📚 International A-Level Physics PH04 Unit 4: Formula Derivations | 国际A-Level物理PH04第四单元：公式推导
Unit 4 of the International A-Level Physics specification (PH04) covers mechanics and materials, oscillations and waves. A deep understanding of how the key equations are derived is essential for mastering the concepts and scoring well on exam questions. This article walks you through the foundational derivations step by step, linking circular motion to simple harmonic motion, mass-spring systems, and pendulums.
国际A-Level物理第四单元(PH04)涵盖力学与材料、振动与波动。深刻理解关键公式的推导过程对于掌握这些概念并在考试中取得高分至关重要。本文将逐步带你走完这些基础推导，将圆周运动与简谐运动、弹簧振子和单摆联系起来。
1. Centripetal Acceleration Derivation | 向心加速度推导
An object moving with constant speed v in a circle of radius r changes direction continuously, giving rise to a centripetal acceleration directed towards the centre. Consider two velocity vectors separated by a small angle Δθ; the change in velocity Δv can be approximated as v Δθ. Dividing by time Δt gives acceleration a = v (Δθ/Δt). Since the angular speed ω is defined as Δθ/Δt, we obtain a = v ω. Substituting v = r ω gives the familiar forms a = v²/r = ω² r.
一个物体以恒定速率v在半径为r的圆上运动，方向持续改变，从而产生指向圆心的向心加速度。考虑两个速度矢量被一个小角Δθ分隔；速度变化Δv可近似为v Δθ。除以时间Δt得加速度a = v (Δθ/Δt)。由于角速度ω定义为Δθ/Δt，得到a = v ω。代入v = r ω得出常见形式a = v²/r = ω² r。
a = v ω → a = v²/r = ω² r
In vector form, the acceleration is a = −ω² r, with the negative sign indicating direction towards the centre. This derivation does not require calculus; it uses the geometry of small angles.
矢量形式为a = −ω² r，负号表示指向圆心。这个推导不需要微积分，运用了小角度的几何关系。
2. Relationship between Linear and Angular Quantities | 线量与角量关系推导
The angular displacement θ (in radians) is defined as the arc length s divided by the radius r: θ = s / r. Differentiating with respect to time: since dθ/dt = ω and ds/dt = v, we have ω = v / r, hence v = r ω. Similarly, tangential acceleration a_t relates to angular acceleration α: a_t = r α. The centripetal acceleration vector is a separate component perpendicular to the velocity.
角位移θ(弧度)定义为弧长s除以半径r：θ = s / r。对时间求导：dθ/dt = ω, ds/dt = v，故ω = v / r，因此v = r ω。类似地，切向加速度a_t与角加速度α的关系为a_t = r α。向心加速度矢量是垂直于速度的另一分量。
- v = r ω – scalar relation, valid for instantaneous speeds
- a_t = r α – for changing angular speed
- a_c = v²/r = r ω² – directed radially inwards
- v = r ω – 标量关系，适用于瞬时速率
- a_t = r α – 用于角速度变化时
- a_c = v²/r = r ω² – 方向径向向内
3. Simple Harmonic Motion as a Projection of Circular Motion | 简谐运动作为圆周运动的投影
A particle moving uniformly in a circle, when viewed edge‑on, appears to oscillate back and forth along a diameter. This projection is simple harmonic motion (SHM). If the circle has radius A and angular speed ω, the displacement x from the equilibrium position is the horizontal component of the radius vector: x = A cos(ω t) or x = A sin(ω t) depending on the starting phase.
一个在圆周上匀速运动的粒子，从侧面看去，表现为沿直径来回振荡。这个投影就是简谐运动(SHM)。如果圆周半径为A、角速度为ω，则偏离平衡位置的位移x是半径矢量的水平分量：x = A cos(ω t)或x = A sin(ω t)，取决于起始相位。
x = A sin(ω t + φ)
Using this geometrical link, the velocity and acceleration of the projected motion can be derived from the tangential velocity and centripetal acceleration of the uniform circular motion.
利用这种几何联系，可以从匀速圆周运动的切向速度和向心加速度导出投影运动的速度和加速度。
4. Deriving the Displacement Equation x = A sin(ω t) | 位移方程x = A sin(ω t)推导
Start with the projection of the position vector on the y‑axis (or x-axis) of a circle of radius A. Suppose at t = 0 the particle is at the equilibrium position and moving in the positive direction. Then the angle rotated from the reference axis is ω t. The vertical projection gives x = A sin(ω t). This is the standard SHM displacement when oscillations begin from the equilibrium with maximum upward velocity.
从半径为A的圆的位置矢量在y轴(或x轴)上的投影出发。假设t = 0时粒子在平衡位置并向正方向运动，则从参考轴转过的角度为ω t。垂直投影给出x = A sin(ω t)。这是从平衡位置开始以最大向上速度振荡时的标准简谐位移表达式。
If instead the particle starts at maximum displacement A, the projection is x = A cos(ω t). The constant A is the amplitude, and ω is the angular frequency.
如果粒子从最大位移A处开始，投影为x = A cos(ω t)。常数A为振幅，ω为角频率。
5. Velocity in SHM | 简谐运动中的速度
From the projection, the velocity of the oscillating mass is the horizontal component of the circular motion’s tangential velocity (v_c = A ω). Therefore, v = ± A ω cos(ω t) when x = A sin(ω t). Using the identity cos²(ω t) = 1 − sin²(ω t), we obtain v = ± ω √(A² − x²). The sign depends on direction; the speed is maximum (v_max = ω A) when x = 0 and zero at the extremes.
由投影关系可知，振荡质量的速度是圆周运动切向速度(v_c = A ω)的水平分量。因此，当x = A sin(ω t)时，v = ± A ω cos(ω t)。利用恒等式cos²(ω t) = 1 − sin²(ω t)，得到v = ± ω √(A² − x²)。正负号取决于方向；当x = 0时速率最大(v_max = ω A)，在端点处为零。
v = ± ω √(A² − x²)
This relationship is extremely useful for linking velocity to displacement without knowing time explicitly.
这个关系在不需要明确时间的情况下将速度与位移联系起来，非常有用。
6. Acceleration in SHM and the Defining Equation | 简谐运动的加速度及定义式
The acceleration a of the projected motion is the horizontal component of the centripetal acceleration a_c = −ω² r (with r = A). Thus a = −ω² (A sin(ω t)) = −ω² x. This is the hallmark of SHM: acceleration is directly proportional to displacement from equilibrium and always directed towards the equilibrium position.
投影运动的加速度a是向心加速度a_c = −ω² r (其中r = A)的水平分量。因此a = −ω² (A sin(ω t)) = −ω² x。这是简谐运动的标志：加速度与偏离平衡位置的位移成正比，且总是指向平衡位置。
a = −ω² x
The negative sign indicates restoring force nature. Deriving this from Newton’s second law yields the differential equation d²x/dt² = −ω² x, whose general solution is a sinusoidal function.
负号表示恢复力的性质。从牛顿第二定律推导可得微分方程d²x/dt² = −ω² x，其通解为正弦函数。
7. Maximum Acceleration and Its Significance | 最大加速度及其意义
From a = −ω² x, the magnitude of acceleration is greatest when displacement is at the amplitude A: a_max = ω² A. In terms of frequency f, ω = 2π f, so a_max = (2π f)² A = 4π² f² A. This is important for designing structures to withstand vibrations; the acceleration can be many times g.
由a = −ω² x，当位移等于振幅A时，加速度大小最大：a_max = ω² A。用频率f表示，ω = 2π f，故a_max = (2π f)² A = 4π² f² A。这对于设计承受振动的结构很重要；加速度可能是重力加速度的多倍。
- At extreme points: |a| = a_max, v = 0
- At equilibrium: a = 0, |v| = v_max = ω A
- 在端点：|a| = a_max, v = 0
- 在平衡位置：a = 0, |v| = v_max = ω A
8. Mass-Spring System: Deriving T = 2π√(m/k) | 弹簧振子：周期T = 2π√(m/k)推导
For a mass m attached to a spring of force constant k, Hooke’s law gives the restoring force F = −k x. Using Newton’s second law F = m a, we have m a = −k x → a = −(k/m) x. Comparing with the SHM defining equation a = −ω² x, we identify ω² = k/m. Thus the angular frequency is ω = √(k/m). The period T is related by ω = 2π/T, leading to T = 2π/ω = 2π √(m/k).
对于连接在劲度系数为k的弹簧上的质量m，胡克定律给出恢复力F = −k x。运用牛顿第二定律F = m a，得m a = −k x → a = −(k/m) x。与简谐运动定义式a = −ω² x对比，得出ω² = k/m，故角频率ω = √(k/m)。周期T通过ω = 2π/T关联，得到T = 2π/ω = 2π √(m/k)。
T = 2π √(m/k)
This assumes a massless spring and no damping. The frequency f = 1/T = (1/2π) √(k/m). This derivation underlines why stiffer springs (larger k) give higher frequencies and larger masses give lower frequencies.
这假定弹簧质量为零且无阻尼。频率f = 1/T = (1/2π) √(k/m)。这个推导说明了为什么较硬的弹簧(较大的k)产生较高频率，而较大的质量导致较低频率。
9. Simple Pendulum: Deriving T = 2π√(l/g) | 单摆：周期T = 2π√(l/g)推导
For a simple pendulum (a point mass on a light, inextensible string of length l), the restoring force when displaced by a small angle θ is approximately −m g θ. The tangential acceleration a_t relates to angular acceleration α = a_t / l. Using τ = I α, the torque about the pivot is −m g l sinθ. For small angles sinθ ≈ θ, giving the equation I α = −m g l θ. For a point mass I = m l², so m l² α = −m g l θ → α = −(g/l) θ.
对于单摆(质点系于长度为l的轻质不可伸长的绳上)，当偏离一个小角度θ时，恢复力近似为−m g θ。切向加速度a_t与角加速度α的关系为α = a_t / l。利用τ = I α，绕支点的力矩为−m g l sinθ。小角度下sinθ ≈ θ，得到I α = −m g l θ。对于质点I = m l²，故m l² α = −m g l θ → α = −(g/l) θ。
This is the angular equivalent of a = −ω² x, so ω² = g/l. Hence ω = √(g/l) and the period T = 2π/ω = 2π √(l/g).
这是a = −ω² x的角量形式，故ω² = g/l。因此ω = √(g/l)，周期T = 2π/ω = 2π √(l/g)。
T = 2π √(l/g)
The period is independent of the mass and, for small amplitudes, independent of the amplitude itself — this is the isochronous property of a pendulum.
周期与质量无关，且对于小振幅来说与振幅本身也无关——这是单摆的等时性。
10. Energy in Simple Harmonic Motion | 简谐运动中的能量
In SHM, the total mechanical energy is conserved if no damping acts. The kinetic energy is KE = ½ m v² = ½ m ω² (A² − x²). The potential energy stored (for a spring) is PE = ½ k x² = ½ m ω² x² (since k = m ω²). Adding them gives total energy E_total = ½ m ω² A² = ½ k A², a constant. Energy continuously transforms between kinetic and potential forms.
在简谐运动中，若无阻尼作用，总机械能守恒。动能为KE = ½ m v² = ½ m ω² (A² − x²)。储存的势能(对于弹簧)为PE = ½ k x² = ½ m ω² x² (因k = m ω²)。两者相加得总能量E_total = ½ m ω² A² = ½ k A²，为常数。能量在动能与势能之间连续转换。
E_total = ½ m ω² A² = ½ k A²
At x = 0, PE = 0 and KE is maximum; at x = ±A, KE = 0 and PE is maximum. The energy relationship provides an alternative way to derive v = ± ω √(A² − x²).
当x = 0时，PE = 0，KE最大；当x = ±A时，KE = 0，PE最大。能量关系提供了推导v = ± ω √(A² − x²)的另一种途径。
11. Summary of Key Derived Equations | 关键推导公式总结
| Concept | Formula | Notes |
|---|---|---|
| Centripetal acceleration | a = v²/r = r ω² | Always directed to centre |
| Tangential speed | v = r ω | Valid in rad s⁻¹ |
| SHM displacement | x = A sin(ω t) or x = A cos(ω t) | Depends on phase at t=0 |
| SHM velocity | v = ± ω √(A² − x²) | v_max = ω A |
| SHM acceleration | a = −ω² x | Defining equation |
| Mass-spring period | T = 2π √(m/k) | For ideal spring |
| Simple pendulum period | T = 2π √(l/g) | Small angles only |
| Total energy in SHM | E = ½ m ω² A² | Constant if undamped |
Remember that all angular quantities must be in radians when using these relationships. The derivations shown here emphasise the unity between circular motion and oscillations, which is central to PH04.
请记住，在使用这些关系时所有角度量必须以弧度为单位。这里给出的推导强调了圆周运动与振动之间的统一，这是PH04的核心。
12. Exam Tips for Derivation Questions | 推导题的考试技巧
In exam questions that ask you to derive these equations, it is vital to show clear steps: define variables, state the relevant law (Newton’s second law, Hooke’s law, small-angle approximation), and make the link to the SHM defining equation a = −ω² x. Many marks are awarded for explaining why the approximation is valid (e.g., sinθ ≈ θ for θ < 10°). Practise drawing diagrams that relate circular motion radius to amplitude, and projecting vectors.
在要求推导这些公式的考试题中，务必展示清晰的步骤：定义变量，陈述相关定律(牛顿第二定律、胡克定律、小角度近似)，并与简谐运动定义式a = −ω² x建立联系。很多分数是赋予解释为什么近似有效(例如当θ < 10°时 sinθ ≈ θ)。练习绘制将圆周运动半径与振幅联系起来的图，以及投影矢量。
For energy derivations, starting from v = ± ω √(A² − x²) and substituting into KE + PE = constant is a reliable approach. Always check that your final expression has correct dimensions.
对于能量推导，从v = ± ω √(A² − x²)出发并代入KE + PE = 常数是一种可靠的方法。务必检查最终表达式的量纲是否正确。
Published by TutorHao | Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply