📚 PH03 Insert Formula Derivation | PH03 资料册公式推导
The PH03 practical skills paper for International A-Level Physics often includes an insert with experimental data and prompts for formula derivation. Mastering the ability to derive key relationships from first principles and known equations is essential for success. This article covers the most common derivations, from free-fall and pendulum mechanics to resistivity and error propagation, with step-by-step explanations.
国际A-Level物理的PH03实验技能考试常包含一个资料册，提供实验数据和推导公式的提示。掌握从基本原理和已知方程推导关键关系的能力至关重要。本文涵盖最常见的推导，从自由落体和单摆力学到电阻率及误差传播，提供逐步解释。
1. The Role of the PH03 Insert | PH03 资料册的作用
The PH03 insert provides experimental context, raw data, and sometimes partially derived expressions. You are expected to complete or derive final formulas for quantities like g, resistivity, or internal resistance. Understanding the underlying physical model is the first step before rearranging equations and substituting variables.
PH03 资料册提供实验情境、原始数据，有时是部分推导的表达式。你需要补全或推导出最终公式，如 g、电阻率或内阻。在整理方程和代入变量之前，理解其物理模型是第一步。
2. Deriving g from Free Fall Data | 从自由落体数据推导 g
For a free-fall experiment where an object is dropped from rest, the distance fallen h is related to time t by the kinematic equation h = ½ g t². Rearranging for g gives g = 2h / t². If a graph of h against t² is plotted, the slope is ½ g, so g can be found from g = 2 × slope.
对于物体从静止下落的自由落体实验，下落距离 h 与时间 t 的关系由运动学方程 h = ½ g t² 给出。整理得到 g = 2h / t²。若绘制 h 对 t² 的图，斜率为 ½ g，故可通过 g = 2 × 斜率求出 g。
h = ½ g t² → g = 2h / t²
Often the insert provides a table of h and t; you calculate t² and plot the graph to determine the slope. Remember to consider starting from zero and include error bars.
资料册常提供 h 和 t 的数据表；你计算 t² 并作图确定斜率。记住考虑从零开始并包含误差棒。
3. Pendulum: Deriving g from Period and Length | 单摆：从周期与摆长推导 g
For a simple pendulum with small amplitude (less than about 10°), the period T is given by T = 2π √(l/g). Squaring both sides yields T² = 4π² l / g. Hence g = 4π² l / T². A graph of T² against l then has gradient equal to 4π² / g, allowing g to be calculated as g = 4π² / gradient.
对于小振幅（约 10° 以下）的单摆，周期 T = 2π √(l/g)。两边平方得 T² = 4π² l / g，因此 g = 4π² l / T²。T² 对 l 图的斜率等于 4π² / g，从而 g = 4π² / 斜率。
T = 2π √(l/g) → T² = 4π² l / g → g = 4π² l / T²
In the insert, you might see columns for l and T; you must derive g from the mean of multiple trials or through graphical analysis. Pay attention to measuring l from the pivot to the centre of mass of the bob.
在资料册中，你可能会看到 l 与 T 的列；你需要通过多次试验的平均值或图像分析求出 g。注意测量 l 应从摆轴到摆锤质心。
4. Electrical Resistivity from Wire Measurements | 由导线测量推导电阻率
The resistance R of a uniform wire is related to its resistivity ρ, length L, and cross-sectional area A by R = ρ L / A. The area can be expressed in terms of diameter d: A = π d² / 4. Substituting gives ρ = R A / L = R π d² / (4 L). When varying L and recording R, a graph of R versus L has slope = ρ / A, so ρ = slope × A.
均匀导线的电阻 R 与电阻率 ρ、长度 L 和横截面积 A 的关系为 R = ρ L / A。面积可用直径 d 表示为 A = π d² / 4。代入得 ρ = R A / L = R π d² / (4 L)。当改变 L 并记录 R 时，R–L 图的斜率 = ρ / A，故 ρ = 斜率 × A。
ρ = R A / L = (R π d²) / (4 L)
This derivation appears frequently in PH03 inserts; you must be able to rearrange for ρ and use the gradient of the linear graph. Also remember that diameter d should be measured at several points along the wire to obtain an average.
该推导经常出现在 PH03 资料册中；你必须能够整理出 ρ 并利用线性图的斜率。还需记住，直径 d 应沿导线多点测量以获取平均值。
5. Determining EMF and Internal Resistance | 测定电动势与内阻
The terminal voltage V of a cell supplying a current I is V = E – I r, where E is the electromotive force and r the internal resistance. Recording V for different I values and plotting V against I gives a straight line with y-intercept = E and gradient = –r.
电池供给电流 I 时的端电压 V = E – I r，其中 E 为电动势，r 为内阻。记录不同 I 下的 V 并绘制 V–I 图，得到截距 = E、斜率 = –r 的直线。
V = E – I r
Alternatively, from V = I R and I = V/R, you can write E = V + (V/R) r, which rearranges to 1/V = (1/E) + (r/E)(1/R). A graph of 1/V against 1/R yields intercept 1/E and slope r/E. The insert may ask you to derive this linear form.
或者，由 V = I R 和 I = V/R，可写出 E = V + (V/R) r，整理得 1/V = (1/E) + (r/E)(1/R)。1/V 对 1/R 图的截距为 1/E，斜率为 r/E。资料册可能要求你推导此线性形式。
6. Young’s Modulus Derivation | 杨氏模量推导
Young’s modulus E_young is defined as stress divided by strain: E_young = (F/A) / (ΔL/L) = F L / (A ΔL). For a loaded wire, F = m g, and A = π d²/4, leading to E_young = 4 m g L / (π d² ΔL). The slope of a stress–strain graph equals E_young directly.
杨氏模量 E_young 定义为应力除以应变：E_young = (F/A) / (ΔL/L) = F L / (A ΔL)。对于加载的金属丝，F = m g, A = π d²/4，得 E_young = 4 m g L / (π d² ΔL)。应力–应变图的斜率直接等于 E_young。
E_young = F L / (A ΔL) = 4 m g L / (π d² ΔL)
In PH03, you might be provided with mass m and extension ΔL data; you then calculate stress and strain and plot the graph. The linear region gradient yields Young’s modulus; careful unit handling is crucial.
在 PH03 中，你可能会得到质量 m 和伸长量 ΔL 的数据；随后计算应力、应变并作图。线性区域的斜率给出杨氏模量；单位处理要格外小心。
7. Logarithmic Derivation for Exponential Relationships | 指数关系的对数推导
When a quantity decays exponentially, e.g. capacitor discharge current I = I₀ e^(–t/τ), taking natural logarithms linearises the relationship: ln I = ln I₀ – t/τ. A plot of ln I against t gives gradient –1/τ and intercept ln I₀. This technique works for any y = A e^(k x).
当某量呈指数衰减时，例如电容放电电流 I = I₀ e^(–t/τ)，取自然对数使其线性化：ln I = ln I₀ – t/τ。作 ln I 对 t 的图，斜率为 –1/τ，截距为 ln I₀。此方法适用于任何 y = A e^(k x) 形式。
ln I = ln I₀ – t/τ
The insert may include exponential data and ask you to derive the time constant τ. You must be able to transform the equation, identify the gradient, and convert back to τ. For a cooling curve T – T_room = (T₀ – T_room) e^(–kt), the same method applies.
资料册可能包含指数数据，要求推导时间常数 τ。你必须会转换方程、识别斜率并换回 τ。对于冷却曲线 T – T_室温 = (T₀ – T_室温) e^(–kt)，方法相同。
8. Uncertainty Propagation for Derived Quantities | 导出量的不确定度传递
When a quantity Q is calculated from measured variables, absolute and percentage uncertainties combine according to simple rules. If Q = a ± b, absolute uncertainty ΔQ = Δa + Δb. If Q = a × b or a / b, percentage uncertainty in Q = %Δa + %Δb. For Q = a^n, %ΔQ = |n| × %Δa. These rules are essential for error analysis in the PH03 exam.
当量 Q 由测量变量计算时，绝对和百分比不确定度遵循简单规则组合。若 Q = a ± b，绝对不确定度 ΔQ = Δa + Δb。若 Q = a × b 或 a / b，Q 的百分比不确定度 = %Δa + %Δb。对于 Q = a^n，%ΔQ = |n| × %Δa。这些规则对 PH03 考试中的误差分析至关重要。
| Operation Q | Uncertainty ΔQ or %ΔQ |
|---|---|
| Q = a + b or a – b | ΔQ = Δa + Δb |
| Q = a × b or a / b | %ΔQ = %Δa + %Δb |
| Q = a^n | %ΔQ = |n| × %Δa |
For example, in the free-fall case g = 2h/t², the exponent of h is 1, t is –2, so %Δg = %Δh + 2 × %Δt. Always use the magnitude of exponents.
例如，自由落体 g = 2h/t² 中，h 的指数为 1，t 为 –2，故 %Δg = %Δh + 2 × %Δt。务必使用指数的绝对值。
9. Combining Powers and Roots in Error Propagation | 误差分析中幂与根的组合
Generalising, if Q = k a^m b^n, then the percentage uncertainty is %ΔQ = |m| %Δa + |n| %Δb. This covers any combination of multiplication, division, powers, and roots, making it a powerful shortcut for complex derivations such as g = 4π² l / T², where %Δg = %Δl + 2 %ΔT.
推广而言，若 Q = k a^m b^n，则百分比不确定度 %ΔQ = |m| %Δa + |n| %Δb。这涵盖乘、除、幂和根的任意组合，对于复杂推导 (如 g = 4π² l / T²，%Δg = %Δl + 2 %ΔT) 是一个强有力的快捷方法。
%ΔQ = |m| %Δa + |n| %Δb
Note that constants (k, π, etc.) contribute no uncertainty. You only need the fractional uncertainties of the measured quantities. This is frequently tested in the Insert where you need to propagate errors from raw data to the final derived value.
注意常数 (k, π 等) 不贡献不确定度。只需测量量的相对不确定度。这在资料册中经常考查，你需要将原始数据误差传递到最终导出值。
10. Worked Example: Deriving g with Full Uncertainty | 实例：g 的完整推导及不确定度
A student drops an object from rest. She records height h = 1.00 ± 0.01 m and time t = 0.45 ± 0.01 s. First, calculate g = 2h / t² = 2 × 1.00 / (0.45)² = 9.88 m s⁻². Next, find percentage uncertainties: %Δh = (0.01 / 1.00) × 100 = 1.0%; %Δt = (0.01 / 0.45) × 100 ≈ 2.22%.
某学生从静止释放一物体。她记录高度 h = 1.00 ± 0.01 m，时间 t = 0.45 ± 0.01 s。首先，计算 g = 2h / t² = 2 × 1.00 / (0.45)² = 9.88 m s⁻²。接着，求百分比不确定度：%Δh = 1.0%; %Δt ≈ 2.22%。
Since g = 2h t⁻², %Δg = %Δh + 2 × %Δt = 1.0% +
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