Circular and Periodic Motion | 圆周运动与周期运动

📚 Circular and Periodic Motion | 圆周运动与周期运动

Circular motion and periodic (oscillatory) motion are two of the most elegant applications of Newtonian mechanics. In circular motion, an object moves along a circular path with a constant speed but a continually changing velocity direction, requiring a centripetal force. In periodic motion known as simple harmonic motion (SHM), an object oscillates about an equilibrium position, with its acceleration always directed back towards equilibrium and proportional to its displacement. This article examines both topics in depth, covering angular velocity, centripetal acceleration, the defining equation of SHM, energy exchanges, damping, resonance, and real-world applications, all tailored to the AQA International A-level Physics specification.

圆周运动与周期运动（振动）是牛顿力学中最优雅的两个应用。在圆周运动中，物体沿圆形路径以恒定速率运动，但速度方向不断改变，因此需要向心力。而在被称为简谐运动（SHM）的周期运动中，物体围绕平衡位置来回振动，其加速度始终指向平衡位置且与位移成正比。本文将深入探讨这两个课题，涵盖角速度、向心加速度、简谐运动的定义方程、能量转化、阻尼、共振及实际应用，完全针对 AQA 国际 A-level 物理考纲编写。

1. Angular Displacement and Angular Velocity | 角位移与角速度

Angular displacement (θ) is the angle swept out by a radius line joining the object to the centre of the circular path. It is measured in radians (rad), where one complete revolution corresponds to 2π radians. By definition, 2π radians = 360°, so a radian is the angle subtended when the arc length equals the radius of the circle.

角位移（θ）是连接物体与圆心的半径线所扫过的角度，单位为弧度（rad），一整圈对应 2π 弧度。根据定义，2π 弧度 = 360°，因此 1 弧度就是弧长等于半径时所对应的圆心角。

Angular velocity (ω) is defined as the rate of change of angular displacement, ω = Δθ/Δt, with units of rad s⁻¹. For a complete revolution, Δθ = 2π and Δt = T (the period), giving the crucial identity ω = 2π/T. Since frequency f = 1/T, we also write ω = 2πf. Angular velocity is a vector; for objects in uniform circular motion, its direction is perpendicular to the plane of rotation, but at this level we focus on its magnitude.

角速度（ω）定义为角位移的变化率，即 ω = Δθ/Δt，单位为 rad s⁻¹。对于一整圈，Δθ = 2π，Δt = T（周期），因此得到关键恒等式 ω = 2π/T。由于频率 f = 1/T，我们还可写出 ω = 2πf。角速度是矢量；对于匀速圆周运动，其方向垂直于旋转平面，但在本阶段我们主要关注它的大小。

Consider a ceiling fan completing 120 revolutions per minute. Its frequency is 120/60 = 2 Hz, giving T = 0.5 s. The angular velocity is ω = 2π/T = 2π × 2 = 4π ≈ 12.6 rad s⁻¹. This angular velocity is the same for every point on the fan blade, even though points further from the centre move faster in linear terms.

考虑一台每分钟转 120 圈的电风扇。其频率为 120/60 = 2 Hz，因此 T = 0.5 s。角速度为 ω = 2π/T = 2π × 2 = 4π ≈ 12.6 rad s⁻¹。扇叶上每一点的角速度都相同，即使离圆心更远的点在直线速度上移动得更快。

2. Linear Speed and the Relationship v = rω | 线速度与 v = rω 的关系

For an object in circular motion, the linear speed v is the distance travelled along the circumference per unit time. In one period T, the object travels a distance equal to the circumference 2πr, so v = 2πr/T. Substituting ω = 2π/T, we obtain the fundamental relationship:

对于做圆周运动的物体，线速度 v 是单位时间内沿圆周运动的弧长。在一个周期 T 内，物体走过的距离等于圆周长 2πr，因此 v = 2πr/T。代入 ω = 2π/T，得到基本关系式：

v = rω

This equation shows that linear speed increases linearly with distance from the centre. A point on the rim of a wheel of radius 0.4 m rotating at ω = 5 rad s⁻¹ will have v = 0.4 × 5 = 2 m s⁻¹, while a point halfway to the rim (r = 0.2 m) travels at only 1 m s⁻¹. Even though angular velocity is shared by all radii, the linear speed depends on r.

该方程表明，线速度随离中心的距离线性增大。半径为 0.4 m、角速度为 ω = 5 rad s⁻¹ 的轮子，轮缘上一点的速度为 v = 0.4 × 5 = 2 m s⁻¹，而位于半径一半处（r = 0.2 m）的点速度仅为 1 m s⁻¹。尽管角速度对同一物体上所有半径都相同，但线速度取决于 r。

It is essential to understand that velocity is a vector, and even in uniform circular motion (constant speed), the velocity is not constant because its direction changes continuously. The change in velocity gives rise to acceleration, which we examine next.

必须理解速度是矢量，即使在匀速圆周运动中（速率恒定），速度也并非恒定，因为其方向在连续改变。速度的变化产生了加速度，我们接下来将对此进行分析。

3. Centripetal Acceleration | 向心加速度

Although the speed is constant in uniform circular motion, the direction of the velocity vector changes continuously, producing an acceleration directed towards the centre of the circle. This is called the centripetal acceleration, and its magnitude is given by:

虽然在匀速圆周运动中速率是恒定的，但速度矢量的方向在连续改变，从而产生了指向圆心的加速度，这称为向心加速度，其大小为：

a = v²/r = rω² = ωv

The direction of this acceleration is always towards the centre, perpendicular to the instantaneous velocity. This is why, for instance, a satellite in orbit is constantly “falling” towards Earth — without a gravitational pull, it would move in a straight line.

该加速度的方向始终指向圆心，与瞬时速度垂直。这就是为什么轨道上的卫星不断”落向”地球——如果没有引力，它将沿直线运动。

To see why the formula involves v², consider the small time interval Δt during which the object moves through a small angle Δθ. The velocity vector rotates by Δθ, and the change in velocity Δv points towards the centre. For small angles, Δv ≈ vΔθ, so a = Δv/Δt = vΔθ/Δt = vω = v(v/r) = v²/r. This derivation is often examined in A-level questions.

为了理解公式中为何出现 v²，考虑极短时间间隔 Δt 内物体转过小角度 Δθ 的情形。速度矢量旋转 Δθ，速度变化量 Δv 指向圆心。对于小角度，Δv ≈ vΔθ，因此 a = Δv/Δt = vΔθ/Δt = vω = v(v/r) = v²/r。这个推导过程常出现在 A-level 考题中。

4. Centripetal Force and Applications | 向心力及其应用

By Newton’s second law, a net force is required to produce the centripetal acceleration. This centripetal force always acts towards the centre of the circular path and has magnitude:

根据牛顿第二定律，产生向心加速度需要合力。这个向心力始终指向圆周路径的圆心，大小为：

F = mv²/r = mrω²

It is crucial to realise that centripetal force is not a new kind of force; it is the name given to whichever real force (or combination of forces) happens to be pulling the object towards the centre. For a satellite, it is gravity; for a ball on a string, it is tension; for a car rounding a bend, it is friction.

必须认识到，向心力并非一种新的力；它是对恰好将物体拉向圆心的那个真实力（或力的合力）的称呼。对于卫星，向心力是引力；对于绳系小球，向心力是张力；对于转弯的汽车，向心力是摩擦力。

Car on a banked road: When a car travels around a circular bend of radius r at speed v, friction between tyres and road supplies the centripetal force. On an unbanked road, the maximum safe speed is limited by μR, where R = mg is the normal reaction. Setting μmg = mv²/r gives v_max = √(μgr). If the road is banked at angle θ, the horizontal component of the normal reaction contributes mv²/r = N sin θ, allowing higher speeds without relying solely on friction.

倾斜路面上行驶的汽车：当汽车以速度 v 通过半径为 r 的弯道时，轮胎与路面之间的摩擦力提供向心力。在水平路面上，最大安全速度受 μR 限制，其中 R = mg 为正压力。令 μmg = mv²/r 可得 v_max = √(μgr)。如果路面倾斜角为 θ，正压力的水平分量 N sin θ 可贡献向心力 mv²/r，从而允许更高速度而不完全依赖摩擦力。

Vertical circles: At the top of a vertical loop of radius r, a car or roller-coaster requires the centripetal force F = mv²/r to point downwards. At the critical speed v = √(gr), the normal reaction becomes zero and the object is momentarily weightless. Below this speed, the object would fall off the track. At the bottom of the loop, the reaction force must overcome both the centripetal requirement and the weight, so N = mg + mv²/r.

竖直圆周：在半径为 r 的竖直圆环顶部，过山车所需的向心力 F = mv²/r 指向下方。在临界速度 v = √(gr) 时，正压力变为零，物体瞬时处于失重状态。若速度低于此值，物体将脱离轨道。在圆环底部，支持力必须同时克服向心力需求和重力，因此 N = mg + mv²/r。

5. Simple Harmonic Motion: The Defining Equation | 简谐运动：定义方程

Simple harmonic motion (SHM) is a special type of periodic oscillation in which the acceleration of an object is directly proportional to its displacement from a fixed equilibrium position and is always directed towards that equilibrium position. Mathematically:

简谐运动（SHM）是一种特殊的周期振动，其中物体的加速度与其相对固定平衡位置的位移成正比，且始终指向平衡位置。数学表达式为：

a = −ω²x

Here x is the displacement from equilibrium, ω is the angular frequency (units rad s⁻¹), and the negative sign indicates that acceleration opposes displacement. The minus sign is essential — it makes the motion oscillatory rather than runaway. The constant ω² is positive for all oscillating systems described by this equation.

其中 x 是相对平衡位置的位移，ω 是角频率（单位 rad s⁻¹），负号表示加速度与位移方向相反。负号至关重要——它使运动成为振荡而非发散。对于所有满足此方程的振动系统，常数 ω² 均为正值。

Two classic systems obey this equation exactly for small displacements. First, a mass m attached to a spring of force constant k: applying Hooke’s law F = −kx and Newton’s second law F = ma yields ma = −kx, or a = −(k/m)x, hence ω² = k/m. Second, a simple pendulum of length L: the restoring force is the component of weight along the arc, approximately −mg·x/L for small angles, giving a = −(g/L)x and ω² = g/L.

两个经典系统在小位移条件下严格满足此方程。第一，劲度系数为 k 的弹簧上悬挂质量为 m 的物体：应用胡克定律 F = −kx 和牛顿第二定律 F = ma，得到 ma = −kx，即 a = −(k/m)x，因此 ω² = k/m。第二，长度为 L 的单摆：回复力是重力沿弧线的分量，在小角度下近似为 −mg·x/L，得到 a = −(g/L)x，因此 ω² = g/L。

6. SHM Equations: Displacement, Velocity and Acceleration | 简谐运动方程：位移、速度与加速度

For an object starting from maximum displacement A (the amplitude) at t = 0, the solution to the SHM differential equation a = −ω²x is:

对于在 t = 0 时从最大位移 A（振幅）出发的物体，简谐运动微分方程 a = −ω²x 的解为：

x = A cos(ωt)

Differentiating with respect to time gives the velocity:

对时间求导得到速度：

v = −Aω sin(ωt)

Differentiating again gives the acceleration:

再次求导得到加速度：

a = −Aω² cos(ωt) = −ω²x

These equations reveal key features. The velocity is maximum (magnitude Aω) when passing through equilibrium (x = 0, so sin(ωt) = ±1) and zero at the extremes (x = ±A). The acceleration is zero at equilibrium and maximum (magnitude Aω²) at the turning points. There is a phase difference of π/2 between displacement and velocity, and π between displacement and acceleration.

这些方程揭示了关键特征。速度在通过平衡位置时最大（大小为 Aω），在两端极值处为零。加速度在平衡位置为零，在转折点处最大（大小为 Aω²）。位移与速度之间存在 π/2 的相位差，位移与加速度之间存在 π 的相位差。

Eliminating t between the displacement and velocity equations yields the important relationship between velocity and position, independent of time:

在位移方程和速度方程之间消去 t，可得到不依赖时间的重要关系式——速度与位置的关系：

v = ±ω√(A² − x²)

This is extremely useful for finding the speed at any given displacement. For example, if A = 0.05 m and ω = 10 rad s⁻¹, the speed at x = 0.03 m is v = 10 × √(0.05² − 0.03²) = 10 × 0.04 = 0.4 m s⁻¹, while the maximum speed at equilibrium is 0.5 m s⁻¹.

该式在已知位移时求速度极为有用。例如，若 A = 0.05 m、ω = 10 rad s⁻¹，则在 x = 0.03 m 处的速度为 v = 10 × √(0.05² − 0.03²) = 10 × 0.04 = 0.4 m s⁻¹，而在平衡位置的最大速度为 0.5 m s⁻¹。

7. Graphs of SHM: Displacement, Velocity, Acceleration | 简谐运动图像：位移、速度、加速度

Examining the three graphs side by side is essential for understanding phase relationships. The displacement–time graph is a cosine curve starting at +A. The velocity–time graph is a negative sine curve — it starts at zero, becomes increasingly negative, reaching −Aω at t = T/4, then returns to zero at t = T/2 while the displacement reaches −A. The acceleration–time graph is an inverted cosine curve, always in phase opposition (π out of phase) with displacement.

将三幅图并排观察对于理解相位关系至关重要。位移–时间图像是从 +A 开始的余弦曲线。速度–时间图像是负正弦曲线——从零开始逐渐变负，在 t = T/4 时达到 −Aω，然后在 t = T/2（位移到达 −A）时回到零。加速度–时间图像是倒置余弦曲线，始终与位移反相（相位差为 π）。

Quantity	At x = +A	At x = 0 (equilibrium)	At x = −A
Displacement x	Maximum: +A	Zero	Maximum: −A
Velocity v	Zero	Maximum: ±Aω	Zero
Acceleration a	Maximum: −Aω²	Zero	Maximum: +Aω²
Kinetic energy	Zero	Maximum: ½mω²A²	Zero
Potential energy	Maximum: ½mω²A²	Zero	Maximum: ½mω²A²

From the graphs, the period T of the oscillation can be read directly as the time between successive identical points. The amplitude A is the maximum displacement. These two parameters, together with the phase constant, completely determine the motion.

从图像中可以直接读出振荡周期 T（相邻相同状态点之间的时间间隔）和振幅 A（最大位移）。这两个参数连同相位常数完全决定了运动状态。

8. Energy in Simple Harmonic Motion | 简谐运动中的能量

In an ideal undamped oscillator, mechanical energy is conserved. The total energy is the sum of kinetic energy (KE) and potential energy (PE). For a mass–spring system, the potential energy is elastic energy given by ½kx², and the kinetic energy is ½mv². Substituting v² = ω²(A² − x²) and k = mω², we find:

在理想无阻尼振荡器中，机械能守恒。总能量等于动能（KE）与势能（PE）之和。对于弹簧–质量系统，势能为弹性势能 ½kx²，动能为 ½mv²。代入 v² = ω²(A² − x²) 和 k = mω²，可得：

KE = ½mω²(A² − x²), PE = ½mω²x²

Total energy E = KE + PE = ½mω²A². This total is constant and proportional to the square of the amplitude. Doubling the amplitude quadruples the total energy. At equilibrium (x = 0), all energy is kinetic: E = ½mω²A². At the extremes (x = ±A), all energy is potential with kinetic energy zero.

总能量 E = KE + PE = ½mω²A²。该总量恒定且与振幅的平方成正比。振幅加倍，总能量变为四倍。在平衡位置（x = 0）时，全部能量为动能：E = ½mω²A²。在两端（x = ±A）时，全部能量为势能，动能为零。

Energy exchanges are beautifully illustrated by a pendulum. At the highest point of its swing, the bob has maximum gravitational potential energy and zero kinetic energy. As it descends towards equilibrium, PE converts into KE. At the lowest point, speed is maximum and PE is at its minimum. Without friction, this cycle repeats indefinitely, but in practice air resistance gradually dissipates energy — the process we now examine.

能量转化在单摆中得到很好的展示。摆锤在摆动最高点时具有最大重力势能和零动能。当它向平衡位置下落时，势能转化为动能。在最低点，速度最大而势能最小。若无摩擦，这一循环将无限重复；但实际中空气阻力会逐渐耗散能量——这正是我们接下来要讨论的过程。

9. Damping and Resonance | 阻尼与共振

Real oscillating systems lose mechanical energy through friction, air resistance or internal dissipation. This is known as damping. There are three regimes. Light damping: the amplitude decays exponentially with time but the system still oscillates, and the period remains approximately constant. Critical damping: the system returns to equilibrium in the shortest possible time without oscillating — this is the ideal setting for car shock absorbers and door closers. Heavy damping: the system returns to equilibrium very slowly, taking longer than in the critical case, again without oscillating.

真实振荡系统通过摩擦、空气阻力或内耗损失机械能，这称为阻尼。存在三种状态。欠阻尼（轻阻尼）：振幅随时间按指数衰减，但系统仍会振荡，周期近似保持不变。临界阻尼：系统在最短时间内回到平衡位置且不发生振荡——这是汽车减震器和闭门器的理想设置。过阻尼：系统回到平衡位置非常缓慢，比临界情况耗时更长，同样不发生振荡。

Resonance occurs when a periodic driving force is applied to an oscillator at a frequency equal to its natural frequency. The driving force continuously does positive work on the system, causing the amplitude to grow. Without damping, the amplitude would increase without limit. With light damping, the amplitude becomes large but finite; the sharpness of the resonance peak decreases as damping increases. Resonance phenomena include the shattering of a wine glass by a singer’s voice, the violent swaying of the Millennium Bridge in London (fixed by adding dampers), and the tuned circuits in radio receivers.

当周期性驱动力以与系统固有频率相等的频率施加于振荡器时，就会发生共振。驱动力持续对系统做正功，导致振幅增大。若无阻尼，振幅将无限增大。若有轻阻尼，振幅会变得很大但有限；共振峰的尖锐程度随阻尼增大而减小。共振现象包括歌手歌声震碎酒杯、伦敦千禧桥的剧烈摆动（通过加装阻尼器解决）、以及收音机中的调谐电路等。

In the context of resonance, we compare the natural frequency f₀ of an undamped oscillator to the driving frequency f_d. When f_d = f₀, energy transfer is maximised. In AQA questions, this often appears as a graph of amplitude versus driving frequency for different damping levels: the peak occurs at f₀ (slightly lower for heavier damping in some systems), and the width of the peak increases with damping.

在共振的讨论中，我们将无阻尼振荡器的固有频率 f₀ 与驱动频率 f_d 进行比较。当 f_d = f₀ 时，能量传递最大化。在 AQA 考题中，这常以不同阻尼水平下振幅随驱动频率变化的图像出现：峰值出现在 f₀ 处（某些系统中重阻尼下略低），且峰宽随阻尼增大而增加。

10. Applications: Pendulum and Mass–Spring System | 应用：单摆与弹簧振子

The period of a simple pendulum is T = 2π√(L/g), where L is the length from the pivot to the centre of mass of the bob. Notably, the period is independent of the mass of the bob and independent of amplitude for small angles (less than about 10°). This property makes pendulums ideal for timekeeping — a grandfather clock’s pendulum has a fixed period determined solely by its length.

单摆的周期为 T = 2π√(L/g)，其中 L 是从支点到摆锤重心的长度。值得注意的是，周期与摆锤质量无关，且在小角度（小于约 10°）下与振幅无关。这一特性使单摆成为理想的计时工具——落地钟的摆钟周期仅由其摆长决定。

The period of a mass–spring system is T = 2π√(m/k), independent of the amplitude and of g. This means the same mass–spring oscillator behaves identically on Earth, on the Moon or in an accelerating lift. Mass–spring systems are used in wristwatch balance wheels, vehicle suspensions, and as timing elements in precision instruments.

弹簧振子的周期为 T = 2π√(m/k)，与振幅和重力加速度 g 均无关。这意味着同一个弹簧振子在地球、月球或加速电梯中的行为完全相同。弹簧振子系统被用于手表摆轮、车辆悬架以及精密仪器中的计时元件。

A particularly important relationship connects SHM and circular motion: the projection of uniform circular motion onto any straight line is SHM. Imagine a particle orbiting a circle of radius A at angular velocity ω. Its projection onto the x-axis has displacement x = A cos(ωt), velocity v = −Aω sin(ωt), and acceleration a = −ω²x — exactly the defining equations of SHM. Thus a rotating vector (phasor) can represent an oscillator, a technique widely used in physics and engineering.

一个特别重要的联系将 SHM 与圆周运动连接起来：匀速圆周运动在任意直线上的投影就是简谐运动。设想一个质点以角速度 ω 沿半径为 A 的圆运动，它在 x 轴上的投影位移为 x = A cos(ωt)，速度为 v = −Aω sin(ωt)，加速度为 a = −ω²x——与 SHM 的定义方程完全一致。因此，旋转矢量（相量）可以表示一个振荡器，这一方法在物理学和工程学中广泛应用。

In an A-level experiment to determine g, a student measures the period T of a pendulum for several lengths L. Plotting T² against L yields a straight line through the origin with gradient 4π²/g. From the gradient, g can be calculated. Systematic errors such as timing uncertainty in starting and stopping the stopwatch are reduced by timing 20 oscillations rather than one, thereby dividing the timing error by 20.

在测定重力加速度 g 的 A-level 实验中，学生测量不同摆长 L 下单摆的周期 T。以 T² 对 L 作图，得到过原点的直线，斜率为 4π²/g。由斜率即可计算 g。为减小系统误差（如秒表启动和停止的时间不确定性），计时 20 次全振动而非 1 次，从而将计时误差缩小至原来的 1/20。

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Circular and Periodic Motion | 圆周运动与周期运动