📚 IB Physics: Formula Derivation | IB 物理：公式推导
In IB Physics, being able to derive key equations from fundamental principles is a core skill that deepens understanding and prepares you for Paper 1 multiple-choice questions, Paper 2 data analysis, and the internal assessment. This article walks you through the most important derivations, step by step, so you can see how theoretical models connect to the real world.
在 IB 物理中，从基本原理推导关键方程是一项核心能力，它既能加深理解，也有助于应对试卷一的选择题、试卷二的数据分析以及内部评估。本文逐步解析最重要的推导过程，让你看清理论模型与现实世界之间的关联。
1. Deriving the SUVAT Equations | 推导匀变速直线运动方程
quations emerge directly from the definition of acceleration and the area under a velocity– celeration a = (v − u)/t, where u is initial velocity, v is final velocity, and t is time.
四个匀变速直线运动方程来源于加速度的定义以及速度−时间图下的面积。从恒定加速度 a = (v − u)/t 入手，其中 u 是初速度，v 是末速度，t 是时间。
v) t. Substituting v = u + at into this average-velocity expression yields s = ut + ½ at².
整理可得 v = u + at。接下来，位移 s 等于平均速度乘以时间：s = ½ (u + v) t。将 v = u + at 代入这个平均速度表达式，得到 s = ut + ½ at²。
gives s = ½ (u + v) t again. These four equations form the backbone of linear kinematics.
从前两个方程中消去 t 可推出 v² = u² + 2as；消去 a 则再次得到 s = ½ (u + v) t。这四个方程构成了直线运动学的核心。
- v = u + at
- s = ½ (u + v) t
- s = ut + ½ at²
- v² = u² + 2as
Typical derivation flow: a = (v − u)/t → v = u + at → s = ut + ½ at².
典型推导流程：a = (v − u)/t → v = u + at → s = ut + ½ at²。
2. Deriving Centripetal Acceleration a = v²/r | 推导向心加速度 a = v²/r
erentiate twice to find acceleration, or use geometry: in a short time Δt, the velocity vector rotates through angle Δθ.
考虑一个物体以恒定速率 v 在半径为 r 的圆周上运动。在极坐标中，位置矢量为 r(cosθ i + sinθ j)，其中 θ = ωt。对其求两次导数可得到加速度；也可以利用几何方法：在很短的时间 Δt 内，速度矢量转过角度 Δθ。
Δt / r) = v²Δt / r. The acceleration a = Δv / Δt → a = v² / r, directed toward the centre.
速度变化量的大小约为 Δv ≈ vΔθ，弧长 rΔθ = vΔt，因此 Δθ = vΔt / r，进而 Δv = v (vΔt / r) = v²Δt / r。加速度 a = Δv / Δt → a = v² / r，方向指向圆心。
so write a = ω²r. For the IB, be ready to show this derivation with clear vector diagrams.
利用角速度 ω = v/r，还可以写成 a = ω²r。IB 考试中需要能配合清晰的矢量图展示该推导。
a = v² / r = ω²r
3. Newton's Law of Gravitation and g = GM/r² | 万有引力定律与重力加速度 g = GM/r²
Newton' other with force F = GMm / r², where r is their separation. For a small mass m near Earth's surface, the weight mg must equal the gravitational force on m.
牛顿定律指出，两质点 M 和 m 之间的引力为 F = GMm / r²，r 为它们的间距。在地表附近的小质量 m，其重力 mg 必须等于 m 所受的引力。
Thus GMm / R² = mg, where R is Earth' strength equals the gravitational force per unit mass and is independent of the test mass.
因此 GMm / R² = mg，其中 R 为地球半径，M 为地球质量。约去 m 得到 g = GM / R²。这表明重力场强度等于单位质量的引力，且与检验质量无关。
This derivation is fundamental to understanding satellite motion and gravitational potential.
在高度 h 处，g = GM / (R + h)²。这一推导对于理解卫星运动和引力势至关重要。
4. Deriving Escape Velocity v = √2GM/r | 推导逃逸速度 v = √2GM/r
Escape velocity is the minimum speed needed for an object to leave a planet's gravitational influence without further propulsion. At the planet's surface, the total mechanical energy is ½ mv² − GMm/r.
逃逸速度是物体无需后续推进就能脱离行星引力束缚的最小速度。在行星表面，总机械能为 ½ mv² − GMm/r。
ro for the borderline case. By conservation of energy, ½ mv² − GMm/r = 0. Solve for v: v = √2GM/r.
在逃逸的临界情况下，无穷远处的动能和势能都趋于零。根据能量守恒，½ mv² − GMm/r = 0。解出 v：v = √2GM/r。
lanet. For Earth, plugging in M ≈ 5.97×10²⁴ kg and R ≈ 6.37×10⁶ m gives about 11.2 km s⁻¹.
注意 m 被约去了；逃逸速度只依赖于行星的质量和半径。对于地球，代入 M ≈ 5.97×10²⁴ kg，R ≈ 6.37×10⁶ m，得出约 11.2 km s⁻¹。
vesc = √2GM / r
5. Kinetic Theory Derivation of pV = ⅓ N m | 气体动理论推导 pV = ⅓ N m
ide L. Consider one molecule hitting a wall perpendicularly; its momentum change is 2mv_x.
将气体建模为 N 个质量为 m 的相同分子，在一个边长为 L 的立方盒内随机运动。考虑一个分子垂直撞击墙壁，其动量变化为 2mv_x。
The time between collisions with the same wall is 2L/vx, so the average force on that wall is Δp/Δt = (2mv_x) / (2L/vx Summing over all N molecules and dividing by wall area L² gives pressure p = (m / L³) Σ vx².
与同一面墙壁碰撞的间隔时间为 2L/vx，因此对该壁的平均作用力为 Δp/Δt = (2mv_x) / (2L/vx) = mv_x²/L。对所有 N 个分子求和，再除以墙壁面积 L²，得到压强 p = (m / L³) Σ vx²。
Because the motion is random, the mean square speeds are equal:
由于运动是随机的，方向方均值相等：
Connecting
结合理想气体定律将
pV = ⅓ N m
6. Deriving a = −ω²x for Simple Harmonic Motion | 推导简谐运动的 a = −ω²x
Applying Newton’s second law, ma = −kx, so a = −(k/m)x. Defining ω² = k/m gives a = −ω²x.
简谐运动的定义是回复力与位移成正比：F = −kx。应用牛顿第二定律，ma = −kx，因此 a = −(k/m)x。定义 ω² = k/m，得到 a = −ω²x。
x₀ cos(ωt). Taking derivatives verifies that v = ωx₀ cos(ωt) and a = −ω²x₀ sin(ωt) = −ω²x.
这个微分方程的解是 x = x₀ sin(ωt) 或 x = x₀ cos(ωt)。通过求导可以验证 v = ωx₀ cos(ωt) 以及 a = −ω²x₀ sin(ωt) = −ω²x。
y the period T = 2π/ω is independent of amplitude, a hallmark of isochronous oscillations.
该推导揭示了为什么周期 T = 2π/ω 与振幅无关——这是等时振荡的标志。
a = −ω²x, with ω = 2πf
7. Time Period of a Simple Pendulum, T = 2π√l/g | 单摆周期 T = 2π√l/g
mall angles, sinθ ≈ θ = x/l, so F ≈ −mg x/l. This fits the SHM form F = −kx with k = mg/l.
对于摆长为 l 的单摆，切向回复力为 mg sinθ。小角度下 sinθ ≈ θ = x/l，因此 F ≈ −mg x/l。这符合 SHM 的形式 F = −kx，且 k = mg/l。
Using the SHM period formula T = 2π√m/k and substituting k gives T = 2π√(m / (mg/l)) = 2π√l/g. The mass cancels, showing period depends only on length and gravitational field strength.
利用 SHM 周期公式 T = 2π√m/k 并代入 k，得 T = 2π√(m / (mg/l)) = 2π√l/g。质量被约去，表明周期只取决于摆长和重力场强度。
For IB, you should be able to derive this from first principles, referencing the small-angle approximation and the SHM condition a ∝ −x.
IB 要求能从基本原理推导此式，说明小角度近似和 a ∝ −x 的简谐条件。
T = 2π√l/g
8. Doppler Effect Formula for Sound: f' = f (v ± uₒ)/(v ± uₛ) | 声波多普勒效应公式：f' = f (v ± uₒ)/(v ± uₛ)
When a source or observer moves relative to the medium, the observed frequency shifts. Derive separately for moving observer and moving source, then combine.
当波源或观察者相对介质运动时，观测频率会改变。可分别推导观察者运动和波源运动的情况，再合并。
ving toward it at speed uₒ, the relative speed of sound is v + uₒ, so observed frequency f' = (v + uₒ)/λ = f (v + uₒ)/v. Away gives f' = f (v − uₒ)/v.
波源静止、观察者以速度 uₒ 朝向波源运动时，声音的相对速度为 v + uₒ，因此观测频率 f' = (v + uₒ)/λ = f (v + uₒ)/v。远离时 f' = f (v − uₒ)/v。
For a moving source, the wavelength ahead becomes λ' = (v − uₛ)/f, so f' = v/λ' = f v/(v − uₛ) when approaching. Combining both motions yields the general formula: f' = f (v ± uₒ)/(v ± uₛ), with signs chosen appropriately.
对于运动波源，前方的波长变为 λ' = (v − uₛ)/f，从而当接近时 f' = v/λ' = f v/(v − uₛ)。将两种运动结合，得到一般公式：f' = f (v ± uₒ)/(v ± uₛ)，符号需恰当选取。
This equation is fundamental for solving problems involving sirens, radar speed traps, and the redshift of light.
该公式是解决警笛、雷达测速以及光波红移等问题的基础。
f' = f (v ± uₒ) / (v ± uₛ)
9. Single-slit Diffraction Minima: nλ = a sinθ | 单缝衍射暗纹条件：nλ = a sinθ
When monochromatic light passes through a slit of width a, secondary wavelets interfere. Minima occur when wavelets from the top and middle of the slit are out of phase by λ/2, causing cancellation.
单色光通过宽度为 a 的狭缝时，次级子波发生干涉。当从狭缝顶部和中点发出的子波相位差为 λ/2 时，相消干涉形成暗纹。
e generally, dividing the slit into n pairs results in nλ = a sinθ, where n = 1, 2, 3, ...
狭缝上边缘与中心的光线之间的程差为 a/2 sinθ。令其等于 λ/2 得到一级暗纹条件 a sinθ = λ。更一般地，将狭缝分为 n 对子波，可得 nλ = a sinθ，n = 1, 2, 3, ...
This derivation explains the central bright fringe and the rapid decrease in intensity of higher-order maxima.
该推导解释了中央亮纹的形成以及高级次明纹强度迅速下降的原因。
nλ = a sinθ, n ∈ ℕ
10. Deriving the Radioactive Decay Law N = N₀ e−λt | 推导放射性衰变定律 N = N₀ e−λt
roportional to the number of undecayed nuclei: dN/dt = −λN, where λ is the decay constant.
放射性衰变是一个随机过程，其活度 A = −dN/dt 与尚未衰变的原子核数成正比：dN/dt = −λN，λ 为衰变常数。
0 to N at t yields ln(N/N₀) = −λt. Exponentiating gives the exponential decay law N = N₀ e−λt.
分离变量：dN/N = −λ dt。从 t = 0 时 N = N₀ 积分到 t 时刻的 N，得到 ln(N/N₀) = −λt。取指数后得到指数衰变定律 N = N₀ e−λt。
w also appears in capacitor discharge and damped oscillations, highlighting the unifying power of exponential mathematics.
半衰期 T₁/₂ 定义为 N = N₀/2 的时刻，代入得 T₁/₂ = ln 2 / λ。该定律也出现在电容器放电和阻尼振荡中，凸显了指数数学的统一力量。
N = N₀ e−λt, T₁/₂ = ln 2 / λ
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