AS Physics Unit 5 Insert June 2019: Derivations of Key Formulas | AS物理单元5 2019年6月插页公式推导

📚 AS Physics Unit 5 Insert June 2019: Derivations of Key Formulas | AS物理单元5 2019年6月插页公式推导

The June 2019 Unit 5 formula insert for AS Physics provides essential equations covering thermal physics, oscillations, waves and astrophysics. Understanding their derivations goes beyond rote learning — it builds a robust conceptual framework and equips you to tackle unfamiliar problems with confidence. This article walks you through the logic and steps behind the most pivotal formulas from that insert.

2019年6月AS物理单元5的公式插页提供了热力学、振动、波动与天体物理的核心方程。理解这些公式的来龙去脉，远胜于死记硬背——它能帮你构建坚实的概念框架，并自信地应对陌生题目。本文将带你一步步拆解该插页中最重要的公式推导。

1. Kinetic Theory Derivation of pV = 1/3 N m ⟨c²⟩ | 气体动理论推导 pV = 1/3 N m ⟨c²⟩

Imagine a single molecule of mass m moving with velocity components (v_x, v_y, v_z) inside a cube of side L. When it collides elastically with a wall perpendicular to the x-axis, its momentum change is Δp = 2 m v_x. The time between successive collisions with that same wall is Δt = 2L / v_x. Hence the average force on the wall is F = Δp/Δt = (2 m v_x) / (2L / v_x) = m v_x² / L.

设单个分子质量为 m，以速度分量 (v_x, v_y, v_z) 在边长为 L 的立方体内运动。它与垂直于 x 轴的器壁发生弹性碰撞时，动量变化为 Δp = 2 m v_x。先后两次与该壁碰撞的时间间隔为 Δt = 2L / v_x，因此对器壁的平均作用力为 F = Δp/Δt = (2 m v_x) / (2L / v_x) = m v_x² / L。

Pressure from this one molecule on the wall of area A = L² is p_x = F/A = m v_x² / L³ = m v_x² / V, where V = L³. Summing over all N molecules and using the mean square speed in the x-direction ⟨v_x²⟩ gives the total pressure p = N m ⟨v_x²⟩ / V. For random motion, ⟨v²⟩ = ⟨v_x²⟩ + ⟨v_y²⟩ + ⟨v_z²⟩ = 3⟨v_x²⟩, so ⟨v_x²⟩ = ⟨v²⟩/3. Substituting yields the key result pV = 1/3 N m ⟨v²⟩.

该分子对面积为 A = L² 的器壁产生的压强为 p_x = F/A = m v_x² / L³ = m v_x² / V，其中 V = L³。对所有 N 个分子求和，并利用 x 方向的方均速率 ⟨v_x²⟩，可得总压强 p = N m ⟨v_x²⟩ / V。由于随机运动满足 ⟨v²⟩ = ⟨v_x²⟩ + ⟨v_y²⟩ + ⟨v_z²⟩ = 3⟨v_x²⟩，故 ⟨v_x²⟩ = ⟨v²⟩/3。代入即得核心公式 pV = 1/3 N m ⟨v²⟩。

2. Linking Kinetic Theory to Temperature: ½ m ⟨v²⟩ = (3/2) kT | 动能与温度的关系：½ m ⟨v²⟩ = (3/2) kT

The ideal gas law states pV = N k T, where k is the Boltzmann constant. Equating this with the kinetic theory result gives 1/3 N m ⟨v²⟩ = N k T. Cancelling N and multiplying both sides by 3/2 we obtain ½ m ⟨v²⟩ = (3/2) k T. This directly shows that the average translational kinetic energy of a molecule is proportional to the absolute temperature.

理想气体定律为 pV = N k T，其中 k 为玻尔兹曼常数。令其与气体动理论的结果相等，得到 1/3 N m ⟨v²⟩ = N k T。消去 N 并在两边同时乘以 3/2，即得 ½ m ⟨v²⟩ = (3/2) k T。这直接表明分子的平均平动动能与热力学温度成正比。

3. Deriving the Stefan–Boltzmann Law L = σ A T⁴ | 斯忒藩–玻尔兹曼定律 L = σ A T⁴ 的推导

For a black-body cavity, the energy density u of the radiation is a universal function of temperature. Thermodynamic arguments (using radiation pressure and a Carnot cycle) show that u ∝ T⁴, so we write u = a T⁴, where a is a constant. The radiation flux (power per unit area) leaving a small hole is found to be M = (c/4) u, thus M = (c a /4) T⁴. The total luminosity from a surface of area A is therefore L = σ A T⁴, with the Stefan–Boltzmann constant σ = (c a)/4.

对于黑体空腔，辐射的能量密度 u 是温度的普适函数。通过热力学论证（利用辐射压强和卡诺循环）可知 u ∝ T⁴，因此可记作 u = a T⁴，a 为常数。从小孔逸出的辐射通量（单位面积功率）为 M = (c/4) u，即 M = (c a /4) T⁴。故面积为 A 的表面发出的总光度为 L = σ A T⁴，斯忒藩–玻尔兹曼常量 σ = (c a)/4。

4. Simple Harmonic Motion Acceleration: a = –ω² x | 简谐运动的加速度 a = –ω² x

For a particle executing SHM, its displacement can be expressed as x = A cos(ω t) or x = A sin(ω t). Differentiating once gives velocity v = –A ω sin(ω t); differentiating again gives acceleration a = –A ω² cos(ω t) = –ω² x. The minus sign indicates that the acceleration is always directed towards the equilibrium position, proportional to the displacement.

做简谐运动的质点，其位移可写为 x = A cos(ω t) 或 x = A sin(ω t)。求一次导数得速度 v = –A ω sin(ω t)；再求导得加速度 a = –A ω² cos(ω t) = –ω² x。负号表明加速度始终指向平衡位置，且大小与位移成正比。

5. Maximum Speed in SHM: v_max = ω A | 简谐运动的最大速度 v_max = ω A

From the velocity expression v = –A ω sin(ω t), the maximum magnitude occurs when sin(ω t) = ±1, giving v_max = ω A. This can also be found via energy conservation: total energy E = ½ m ω² A² = ½ m v² + ½ m ω² x². Setting x = 0 (equilibrium) eliminates the potential term, yielding ½ m v_max² = ½ m ω² A², so v_max = ω A.

由速度表达式 v = –A ω sin(ω t)，当 sin(ω t) = ±1 时速度幅值最大，得 v_max = ω A。亦可通过能量守恒求得：总能量 E = ½ m ω² A² = ½ m v² + ½ m ω² x²。令 x = 0（平衡位置）势能项为零，得到 ½ m v_max² = ½ m ω² A²，故 v_max = ω A。

6. Deriving the Period of a Mass–Spring System: T = 2π √(m/k) | 质量–弹簧系统周期 T = 2π √(m/k) 的推导

For a mass m on a light spring of force constant k, Hooke’s law provides the restoring force F = –k x. Newton’s second law gives m a = –k x, or a = –(k/m) x. Comparing this with the standard SHM equation a = –ω² x, we identify ω² = k/m, thus ω = √(k/m). Since the period T = 2π/ω, we obtain T = 2π √(m/k).

对于劲度系数为 k 的轻弹簧和质量 m，胡克定律给出回复力 F = –k x。牛顿第二定律给出 m a = –k x，即 a = –(k/m) x。与简谐运动标准方程 a = –ω² x 对比，可得 ω² = k/m，因而 ω = √(k/m)。由周期 T = 2π/ω，得到 T = 2π √(m/k)。

7. Deriving the Period of a Simple Pendulum: T = 2π √(l/g) | 单摆周期 T = 2π √(l/g) 的推导

Consider a simple pendulum of length l with a bob of mass m displaced by a small angle θ. The restoring force along the tangent is –m g sin θ ≈ –m g θ for small angles. The linear acceleration along the arc is a = l (d²θ/dt²). Newton’s second law then gives m l d²θ/dt² = –m g θ, which simplifies to d²θ/dt² = –(g/l) θ. This is SHM with ω² = g/l, so T = 2π/ω = 2π √(l/g).

考虑长为 l 的单摆，摆球质量为 m，偏离小角度 θ。沿切线方向的回复力为 –m g sin θ ≈ –m g θ（小角度近似）。沿弧线的线

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