Complete Physics Formula Derivations | 物理公式推导大全

📚 Complete Physics Formula Derivations | 物理公式推导大全

Understanding how key physics formulas are derived not only deepens comprehension but also builds problem-solving confidence. This article walks you through the derivations of fundamental equations in mechanics, electricity, and waves, presented in clear step-by-step bilingual notes.

理解核心物理公式的推导过程，不仅能加深理解，还能增强解题自信。本文带您一步步推导力学、电学和波动学中的基本方程，以清晰的双语笔记呈现。

1. Equations of Motion with Constant Acceleration | 匀加速直线运动公式

For an object moving with constant acceleration a, the velocity changes uniformly. If initial velocity is u and final velocity v after time t, then by definition a = (v – u)/t, giving v = u + at.

对于匀加速直线运动，加速度 a 恒定。若初速度为 u，经时间 t 后末速度为 v，由定义 a = (v – u)/t，可得 v = u + at。

v = u + at

Displacement s is the area under the velocity-time graph. Since velocity changes linearly, average velocity = (u + v)/2. Hence s = ½(u + v)t. Substituting v from the first equation yields s = ut + ½at².

位移 s 等于速度-时间图下的面积。由于速度线性变化，平均速度 = (u + v)/2，故 s = ½(u + v)t。代入 v = u + at 得 s = ut + ½at²。

s = ut + ½at²

To eliminate time, solve for t from v = u + at and substitute into s = ½(u + v)t: t = (v – u)/a, so s = ½(u + v)(v – u)/a = (v² – u²)/(2a). Rearranging gives v² = u² + 2as.

消去时间 t：由 v = u + at 得 t = (v – u)/a，代入 s = ½(u + v)t 得 s = (v² – u²)/(2a)，整理得 v² = u² + 2as。

v² = u² + 2as

2. Newton’s Second Law and Impulse | 牛顿第二定律与冲量

Newton’s second law states that net force F equals rate of change of momentum: F = dp/dt. For constant mass, p = mv, so dp/dt = m dv/dt = ma. Thus F = ma.

牛顿第二定律表明合外力等于动量变化率：F = dp/dt。当质量恒定时，p = mv，故 dp/dt = m dv/dt = ma，因此 F = ma。

Impulse J is the integral of force over time: J = ∫F dt. Since F = dp/dt, we have J = ∫dp = Δp = m(v – u). Impulse equals change in momentum.

冲量 J 是力对时间的积分：J = ∫F dt。因 F = dp/dt，故 J = ∫dp = Δp = m(v – u)。冲量等于动量的变化。

3. Work-Energy Theorem | 功能定理

Work done by a constant force F acting over displacement d with angle θ is W = F d cosθ. If the force causes acceleration, using F = ma and kinematics v² = u² + 2as (where s = d and a along displacement), we get W = mad = m (v² – u²)/2 = ½mv² – ½mu². Hence, net work equals change in kinetic energy.

恒力 F 作用位移 d，夹角 θ 时做功 W = F d cosθ。若力产生加速度，使用 F = ma 和运动学 v² = u² + 2as（s = d，a 沿位移方向），得 W = mad = m (v² – u²)/2 = ½mv² – ½mu²。因此，合外力做功等于动能变化量。

W = ΔKE = ½mv² – ½mu²

4. Conservation of Momentum | 动量守恒

For two interacting objects, Newton’s third law says F₁₂ = -F₂₁. If the interaction time is Δt, impulses are equal and opposite: F₁₂ Δt = -F₂₁ Δt. This implies Δp₁ = -Δp₂. Therefore, total momentum before interaction equals total momentum after: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.

两个相互作用的物体，由牛顿第三定律 F₁₂ = -F₂₁，若作用时间 Δt 相同，冲量等大反向：F₁₂ Δt = -F₂₁ Δt，即 Δp₁ = -Δp₂。所以系统总动量在作用前后守恒：m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂。

5. Centripetal Acceleration | 向心加速度

An object moving at constant speed v in a circle of radius r travels an arc length s = rθ. The velocity vector changes direction by θ over time Δt. The change in velocity Δv = vθ (for small θ). The acceleration directed toward the centre has magnitude a = Δv/Δt = v(θ/Δt) = vω, where ω = θ/Δt is angular speed. Since v = ωr, we obtain a = v²/r = ω²r.

物体以恒定速率 v 在半径 r 的圆周上运动，弧长 s = rθ。经历 Δt 时间，速度方向转过角度 θ。速度变化量大小 Δv = vθ（当 θ 很小时）。向心加速度大小 a = Δv/Δt = v(θ/Δt) = vω，ω 为角速度。又 v = ωr，可得 a = v²/r = ω²r。

a = v²/r = ω²r

6. Newton’s Law of Gravitation and Kepler’s Third Law | 万有引力定律与开普勒第三定律

For a planet in a circular orbit, gravitational force provides centripetal force: G M m / r² = m v² / r. The orbital speed v = 2πr / T, where T is period. Substituting gives G M / r = 4π² r² / T², so T² = (4π² / GM) r³. Thus T² ∝ r³, Kepler’s third law.

对于圆轨道上的行星，万有引力提供向心力：G M m / r² = m v² / r。轨道速度 v = 2πr / T，T 为周期。代入得 G M / r = 4π² r² / T²，因此 T² = (4π² / GM) r³，即 T² ∝ r³，开普勒第三定律。

T² = (4π² / GM) r³

7. Simple Harmonic Motion (SHM) | 简谐运动

For a mass-spring system, Hooke’s law gives restoring force F = -k x. By Newton’s second law, m a = -k x, so a = -(k/m) x. Defining ω² = k/m yields the SHM equation a = -ω² x. The solution is x = A sin(ωt + φ), with velocity v = ωA cos(ωt + φ) and acceleration a = -ω²A sin(ωt + φ) = -ω² x.

对弹簧振子，由胡克定律回复力 F = -k x，根据牛顿第二定律 m a = -k x，得 a = -(k/m) x。令 ω² = k/m，即得简谐运动方程 a = -ω² x。其解为 x = A sin(ωt + φ)，速度 v = ωA cos(ωt + φ)，加速度 a = -ω²A sin(ωt + φ) = -ω² x。

a = -ω² x

8. Resistance and Resistivity | 电阻与电阻率

The resistance R of a uniform conductor is defined as R = V / I. Using electric field E = V / L and current density J = I / A, and Ohm’s microscopic form J = σE (σ = 1/ρ), we have I/A = (1/ρ)(V/L). Rearranging yields V/I = ρL/A, so R = ρL/A.

均匀导体的电阻 R 定义为 R = V / I。利用电场 E = V / L，电流密度 J = I / A，以及欧姆定律的微观形式 J = σE（σ = 1/ρ），得 I/A = (1/ρ)(V/L)。整理后 V/I = ρL/A，因此 R = ρL/A。

R = ρL / A

9. Energy Stored in a Capacitor | 电容器储存的能量

During charging, the work required to move a small charge dq against the potential difference V is dW = V dq. Since V = q/C, total work W = ∫₀^Q (q/C) dq = ½ Q²/C. Using Q = CV, this equals ½ QV = ½ CV². This work is stored as electrical potential energy.

充电过程中，移动微小电荷 dq 需克服电势差 V 做功 dW = V dq。因 V = q/C，总功 W = ∫₀^Q (q/C) dq = ½ Q²/C。利用 Q = CV，可得 ½ QV = ½ CV²。这部分功以电势能形式储存。

E = ½ CV² = ½ QV = ½ Q²/C

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Complete Physics Formula Derivations | 物理公式推导大全