PH04 Insert Formula Derivations | PH04插页公式推导

📚 PH04 Insert Formula Derivations | PH04插页公式推导

The PH04 Insert provided in International A-Level Physics Unit 4 examinations contains a selection of key formulas. Understanding their derivations not only helps in memorizing them but also deepens your grasp of physical principles. This article guides you through the step-by-step derivations of the most important equations from the insert.

在国际A-Level物理第四单元考试中提供的PH04插页包含了一系列关键公式。理解它们的推导不仅有助于记忆，还能加深对物理原理的掌握。本文将带你逐步推导插页中最重要的方程。

1. Linear Velocity and Angular Velocity | 线速度与角速度

rθ. The linear speed v is the rate of change of arc length: v = s / t = r (Δθ / Δt) = rω.

角速度ω定义为角位移的变化率：ω = Δθ / Δt。对于在半径为r的圆周上运动的物体，通过的弧长为s = rθ。线速度v是弧长的变化率：v = s / t = r (Δθ / Δt) = rω。

v = ωr

2. Centripetal Acceleration | 向心加速度

² Δt / r. Hence the acceleration a = Δv / Δt = v² / r. Substituting v = ωr gives a = ω² r.

考虑一个物体以恒定速率v做圆周运动。在短时间Δt内，速度方向转过Δθ。速度变化的大小近似为Δv = v Δθ（小角度）。因为Δθ = (v Δt) / r，所以Δv = v² Δt / r。因此加速度a = Δv / Δt = v² / r。代入v = ωr可得a = ω² r。

a = v² / r = ω² r

3. Centripetal Force | 向心力

According to Newton' e required to keep an object of mass m moving in a circular path is F = m v² / r = m ω² r.

根据牛顿第二定律，产生向心加速度的合力为F = m a。因此，使质量为m的物体保持圆周运动所需的向心力为F = m v² / r = m ω² r。

F = m v² / r = m ω² r

4. Acceleration in Simple Harmonic Motion | 简谐运动的加速度

is directly proportional to displacement and always points towards the equilibrium position.

在简谐运动中，回复力与离开平衡位置的位移成正比且方向相反：F = -k x，其中k为力常数。由F = m a可得a = -(k/m) x。定义ω² = k/m，则特征方程为a = -ω² x。这表明加速度与位移成正比且总指向平衡位置。

a = -ω² x

5. Displacement–Time Equation for SHM | 简谐运动的位移-时间方程

ing x = A cos ωt. If starting from equilibrium moving in positive direction, x = A sin ωt.

微分方程d²x/dt² = -ω² x的通解为x = A cos(ωt + φ)，其中A为振幅，φ为相位常数。选择初条件为最大正向位移（t=0时x = A），则φ = 0，得到x = A cos ωt。若从平衡位置向正向运动开始，则x = A sin ωt。

x = A cos (ωt) or x = A sin (ωt)

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6. Velocity–Displacement Relation in SHM | 简谐运动的速度-位移关系

. Kinetic energy is total energy minus potential: ½ m v² = ½ m ω² (A² - x²). Thus, v = ± ω √A² - x². The sign depends on the direction of motion.

无阻尼简谐运动系统的总能量守恒：E = ½ k A² = ½ m ω² A²。在任一位移x处，势能为½ k x² = ½ m ω² x²。动能为总能量减去势能：½ m v² = ½ m ω² (A² - x²)。因此，v = ± ω √A² - x²。符号取决于运动方向。

v = ± ω √A² - x²

7. Period of a Mass–Spring System | 弹簧振子的周期

From the definitions, ω = √k/m for a mass-spring system. Since ω = 2π / T, we have 2π / T = √k/m. Rearranging gives the period T = 2π √m/k. This shows that the period depends only on the mass and spring constant, not on the amplitude.

由定义，弹簧振子的角频率ω = √k/m。由于ω = 2π / T，我们有2π / T = √k/m。整理得周期T = 2π √m/k。这表明周期只取决于质量和弹簧常数，与振幅无关。

T = 2π √m / k

8. Period of a Simple Pendulum | 单摆的周期

is the horizontal displacement. Comparing with F = -k x, the effective force constant is k_eff = mg/L. Therefore, ω = √k_eff / m = √g/L. The period is T = 2π / ω = 2π √L/g.

对于长度为L的单摆，小角度回复力为F = -mg sinθ ≈ -mg (x/L)，其中x为水平位移。与F = -k x对比，有效力常数为k_eff = mg/L。因此，ω = √k_eff / m = √g/L。周期为T = 2π / ω = 2π √L/g。

T = 2π √L / g

9. Electric Field Due to a Point Charge | 点电荷的电场强度

Coulomb' s the force per unit positive test charge: E = F / q. Hence, E = k Q / r² = Q / (4πε₀ r²).

库仑定律给出

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