (2 sf)
例 1.8 / Example 1.8
圆柱体半径 $r = 2.50 \pm 0.05$ cm，高 $h = 10.0 \pm 0.2$ cm。求体积及其不确定度。
Cylinder radius $r = 2.50 \pm 0.05$ cm, height $h = 10.0 \pm 0.2$ cm. Find volume and its uncertainty.
解题过程 / Solution
$V = \pi r^2 h = \pi (2.50)^2 (10.0) = 196$ cm$^3$
$\Delta V/V = 2(\Delta r/r) + \Delta h/h = 2(0.05/2.50) + 0.2/10.0 = 0.04 + 0.02 = 0.06$
$\Delta V = 196 imes 0.06 = 11.8$ cm$^3$ ≈ 12 cm$^3$
$V = 196 \pm 12$ cm$^3$ 或 $V = 1.96 imes 10^2 \pm 0.12 imes 10^2$ cm$^3$
注意$r^2$ 贡献了 2 倍相对不确定度——半径的精确测量至关重要。
$\Delta V/V = 2(\Delta r/r) + \Delta h/h = 2(0.05/2.50) + 0.2/10.0 = 0.04 + 0.02 = 0.06$
$\Delta V = 196 imes 0.06 = 11.8$ cm$^3$ ≈ 12 cm$^3$
$V = 196 \pm 12$ cm$^3$ 或 $V = 1.96 imes 10^2 \pm 0.12 imes 10^2$ cm$^3$
注意$r^2$ 贡献了 2 倍相对不确定度——半径的精确测量至关重要。
$V = \pi r^2 h = \pi (2.50)^2 (10.0) = 196$ cm$^3$
$\Delta V/V = 2(\Delta r/r) + \Delta h/h = 2(0.05/2.50) + 0.2/10.0 = 0.04 + 0.02 = 0.06$
$\Delta V = 196 imes 0.06 = 11.8$ cm$^3$ ≈ 12 cm$^3$
$V = 196 \pm 12$ cm$^3$ or $V = 1.96 imes 10^2 \pm 0.12 imes 10^2$ cm$^3$
Note: $r^2$ contributes 2x the relative uncertainty — precise radius measurement is critical.
$\Delta V/V = 2(\Delta r/r) + \Delta h/h = 2(0.05/2.50) + 0.2/10.0 = 0.04 + 0.02 = 0.06$
$\Delta V = 196 imes 0.06 = 11.8$ cm$^3$ ≈ 12 cm$^3$
$V = 196 \pm 12$ cm$^3$ or $V = 1.96 imes 10^2 \pm 0.12 imes 10^2$ cm$^3$
Note: $r^2$ contributes 2x the relative uncertainty — precise radius measurement is critical.
1.5 图像法 / Graphing
IB 1.2IB 实验题必考
绘图原则 / Graphing Principles
$$ ext{最佳拟合直线}\quad y = mx + c$$
IB 物理实验报告和 Paper 3 中，正确作图是核心技能。每一张图都需要：合适的坐标轴刻度、正确标注变量和单位、数据点带误差棒、最佳拟合线、计算斜率与截距。
In IB Physics lab reports and Paper 3, proper graphing is a core skill. Every graph needs: appropriate axis scales, correctly labeled variables with units, data points with error bars, line of best fit, and calculation of slope and intercept.
📐 详细步骤 / Detailed Steps
IB 作图完整流程：
① 选择坐标轴：自变量 x，因变量 y。每个轴占图幅的 50% 以上。
② 刻度与分度：选择方便读取的刻度（1, 2, 5 的倍数），使数据点均匀分布在图中。
③ 标签与单位：标注"物理量/单位"，如 $T^2$/s$^2$。
④ 画数据点：用小叉 × 或小点 • （直径 ≤ 1 mm）。
⑤ 误差棒：每个点画垂直误差棒（$\pm \Delta y$），如有需要也画水平误差棒（$\pm \Delta x$）。
⑥ 最佳拟合线：直线（或光滑曲线）通过尽可能多的误差棒，两侧点数大致相等。
⑦ 斜率计算：用大三角形法，$ ext{斜率} = \Delta y / \Delta x$，三角形面积 ≥ 图幅一半。
⑧ 截距：延长直线到与 y 轴相交。
① 选择坐标轴：自变量 x，因变量 y。每个轴占图幅的 50% 以上。
② 刻度与分度：选择方便读取的刻度（1, 2, 5 的倍数），使数据点均匀分布在图中。
③ 标签与单位：标注"物理量/单位"，如 $T^2$/s$^2$。
④ 画数据点：用小叉 × 或小点 • （直径 ≤ 1 mm）。
⑤ 误差棒：每个点画垂直误差棒（$\pm \Delta y$），如有需要也画水平误差棒（$\pm \Delta x$）。
⑥ 最佳拟合线：直线（或光滑曲线）通过尽可能多的误差棒，两侧点数大致相等。
⑦ 斜率计算：用大三角形法，$ ext{斜率} = \Delta y / \Delta x$，三角形面积 ≥ 图幅一半。
⑧ 截距：延长直线到与 y 轴相交。
Complete IB graphing procedure:
① Choose axes: independent variable on x, dependent on y. Each axis should use ≥50% of the graph paper.
② Scale & divisions: Choose easy-to-read scales (multiples of 1, 2, 5) so data fills the graph.
③ Labels & units: Label as "Quantity/unit", e.g., $T^2$/s$^2$.
④ Plot points: Use small × or • (≤ 1 mm diameter).
⑤ Error bars: Vertical error bars ($\pm \Delta y$) on each point; horizontal ($\pm \Delta x$) if needed.
⑥ Best-fit line: Straight line (or smooth curve) through as many error bars as possible, with roughly equal points on each side.
⑦ Slope calculation: Use a large triangle, slope $= \Delta y / \Delta x$, triangle area ≥ half the graph.
⑧ Intercept: Extend the line to the y-axis.
① Choose axes: independent variable on x, dependent on y. Each axis should use ≥50% of the graph paper.
② Scale & divisions: Choose easy-to-read scales (multiples of 1, 2, 5) so data fills the graph.
③ Labels & units: Label as "Quantity/unit", e.g., $T^2$/s$^2$.
④ Plot points: Use small × or • (≤ 1 mm diameter).
⑤ Error bars: Vertical error bars ($\pm \Delta y$) on each point; horizontal ($\pm \Delta x$) if needed.
⑥ Best-fit line: Straight line (or smooth curve) through as many error bars as possible, with roughly equal points on each side.
⑦ Slope calculation: Use a large triangle, slope $= \Delta y / \Delta x$, triangle area ≥ half the graph.
⑧ Intercept: Extend the line to the y-axis.
线性化 / Linearization
$$y = kx^n \;\Rightarrow\; \ln y = \ln k + n \ln x$$
非线性关系通过取对数转化为线性关系，便于分析。$n$ 是幂指数（斜率），$\ln k$ 是截距。
Non-linear relationships can be transformed into linear form by taking logarithms. $n$ is the exponent (slope), $\ln k$ is the intercept.
📐 线性化方法 / Linearization Methods
常见线性化变换：
① 幂律关系 $y = kx^n$ → $\ln y = \ln k + n \ln x$
• 作图 $\ln y$ vs $\ln x$，斜率 = $n$，截距 = $\ln k$
② 指数关系 $y = Ae^{bx}$ → $\ln y = \ln A + bx$
• 作图 $\ln y$ vs $x$，斜率 = $b$，截距 = $\ln A$
③ 反比关系 $y = k/x$ → $y = k(1/x)$
• 作图 $y$ vs $1/x$，斜率 = $k$
④ 平方反比 $y = k/x^2$ → $y = k(1/x^2)$
• 作图 $y$ vs $1/x^2$，斜率 = $k$
⑤ 周期-摆长 $T = 2\pi\sqrt{L/g}$ → $T^2 = (4\pi^2/g)L$
• 作图 $T^2$ vs $L$，斜率 = $4\pi^2/g$，可反推 $g$
① 幂律关系 $y = kx^n$ → $\ln y = \ln k + n \ln x$
• 作图 $\ln y$ vs $\ln x$，斜率 = $n$，截距 = $\ln k$
② 指数关系 $y = Ae^{bx}$ → $\ln y = \ln A + bx$
• 作图 $\ln y$ vs $x$，斜率 = $b$，截距 = $\ln A$
③ 反比关系 $y = k/x$ → $y = k(1/x)$
• 作图 $y$ vs $1/x$，斜率 = $k$
④ 平方反比 $y = k/x^2$ → $y = k(1/x^2)$
• 作图 $y$ vs $1/x^2$，斜率 = $k$
⑤ 周期-摆长 $T = 2\pi\sqrt{L/g}$ → $T^2 = (4\pi^2/g)L$
• 作图 $T^2$ vs $L$，斜率 = $4\pi^2/g$，可反推 $g$
Common linearization transformations:
① Power law $y = kx^n$ → $\ln y = \ln k + n \ln x$
• Plot $\ln y$ vs $\ln x$, slope = $n$, intercept = $\ln k$
② Exponential $y = Ae^{bx}$ → $\ln y = \ln A + bx$
• Plot $\ln y$ vs $x$, slope = $b$, intercept = $\ln A$
③ Inverse $y = k/x$ → $y = k(1/x)$
• Plot $y$ vs $1/x$, slope = $k$
④ Inverse square $y = k/x^2$ → $y = k(1/x^2)$
• Plot $y$ vs $1/x^2$, slope = $k$
⑤ Period-length $T = 2\pi\sqrt{L/g}$ → $T^2 = (4\pi^2/g)L$
• Plot $T^2$ vs $L$, slope = $4\pi^2/g$, can solve for $g$
① Power law $y = kx^n$ → $\ln y = \ln k + n \ln x$
• Plot $\ln y$ vs $\ln x$, slope = $n$, intercept = $\ln k$
② Exponential $y = Ae^{bx}$ → $\ln y = \ln A + bx$
• Plot $\ln y$ vs $x$, slope = $b$, intercept = $\ln A$
③ Inverse $y = k/x$ → $y = k(1/x)$
• Plot $y$ vs $1/x$, slope = $k$
④ Inverse square $y = k/x^2$ → $y = k(1/x^2)$
• Plot $y$ vs $1/x^2$, slope = $k$
⑤ Period-length $T = 2\pi\sqrt{L/g}$ → $T^2 = (4\pi^2/g)L$
• Plot $T^2$ vs $L$, slope = $4\pi^2/g$, can solve for $g$
斜率与截距的不确定度 / Uncertainty in Slope and Intercept
$$ ext{斜率不确定度} = rac{m_{\max} - m_{\min}}{2}$$
IB 中常用最大最小斜率法：通过误差棒画最大可能斜率线（$m_{\max}$）和最小可能斜率线（$m_{\min}$），不确定度为其差值的一半。
IB uses the max-min slope method: draw lines of maximum possible slope ($m_{\max}$) and minimum possible slope ($m_{\min}$) through the error bars. Uncertainty is half the difference.
📐 最大最小斜率法 / Max-Min Slope Method
最大最小斜率法步骤：
① 画最佳拟合线，计算斜率 $m_{ ext{best}}$
② 画一条尽可能陡但仍通过所有误差棒（或其端点）的线 → $m_{\max}$
③ 画一条尽可能平缓但仍通过所有误差棒（或其端点）的线 → $m_{\min}$
④ $\Delta m = (m_{\max} - m_{\min})/2$
⑤ 结果：$m = m_{ ext{best}} \pm \Delta m$
截距不确定度：使用同样的最大最小线，读取各自的 y 轴截距：$\Delta c = (c_{\max} - c_{\min})/2$。
① 画最佳拟合线，计算斜率 $m_{ ext{best}}$
② 画一条尽可能陡但仍通过所有误差棒（或其端点）的线 → $m_{\max}$
③ 画一条尽可能平缓但仍通过所有误差棒（或其端点）的线 → $m_{\min}$
④ $\Delta m = (m_{\max} - m_{\min})/2$
⑤ 结果：$m = m_{ ext{best}} \pm \Delta m$
截距不确定度：使用同样的最大最小线，读取各自的 y 轴截距：$\Delta c = (c_{\max} - c_{\min})/2$。
Max-min slope method steps:
① Draw the best-fit line, calculate slope $m_{ ext{best}}$
② Draw the steepest line that still passes through all error bars (or their endpoints) → $m_{\max}$
③ Draw the shallowest line that still passes through all error bars → $m_{\min}$
④ $\Delta m = (m_{\max} - m_{\min})/2$
⑤ Result: $m = m_{ ext{best}} \pm \Delta m$
Intercept uncertainty: Use the same max/min lines, read their y-intercepts: $\Delta c = (c_{\max} - c_{\min})/2$.
① Draw the best-fit line, calculate slope $m_{ ext{best}}$
② Draw the steepest line that still passes through all error bars (or their endpoints) → $m_{\max}$
③ Draw the shallowest line that still passes through all error bars → $m_{\min}$
④ $\Delta m = (m_{\max} - m_{\min})/2$
⑤ Result: $m = m_{ ext{best}} \pm \Delta m$
Intercept uncertainty: Use the same max/min lines, read their y-intercepts: $\Delta c = (c_{\max} - c_{\min})/2$.
🎯 典型应用 / Applications：① 弹簧常数实验——$F = kx$，作 $F$ vs $x$ 图，斜率 = $k$；② 单摆求 $g$——$T^2$ vs $L$ 图，斜率 = $4\pi^2/g$；③ 欧姆定律——$V = IR$，$V$ vs $I$ 图，斜率 = $R$；④ 放射性衰变——$\ln N$ vs $t$ 图，斜率 = $-\lambda$（衰变常数）；⑤ 光电效应——$K_{\max}$ vs $f$ 图，斜率 = $h$（普朗克常数），截距 = $-\phi$（逸出功）。
🎯 Typical Applications: ① Spring constant — $F = kx$, plot $F$ vs $x$, slope = $k$; ② Pendulum $g$ — $T^2$ vs $L$ plot, slope = $4\pi^2/g$; ③ Ohm's law — $V = IR$, $V$ vs $I$, slope = $R$; ④ Radioactive decay — $\ln N$ vs $t$, slope = $-\lambda$; ⑤ Photoelectric effect — $K_{\max}$ vs $f$, slope = $h$ (Planck's constant), intercept = $-\phi$ (work function).
例 1.9 / Example 1.9
单摆实验数据：$L = 0.40, 0.60, 0.80, 1.00, 1.20$ m，周期 $T = 1.27, 1.55, 1.80, 2.01, 2.20$ s。(a) 作 $T^2$ vs $L$ 图。(b) 求斜率及 $g$ 值。
Pendulum data: $L = 0.40, 0.60, 0.80, 1.00, 1.20$ m, periods $T = 1.27, 1.55, 1.80, 2.01, 2.20$ s. (a) Plot $T^2$ vs $L$. (b) Find slope and calculate $g$.
解题过程 / Solution
$T^2 = 1.61, 2.40, 3.24, 4.04, 4.84$ s$^2$
作 $T^2$ vs $L$ 图，最佳拟合线斜率：$m \approx 4.04$ s$^2$/m
$T^2 = (4\pi^2/g)L$，因此 $g = 4\pi^2/m = 4\pi^2/4.04 = 9.77$ m/s$^2$
与标准值 $g = 9.81$ m/s$^2$ 非常接近。百分比误差：$|9.77-9.81|/9.81 = 0.4\%$。
作 $T^2$ vs $L$ 图，最佳拟合线斜率：$m \approx 4.04$ s$^2$/m
$T^2 = (4\pi^2/g)L$，因此 $g = 4\pi^2/m = 4\pi^2/4.04 = 9.77$ m/s$^2$
与标准值 $g = 9.81$ m/s$^2$ 非常接近。百分比误差：$|9.77-9.81|/9.81 = 0.4\%$。
$T^2 = 1.61, 2.40, 3.24, 4.04, 4.84$ s$^2$
Plot $T^2$ vs $L$, best-fit slope: $m \approx 4.04$ s$^2$/m
$T^2 = (4\pi^2/g)L$, so $g = 4\pi^2/m = 4\pi^2/4.04 = 9.77$ m/s$^2$
Very close to standard $g = 9.81$ m/s$^2$. Percentage error: $|9.77-9.81|/9.81 = 0.4\%$.
Plot $T^2$ vs $L$, best-fit slope: $m \approx 4.04$ s$^2$/m
$T^2 = (4\pi^2/g)L$, so $g = 4\pi^2/m = 4\pi^2/4.04 = 9.77$ m/s$^2$
Very close to standard $g = 9.81$ m/s$^2$. Percentage error: $|9.77-9.81|/9.81 = 0.4\%$.
例 1.10 / Example 1.10
电容放电实验测得 $V(t)$ 数据：$t=0$ s 时 $V=6.0$ V，$t=5$ s 时 $V=2.2$ V。假设 $V = V_0 e^{-t/RC}$，求时间常数 $RC$。
Capacitor discharge data: at $t=0$ s, $V=6.0$ V; at $t=5$ s, $V=2.2$ V. Given $V = V_0 e^{-t/RC}$, find the time constant $RC$.
解题过程 / Solution
由 $V = V_0 e^{-t/RC}$，取自然对数：$\ln V = \ln V_0 - t/RC$
代入数据：$\ln(2.2) = \ln(6.0) - 5/RC$
$0.788 = 1.792 - 5/RC$
$5/RC = 1.004$
$RC = 4.98$ s。
如果作图 $\ln V$ vs $t$，斜率即为 $-1/RC$。
代入数据：$\ln(2.2) = \ln(6.0) - 5/RC$
$0.788 = 1.792 - 5/RC$
$5/RC = 1.004$
$RC = 4.98$ s。
如果作图 $\ln V$ vs $t$，斜率即为 $-1/RC$。
From $V = V_0 e^{-t/RC}$, take natural log: $\ln V = \ln V_0 - t/RC$
Substitute: $\ln(2.2) = \ln(6.0) - 5/RC$
$0.788 = 1.792 - 5/RC$
$5/RC = 1.004$
$RC = 4.98$ s.
Plotting $\ln V$ vs $t$ gives slope $= -1/RC$.
Substitute: $\ln(2.2) = \ln(6.0) - 5/RC$
$0.788 = 1.792 - 5/RC$
$5/RC = 1.004$
$RC = 4.98$ s.
Plotting $\ln V$ vs $t$ gives slope $= -1/RC$.
🏆 挑战题 / Challenge Problems
多概念综合 / Multi-concept难度 / Hard
以下题目需要综合运用本章的多个知识点。建议先复习所有内容后再尝试。
These problems require integrating multiple concepts from this chapter. Review all sections before attempting.
挑战 1 / Challenge 1 综合不确定度分析 / Comprehensive Uncertainty Analysis
在单摆测 $g$ 的实验中，测得以下数据：$L = 0.500 \pm 0.005$ m，10 次全振动的时间 $t = 14.18 \pm 0.05$ s。求 $g$ 及其不确定度。哪个量的不确定度对结果影响更大？
In a pendulum experiment to find $g$: $L = 0.500 \pm 0.005$ m, time for 10 oscillations $t = 14.18 \pm 0.05$ s. Find $g$ and its uncertainty. Which quantity contributes more to the overall uncertainty?
解题过程 / Solution
$T = t/10 = 1.418$ s
$\Delta T = \Delta t/10 = 0.005$ s
$T = 1.418 \pm 0.005$ s
$g = 4\pi^2 L / T^2 = 4\pi^2 \times 0.500 / (1.418)^2 = 9.81$ m/s$^2$
$\Delta g/g = \Delta L/L + 2(\Delta T/T)$
$= 0.005/0.500 + 2(0.005/1.418)$
$= 0.010 + 0.0071 = 0.0171$
$\Delta g = 9.81 \times 0.0171 = 0.168$ m/s$^2$
$g = 9.81 \pm 0.17$ m/s$^2$（3 sf）
结论：$\Delta L/L = 1.0\%$，$2\Delta T/T = 0.71\%$，长度测量的不确定度贡献更大。应当用更精确的仪器测量摆长。
$\Delta T = \Delta t/10 = 0.005$ s
$T = 1.418 \pm 0.005$ s
$g = 4\pi^2 L / T^2 = 4\pi^2 \times 0.500 / (1.418)^2 = 9.81$ m/s$^2$
$\Delta g/g = \Delta L/L + 2(\Delta T/T)$
$= 0.005/0.500 + 2(0.005/1.418)$
$= 0.010 + 0.0071 = 0.0171$
$\Delta g = 9.81 \times 0.0171 = 0.168$ m/s$^2$
$g = 9.81 \pm 0.17$ m/s$^2$（3 sf）
结论：$\Delta L/L = 1.0\%$，$2\Delta T/T = 0.71\%$，长度测量的不确定度贡献更大。应当用更精确的仪器测量摆长。
$T = t/10 = 1.418$ s
$\Delta T = \Delta t/10 = 0.005$ s
$T = 1.418 \pm 0.005$ s
$g = 4\pi^2 L / T^2 = 4\pi^2 \times 0.500 / (1.418)^2 = 9.81$ m/s$^2$
$\Delta g/g = \Delta L/L + 2(\Delta T/T)$
$= 0.005/0.500 + 2(0.005/1.418)$
$= 0.010 + 0.0071 = 0.0171$
$\Delta g = 9.81 \times 0.0171 = 0.168$ m/s$^2$
$g = 9.81 \pm 0.17$ m/s$^2$ (3 sf)
Conclusion: $\Delta L/L = 1.0\%$, $2\Delta T/T = 0.71\%$ — length measurement contributes more. Use a more precise instrument for length.
$\Delta T = \Delta t/10 = 0.005$ s
$T = 1.418 \pm 0.005$ s
$g = 4\pi^2 L / T^2 = 4\pi^2 \times 0.500 / (1.418)^2 = 9.81$ m/s$^2$
$\Delta g/g = \Delta L/L + 2(\Delta T/T)$
$= 0.005/0.500 + 2(0.005/1.418)$
$= 0.010 + 0.0071 = 0.0171$
$\Delta g = 9.81 \times 0.0171 = 0.168$ m/s$^2$
$g = 9.81 \pm 0.17$ m/s$^2$ (3 sf)
Conclusion: $\Delta L/L = 1.0\%$, $2\Delta T/T = 0.71\%$ — length measurement contributes more. Use a more precise instrument for length.
挑战 2 / Challenge 2 对数-对数图的线性化 / Log-Log Linearization
某行星的卫星轨道数据表明，轨道周期 $T$（天）与轨道半径 $r$（$10^6$ km）满足 $T = k r^n$。测得数据：$r = 2.0, 3.0, 4.0, 6.0, 8.0$（$ imes 10^6$ km），对应 $T = 1.5, 2.8, 4.3, 7.9, 12.2$ 天。作 $\log T$ vs $\log r$ 图，求 $n$ 和 $k$。
A planet's moon data shows orbital period $T$ (days) vs radius $r$ ($10^6$ km) follows $T = k r^n$. Data: $r = 2.0, 3.0, 4.0, 6.0, 8.0$ ($ imes 10^6$ km), $T = 1.5, 2.8, 4.3, 7.9, 12.2$ days. Plot $\log T$ vs $\log r$, find $n$ and $k$.
解题过程 / Solution
取对数：$\log r = 0.30, 0.48, 0.60, 0.78, 0.90$
$\log T = 0.18, 0.45, 0.63, 0.90, 1.09$
作 $\log T$ vs $\log r$ 图，最佳拟合线斜率 $n$：$n = \Delta(\log T)/\Delta(\log r)$
$n \approx (1.09 - 0.18)/(0.90 - 0.30) = 0.91/0.60 = 1.52$
开普勒第三定律预言 $n = 1.5$，结果非常吻合。
截距 $\log k$：$\log T$ vs $\log r$ 的 y 轴截距（$\log r=0$ 处）
代入 $\log r=0.30$ 时 $\log T=0.18$：
$0.18 = \log k + 1.52(0.30)$
$\log k = 0.18 - 0.456 = -0.276$
$k = 10^{-0.276} = 0.53$
验证：$T = 0.53 \times r^{1.5}$，检查 $r=4.0$ 时 $T=0.53 \times 8 = 4.24$ 天 ✓
$\log T = 0.18, 0.45, 0.63, 0.90, 1.09$
作 $\log T$ vs $\log r$ 图，最佳拟合线斜率 $n$：$n = \Delta(\log T)/\Delta(\log r)$
$n \approx (1.09 - 0.18)/(0.90 - 0.30) = 0.91/0.60 = 1.52$
开普勒第三定律预言 $n = 1.5$，结果非常吻合。
截距 $\log k$：$\log T$ vs $\log r$ 的 y 轴截距（$\log r=0$ 处）
代入 $\log r=0.30$ 时 $\log T=0.18$：
$0.18 = \log k + 1.52(0.30)$
$\log k = 0.18 - 0.456 = -0.276$
$k = 10^{-0.276} = 0.53$
验证：$T = 0.53 \times r^{1.5}$，检查 $r=4.0$ 时 $T=0.53 \times 8 = 4.24$ 天 ✓
Take logs: $\log r = 0.30, 0.48, 0.60, 0.78, 0.90$
$\log T = 0.18, 0.45, 0.63, 0.90, 1.09$
Plot $\log T$ vs $\log r$, best-fit slope $n$: $n = \Delta(\log T)/\Delta(\log r)$
$n \approx (1.09 - 0.18)/(0.90 - 0.30) = 0.91/0.60 = 1.52$
Kepler's Third Law predicts $n = 1.5$ — excellent agreement.
Intercept $\log k$: y-intercept of $\log T$ vs $\log r$ (at $\log r=0$)
Using $\log r=0.30$ where $\log T=0.18$:
$0.18 = \log k + 1.52(0.30)$
$\log k = 0.18 - 0.456 = -0.276$
$k = 10^{-0.276} = 0.53$
Check: $T = 0.53 \times r^{1.5}$, at $r=4.0$: $T=0.53 \times 8 = 4.24$ days ✓
$\log T = 0.18, 0.45, 0.63, 0.90, 1.09$
Plot $\log T$ vs $\log r$, best-fit slope $n$: $n = \Delta(\log T)/\Delta(\log r)$
$n \approx (1.09 - 0.18)/(0.90 - 0.30) = 0.91/0.60 = 1.52$
Kepler's Third Law predicts $n = 1.5$ — excellent agreement.
Intercept $\log k$: y-intercept of $\log T$ vs $\log r$ (at $\log r=0$)
Using $\log r=0.30$ where $\log T=0.18$:
$0.18 = \log k + 1.52(0.30)$
$\log k = 0.18 - 0.456 = -0.276$
$k = 10^{-0.276} = 0.53$
Check: $T = 0.53 \times r^{1.5}$, at $r=4.0$: $T=0.53 \times 8 = 4.24$ days ✓
挑战 3 / Challenge 3 电阻率测量的不确定度 / Resistivity Uncertainty
一圆柱形导线的电阻由 $R = \rho L/A$ 给出，其中 $A = \pi d^2/4$。测量值：$L = 2.50 \pm 0.05$ m，$d = 0.50 \pm 0.01$ mm，$R = 0.680 \pm 0.005$ $\Omega$。求电阻率 $\rho$ 及其不确定度。
A cylindrical wire's resistance is $R = \rho L/A$, where $A = \pi d^2/4$. Measurements: $L = 2.50 \pm 0.05$ m, $d = 0.50 \pm 0.01$ mm, $R = 0.680 \pm 0.005$ $\Omega$. Find resistivity $\rho$ and its uncertainty.
解题过程 / Solution
$
ho = R \cdot A / L = R \cdot (\pi d^2/4) / L$
$d = 0.50 \times 10^{-3}$ m，$\Delta d = 0.01 \times 10^{-3}$ m
$A = \pi (0.50 \times 10^{-3})^2 / 4 = 1.963 \times 10^{-7}$ m$^2$
$ ho = 0.680 \times 1.963 \times 10^{-7} / 2.50 = 5.34 \times 10^{-8}$ $\Omega \cdot$ m
$\Delta ho/\rho = \Delta R/R + 2(\Delta d/d) + \Delta L/L$
$= 0.005/0.680 + 2(0.01/0.50) + 0.05/2.50$
$= 0.00735 + 0.04 + 0.02 = 0.06735$
$\Delta ho = 5.34 \times 10^{-8} \times 0.06735 = 3.6 \times 10^{-9}$ $\Omega \cdot$ m
$ ho = (5.3 \pm 0.4) \times 10^{-8}$ $\Omega \cdot$ m
分析：$d$ 的不确定度贡献最大（4%！），因为 $d$ 出现在平方项中且本身精度低（相对不确定度 2%）。用千分尺测 $d$ 可改善。
$d = 0.50 \times 10^{-3}$ m，$\Delta d = 0.01 \times 10^{-3}$ m
$A = \pi (0.50 \times 10^{-3})^2 / 4 = 1.963 \times 10^{-7}$ m$^2$
$ ho = 0.680 \times 1.963 \times 10^{-7} / 2.50 = 5.34 \times 10^{-8}$ $\Omega \cdot$ m
$\Delta ho/\rho = \Delta R/R + 2(\Delta d/d) + \Delta L/L$
$= 0.005/0.680 + 2(0.01/0.50) + 0.05/2.50$
$= 0.00735 + 0.04 + 0.02 = 0.06735$
$\Delta ho = 5.34 \times 10^{-8} \times 0.06735 = 3.6 \times 10^{-9}$ $\Omega \cdot$ m
$ ho = (5.3 \pm 0.4) \times 10^{-8}$ $\Omega \cdot$ m
分析：$d$ 的不确定度贡献最大（4%！），因为 $d$ 出现在平方项中且本身精度低（相对不确定度 2%）。用千分尺测 $d$ 可改善。
$
ho = R \cdot A / L = R \cdot (\pi d^2/4) / L$
$d = 0.50 \times 10^{-3}$ m, $\Delta d = 0.01 \times 10^{-3}$ m
$A = \pi (0.50 \times 10^{-3})^2 / 4 = 1.963 \times 10^{-7}$ m$^2$
$ ho = 0.680 \times 1.963 \times 10^{-7} / 2.50 = 5.34 \times 10^{-8}$ $\Omega \cdot$ m
$\Delta ho/\rho = \Delta R/R + 2(\Delta d/d) + \Delta L/L$
$= 0.005/0.680 + 2(0.01/0.50) + 0.05/2.50$
$= 0.00735 + 0.04 + 0.02 = 0.06735$
$\Delta ho = 5.34 \times 10^{-8} \times 0.06735 = 3.6 \times 10^{-9}$ $\Omega \cdot$ m
$ ho = (5.3 \pm 0.4) \times 10^{-8}$ $\Omega \cdot$ m
Analysis: $d$ contributes the most (4%!), because it's squared and already has low precision (2% relative). Use a micrometer for $d$.
$d = 0.50 \times 10^{-3}$ m, $\Delta d = 0.01 \times 10^{-3}$ m
$A = \pi (0.50 \times 10^{-3})^2 / 4 = 1.963 \times 10^{-7}$ m$^2$
$ ho = 0.680 \times 1.963 \times 10^{-7} / 2.50 = 5.34 \times 10^{-8}$ $\Omega \cdot$ m
$\Delta ho/\rho = \Delta R/R + 2(\Delta d/d) + \Delta L/L$
$= 0.005/0.680 + 2(0.01/0.50) + 0.05/2.50$
$= 0.00735 + 0.04 + 0.02 = 0.06735$
$\Delta ho = 5.34 \times 10^{-8} \times 0.06735 = 3.6 \times 10^{-9}$ $\Omega \cdot$ m
$ ho = (5.3 \pm 0.4) \times 10^{-8}$ $\Omega \cdot$ m
Analysis: $d$ contributes the most (4%!), because it's squared and already has low precision (2% relative). Use a micrometer for $d$.
挑战 4 / Challenge 4 实验设计与误差分析 / Experimental Design & Error Analysis
学生要做自由落体实验测量 $g$，有三种方案：(a) 用 2 m 直尺和秒表测下落时间；(b) 用光电门测通过两个光栅的时间差；(c) 用频闪照相分析位移-时间关系。请分析每种方案的误差来源及其大小。哪种方案精度最高？
A student plans to measure $g$ via free fall with three methods: (a) 2 m ruler + stopwatch; (b) photogates measuring time between two gates; (c) stroboscopic photography analyzing displacement-time. Analyze error sources and magnitude for each. Which is most precise?
解题过程 / Solution
方案 (a) 直尺+秒表：
• $h = 2.00 \pm 0.005$ m，$t = \sqrt{2h/g} \approx 0.639$ s
• 人手反应时间 $\approx \pm 0.2$ s！$\Delta t/t \approx 31\%$
• $g = 2h/t^2$，$\Delta g/g = 0.005/2.00 + 2(0.2/0.639) = 0.0025 + 0.626 = 63\%$
• 人反应时间主导误差，精度极低。
方案 (b) 光电门：
• 光电门间距 $d = 1.00 \pm 0.005$ m
• 电子计时精度 $\pm 0.001$ s，$t \approx 0.452$ s
• $\Delta g/g = 0.005/1.00 + 2(0.001/0.452) = 0.005 + 0.0044 = 0.94\%$
• 精度高，~1%
方案 (c) 频闪照相：
• 频闪频率 50 Hz → $\Delta t = \pm 0.01$ s
• $s = \frac12gt^2$，取多帧数据拟合 $s$ vs $t^2$ 图
• 利用多个数据点拟合，斜率求 $g$，有效减小随机误差
• 精度 ~2%，但能发现系统误差（如空气阻力导致 $g$ 偏小）
结论：光电门方案精度最高，反应时间是最关键的误差来源。
• $h = 2.00 \pm 0.005$ m，$t = \sqrt{2h/g} \approx 0.639$ s
• 人手反应时间 $\approx \pm 0.2$ s！$\Delta t/t \approx 31\%$
• $g = 2h/t^2$，$\Delta g/g = 0.005/2.00 + 2(0.2/0.639) = 0.0025 + 0.626 = 63\%$
• 人反应时间主导误差，精度极低。
方案 (b) 光电门：
• 光电门间距 $d = 1.00 \pm 0.005$ m
• 电子计时精度 $\pm 0.001$ s，$t \approx 0.452$ s
• $\Delta g/g = 0.005/1.00 + 2(0.001/0.452) = 0.005 + 0.0044 = 0.94\%$
• 精度高，~1%
方案 (c) 频闪照相：
• 频闪频率 50 Hz → $\Delta t = \pm 0.01$ s
• $s = \frac12gt^2$，取多帧数据拟合 $s$ vs $t^2$ 图
• 利用多个数据点拟合，斜率求 $g$，有效减小随机误差
• 精度 ~2%，但能发现系统误差（如空气阻力导致 $g$ 偏小）
结论：光电门方案精度最高，反应时间是最关键的误差来源。
Method (a) Ruler + stopwatch:
• $h = 2.00 \pm 0.005$ m, $t = \sqrt{2h/g} \approx 0.639$ s
• Human reaction time $\approx \pm 0.2$ s! $\Delta t/t \approx 31\%$
• $g = 2h/t^2$, $\Delta g/g = 0.005/2.00 + 2(0.2/0.639) = 0.0025 + 0.626 = 63\%$
• Reaction time dominates — extremely low precision.
Method (b) Photogates:
• Gate spacing $d = 1.00 \pm 0.005$ m
• Electronic timer $\pm 0.001$ s, $t \approx 0.452$ s
• $\Delta g/g = 0.005/1.00 + 2(0.001/0.452) = 0.005 + 0.0044 = 0.94\%$
• High precision, ~1%
Method (c) Stroboscopic:
• 50 Hz strobe → $\Delta t = \pm 0.01$ s
• $s = \frac12gt^2$, fit $s$ vs $t^2$ graph with multiple data points
• Multiple points reduce random error through fitting
• Precision ~2%, but can detect systematic errors (air resistance)
Conclusion: Photogates give the highest precision. Reaction time is the critical error source.
• $h = 2.00 \pm 0.005$ m, $t = \sqrt{2h/g} \approx 0.639$ s
• Human reaction time $\approx \pm 0.2$ s! $\Delta t/t \approx 31\%$
• $g = 2h/t^2$, $\Delta g/g = 0.005/2.00 + 2(0.2/0.639) = 0.0025 + 0.626 = 63\%$
• Reaction time dominates — extremely low precision.
Method (b) Photogates:
• Gate spacing $d = 1.00 \pm 0.005$ m
• Electronic timer $\pm 0.001$ s, $t \approx 0.452$ s
• $\Delta g/g = 0.005/1.00 + 2(0.001/0.452) = 0.005 + 0.0044 = 0.94\%$
• High precision, ~1%
Method (c) Stroboscopic:
• 50 Hz strobe → $\Delta t = \pm 0.01$ s
• $s = \frac12gt^2$, fit $s$ vs $t^2$ graph with multiple data points
• Multiple points reduce random error through fitting
• Precision ~2%, but can detect systematic errors (air resistance)
Conclusion: Photogates give the highest precision. Reaction time is the critical error source.
挑战 5 / Challenge 5 综合数据处理 / Comprehensive Data Analysis
$y = kx^n$ 关系中测得：$x = 1.0, 2.0, 3.0, 4.0, 5.0$（$\pm 0.1$），$y = 0.5, 2.8, 7.6, 16.0, 27.9$（$\pm 0.2$）。(a) 用线性化方法求 $n$ 和 $k$ 的最佳估计值。(b) 估算 $n$ 的不确定度。(c) 预测 $x=3.5$ 时 $y$ 的值及不确定度。
For $y = kx^n$ relationship, data: $x = 1.0, 2.0, 3.0, 4.0, 5.0$ ($\pm 0.1$), $y = 0.5, 2.8, 7.6, 16.0, 27.9$ ($\pm 0.2$). (a) Find $n$ and $k$ via linearization. (b) Estimate uncertainty in $n$. (c) Predict $y$ at $x=3.5$ with uncertainty.
解题过程 / Solution
(a) 取自然对数：
$\ln x = 0.00, 0.69, 1.10, 1.39, 1.61$
$\ln y = -0.69, 1.03, 2.03, 2.77, 3.33$
作 $\ln y$ vs $\ln x$ 图：斜率 $n \approx (3.33-(-0.69))/(1.61-0.00) = 4.02/1.61 = 2.50$
截距 $\ln k = -0.69$ → $k = e^{-0.69} = 0.50$
$y \approx 0.50 \, x^{2.5}$
(b) 通过 $\Delta \ln y$ 估算 $\ln y$ 的不确定度：$\Delta(\ln y) \approx \Delta y / y$
例如第一点：$\Delta(\ln y) \approx 0.2/0.5 = 0.40$；第五点：$0.2/27.9 = 0.007$
最大最小斜率法得 $\Delta n \approx 0.08$ → $n = 2.50 \pm 0.08$
(c) $x=3.5$，$\ln x = 1.253$，$\ln y \approx -0.69 + 2.50 \times 1.253 = 2.44$
$y = e^{2.44} = 11.5$
不确定度通过传播公式：$\Delta y/y = n(\Delta x/x) = 2.50(0.1/3.5) = 0.071$
$\Delta y = 11.5 \times 0.071 = 0.82$
$y(x=3.5) = 11.5 \pm 0.8$
$\ln x = 0.00, 0.69, 1.10, 1.39, 1.61$
$\ln y = -0.69, 1.03, 2.03, 2.77, 3.33$
作 $\ln y$ vs $\ln x$ 图：斜率 $n \approx (3.33-(-0.69))/(1.61-0.00) = 4.02/1.61 = 2.50$
截距 $\ln k = -0.69$ → $k = e^{-0.69} = 0.50$
$y \approx 0.50 \, x^{2.5}$
(b) 通过 $\Delta \ln y$ 估算 $\ln y$ 的不确定度：$\Delta(\ln y) \approx \Delta y / y$
例如第一点：$\Delta(\ln y) \approx 0.2/0.5 = 0.40$；第五点：$0.2/27.9 = 0.007$
最大最小斜率法得 $\Delta n \approx 0.08$ → $n = 2.50 \pm 0.08$
(c) $x=3.5$，$\ln x = 1.253$，$\ln y \approx -0.69 + 2.50 \times 1.253 = 2.44$
$y = e^{2.44} = 11.5$
不确定度通过传播公式：$\Delta y/y = n(\Delta x/x) = 2.50(0.1/3.5) = 0.071$
$\Delta y = 11.5 \times 0.071 = 0.82$
$y(x=3.5) = 11.5 \pm 0.8$
(a) Natural logs:
$\ln x = 0.00, 0.69, 1.10, 1.39, 1.61$
$\ln y = -0.69, 1.03, 2.03, 2.77, 3.33$
Plot $\ln y$ vs $\ln x$: slope $n \approx (3.33-(-0.69))/(1.61-0.00) = 4.02/1.61 = 2.50$
Intercept $\ln k = -0.69$ → $k = e^{-0.69} = 0.50$
$y \approx 0.50 \, x^{2.5}$
(b) Estimate $\ln y$ uncertainty: $\Delta(\ln y) \approx \Delta y / y$
First point: $\Delta(\ln y) \approx 0.2/0.5 = 0.40$; Fifth: $0.2/27.9 = 0.007$
Max-min slope gives $\Delta n \approx 0.08$ → $n = 2.50 \pm 0.08$
(c) $x=3.5$, $\ln x = 1.253$, $\ln y \approx -0.69 + 2.50 \times 1.253 = 2.44$
$y = e^{2.44} = 11.5$
Uncertainty: $\Delta y/y = n(\Delta x/x) = 2.50(0.1/3.5) = 0.071$
$\Delta y = 11.5 \times 0.071 = 0.82$
$y(x=3.5) = 11.5 \pm 0.8$
$\ln x = 0.00, 0.69, 1.10, 1.39, 1.61$
$\ln y = -0.69, 1.03, 2.03, 2.77, 3.33$
Plot $\ln y$ vs $\ln x$: slope $n \approx (3.33-(-0.69))/(1.61-0.00) = 4.02/1.61 = 2.50$
Intercept $\ln k = -0.69$ → $k = e^{-0.69} = 0.50$
$y \approx 0.50 \, x^{2.5}$
(b) Estimate $\ln y$ uncertainty: $\Delta(\ln y) \approx \Delta y / y$
First point: $\Delta(\ln y) \approx 0.2/0.5 = 0.40$; Fifth: $0.2/27.9 = 0.007$
Max-min slope gives $\Delta n \approx 0.08$ → $n = 2.50 \pm 0.08$
(c) $x=3.5$, $\ln x = 1.253$, $\ln y \approx -0.69 + 2.50 \times 1.253 = 2.44$
$y = e^{2.44} = 11.5$
Uncertainty: $\Delta y/y = n(\Delta x/x) = 2.50(0.1/3.5) = 0.071$
$\Delta y = 11.5 \times 0.071 = 0.82$
$y(x=3.5) = 11.5 \pm 0.8$
附录：核心公式速查 / Formula Reference
| 知识点 / Topic | 公式 / Formula | 说明 / Notes |
|---|---|---|
| 单位制 / SI Units | 7 个基本单位 / 7 base units | kg, m, s, A, K, mol, cd |
| SI 前缀 / SI prefixes | T $10^{12}$ → G $10^9$ → M $10^6$ → k $10^3$ → m $10^{-3}$ → $\mu$ $10^{-6}$ → n $10^{-9}$ → p $10^{-12}$ | |
| 科学记数法 / Sci. Not. | $a imes 10^n$, $1 \le a < 10$ | 小数点左移 → $n>0$；右移 → $n<0$ |
| 有效数字 / Sig Figs | 结果精度 ≤ 输入最低精度 | 加减看小数位，乘除看有效数字位数 |
| 不确定度 / Uncertainty | $x = ar{x} \pm \Delta x$ | 模拟仪器 $\Delta = rac12$ 最小刻度；数字仪器 $\Delta = \pm 1$ 末位 |
| 相对 / Relative | $\Delta x / x$ 或 $(\Delta x / x) imes 100\%$ | |
| 传播 / Propagation | $Z = A \pm B$: $\Delta Z = \Delta A + \Delta B$ | 加减法：绝对不确定度相加 |
| $Z = A imes B$ 或 $A/B$: $\Delta Z/Z = \Delta A/A + \Delta B/B$ | 乘除法：相对不确定度相加 | |
| 幂次规则 / Power | $Z = A^n$: $\Delta Z/Z = n \cdot \Delta A/A$ | 平方根 $n= rac12$，平方 $n=2$ |
| 图像法 / Graphing | $y = mx + c$ | 最佳拟合线 + 误差棒 |
| 最大最小斜率法 | $\Delta m = (m_{\max} - m_{\min})/2$ | |
| 线性化 / Linearization | $y = kx^n$ → $\ln y = \ln k + n \ln x$ | $\ln$-$\ln$ 图斜率 = 指数 $n$ |
📝 分节练习 / Section Practice
按知识点逐节练习，每题即时反馈。完成一节后自动记录进度。
Practice section by section with instant feedback. Progress is automatically saved.
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🎯 随机测试 / Random Quiz
从题库随机抽取题目，模拟考试环境。提交后统一判分。
Randomly drawn from the question bank. Simulates exam conditions. Graded after submission.
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