📚 Edexcel A-Level Physics Topic 13: Oscillations | 爱德思A-Level物理第13章：振荡
Oscillations are a fundamental part of Edexcel A-Level Physics, bringing together mechanics, energy and wave concepts. In Topic 13 you will explore simple harmonic motion, damping and resonance in depth – and you will need to handle displacement, velocity, acceleration and energy equations confidently.
振荡是爱德思A-Level物理的核心内容之一，它将力学、能量与波动概念融为一体。在第13章中，你将深入学习简谐运动、阻尼和共振，并需要熟练掌握位移、速度、加速度与能量方程。
1. Simple Harmonic Motion (SHM) Basics | 简谐运动基础
SHM occurs when an object moves back and forth about an equilibrium position and its acceleration is always directed towards that position, proportional to the displacement. The defining equation is a &proportional; − x. For true SHM the constant of proportionality is the square of the angular frequency, so we write a = − ω² x.
当物体围绕平衡位置往复运动，且加速度始终指向平衡位置并与位移成正比时，物体做简谐运动。其定义方程为a ∝ − x。对于真正的简谐运动，比例常数是角频率的平方，因此我们写为a = − ω² x。
Angular frequency ω is related to the time period T by ω = 2π / T and to frequency f by ω = 2π f. The SI unit of ω is rad s⁻¹.
角频率 ω 与周期 T 的关系为ω = 2π / T，与频率 f 的关系为ω = 2π f。ω 的SI单位是 rad s⁻¹。
2. Displacement Equation in SHM | 简谐运动的位移方程
Displacement x varies sinusoidally with time. If timing starts when the oscillator is at maximum displacement, we use x = A cos(ω t). If timing starts when it passes equilibrium moving in the positive direction, we use x = A sin(ω t). A is the amplitude – the maximum displacement from equilibrium.
位移 x 随时间呈正弦变化。若从最大位移处开始计时，则用x = A cos(ω t)；若从经过平衡位置向正方向运动时开始计时，则用x = A sin(ω t)。A 为振幅，即离开平衡位置的最大位移。
In exam questions you must identify the phase constant. The general form is x = A cos(ω t + φ) or x = A sin(ω t + φ), where φ is the phase constant determined by initial conditions.
考试中你必须识别相位常数。位移的一般形式为x = A cos(ω t + φ)或x = A sin(ω t + φ)，其中 φ 是由初始条件决定的相位常数。
3. Velocity in SHM | 简谐运动的速度
elocity is the rate of change of displacement. By differentiating x = A sin(ω t) we obtain v = ω A cos(ω t). The velocity can also be expressed in terms of displacement as v = ± ω √A² − x². This shows that speed is maximum when x = 0 and zero when x = ± A.
速度是位移的变化率。对 x = A sin(ω t) 求导可得v = ω A cos(ω t)。速度也可用位移表示为v = ± ω √A² − x²。这表明当 x = 0 时速率最大，当 x = ± A 时速度为零。
The maximum speed is vₘₐₓ = ω A dicates direction: positive when moving left to right, negative when moving right to left.
最大速率为vₘₐₓ = ω A，出现在振子通过平衡位置时。注意 ± 号表示方向：从左向右运动时为正，反之为负。
4. Acceleration in SHM | 简谐运动的加速度
Acceleration is the rate of change of velocity. Differentiating velocity gives a = − ω² A sin(ω t) = − ω² x. This is the hallmark of SHM: acceleration is directly proportional to displacement and always directed towards equilibrium.
加速度是速度的变化率。对速度求导可得a = − ω² A sin(ω t) = − ω² x。这正是简谐运动的标志：加速度与位移成正比，且始终指向平衡位置。
Maximum acceleration occurs at the extremes of motion, where aₘₐₓ = ω² A. At the equilibrium position acceleration is zero, yet speed is greatest – a counter-intuitive but crucial point.
最大加速度出现在运动端点，其值为aₘₐₓ = ω² A。在平衡位置加速度为零，而速率却最大——这一点有违直觉但十分关键。
5. Time Period of a Mass-Spring System | 弹簧振子的周期
k, the restoring force is F = − k x. Using F = m a and a = − ω² x, we find ω² = k/m, hence T = 2π / ω = 2π √m/k. The period depends only on mass and spring constant, not on amplitude.
对于连接在劲度系数为 k 的弹簧上的质量 m，回复力为 F = − k x。利用 F = m a 和 a = − ω² x，得到 ω² = k/m，因此T = 2π / ω = 2π √m/k。周期只取决于质量与弹簧常数，与振幅无关。
This isochronous property (period independent of amplitude) holds for all true SHM systems and is essential for accurate timekeeping devices.
这种等时性（周期与振幅无关）对于所有真正的简谐运动系统都成立，对于精密计时装置至关重要。
6. Time Period of a Simple Pendulum | 单摆的周期
e SHM. The restoring force component is mg sinθ, and for small angles sinθ ≈ θ, leading to ω² = g/L and T = 2π √L/g. Again the period is independent of amplitude provided the angle is small (typically less than about 10°).
长度为 L 的单摆在小角度下做近似的简谐运动。回复力分量为 mg sinθ，小角度下 sinθ ≈ θ，故ω² = g/L，T = 2π √L/g。只要角度足够小（通常小于约10°），周期同样与振幅无关。
Note that the period does not depend on the mass of the pendulum bob. This is a consequence of the equivalence of gravitational and inertial mass.
注意周期与摆球质量无关，这是引力质量与惯性质量等效的结果。
7. Energy Transformations in SHM | 简谐运动中的能量转化
In an isolated SHM system total mechanical energy is conserved. The kinetic energy is K = ½ m v² = ½ m ω² (A² − x²). The potential energy (for a mass-spring system) is U = ½ k x² = ½ m ω² x². The total energy is constant: Etotal = K + U = ½ k A² = ½ m ω² A².
在孤立的简谐运动系统中，总机械能守恒。动能为K = ½ m v² = ½ m ω² (A² − x²)。势能（对于弹簧振子）为U = ½ k x² = ½ m ω² x²。总能量恒定：Etotal = K + U = ½ k A² = ½ m ω² A²。
Energy continuously interchanges between kinetic and potential forms. At maximum displacement energy is all potential; at equilibrium it is all kinetic. Graphs of K and U against displacement or time are symmetrical parabolas or sinusoids.
能量在动能和势能之间不断转化。在最大位移处能量全部为势能；在平衡位置全部为动能。动能、势能随位移或时间的变化图像为对称的抛物线或正弦曲线。
8. Damping: Types and Effects | 阻尼：类型与效应
Real oscillators experience resistive forces that remove energy from the system, causing damping. Light damping leads to a gradual decrease in amplitude but does not change the frequency significantly. Critical damping returns the system to equilibrium in the shortest time without oscillation. Heavy damping produces a slow, non-oscillatory return.
实际的振子会受到阻力作用，使系统能量损失，产生阻尼。轻阻尼使振幅逐渐减小，但对频率影响不大。临界阻尼能使系统在不发生振荡的情况下以最短时间回到平衡位置。重阻尼则产生缓慢的、非振荡性的归位运动。
The damping force is often modelled as proportional to velocity, Fdamp = − b v, where b is the damping constant. The amplitude decays exponentially: A(t) = A₀ e⁻⁽ᵇᐟ²ᵐ⁾ᵗ, provided damping is light.
阻尼力通常模型化为与速度成正比，即 Fdamp = − b v，b 为阻尼常数。在轻阻尼下振幅按指数衰减：A(t) = A₀ e⁻⁽ᵇᐟ²ᵐ⁾ᵗ。
Critical damping occurs when b² = 4 m k. Systems such as car suspension and galvanometers employ near-critical damping to avoid oscillation while ensuring fast response.
当b² = 4 m k时发生临界阻尼。汽车悬架和检流计等系统中采用接近临界阻尼的设计，以在避免振荡的同时保证快速响应。
9. Resonance and Natural Frequency | 共振与固有频率
Every oscillating system has a natural frequency f₀ at which it vibrates freely. When a periodic external force of frequency f is applied, the system undergoes forced oscillations. As f approaches f₀ the amplitude rises sharply – this is resonance.
每个振荡系统都有一个固有频率 f₀，在此频率下系统可自由振动。当施加一个频率为 f 的周期性外力时，系统做受迫振动。当 f 接近 f₀ 时，振幅急剧增大，这就是共振。
The resonance curve (amplitude vs driving frequency) peaks at f₀. The sharpness of the peak depends on damping: the lighter the damping, the taller and narrower the resonance peak. With heavier damping the peak is suppressed and broadened.
共振曲线（振幅–驱动频率图）在 f₀ 处达到峰值。峰形的尖锐程度取决于阻尼：阻尼越轻，共振峰越高越窄；阻尼越重，共振峰越低越宽。
Familiar examples of resonance include a singer shattering a wine glass, the swaying of the Millennium Bridge, and the correct tuning of radio receivers using an LCR circuit. In mechanical systems resonance can be catastrophic, as dramatically demonstrated by the Tacoma Narrows Bridge collapse.
常见的共振例子包括歌手震碎酒杯、千禧桥的晃动以及通过LCR电路调谐无线电接收器。在机械系统中共振可能是灾难性的，塔科马海峡大桥的坍塌就是一个震撼的例子。
10. Real-world Applications and Exam Tips | 实际应用与考试技巧
In Edexcel A-Level exams you will often be asked to describe an experimental method for measuring the period of a pendulum or mass-spring system, analyse datalogger traces of damped motion, or interpret resonance curves. Frequently you must link gradient calculations of a T² vs L graph to the value of g.
在爱德思A-Level考试中，经常会要求描述测量单摆或弹簧振子周期的实验方法、分析阻尼运动的数据记录器迹线，或解读共振曲线。经常需要将 T²–L 图的斜率计算与 g 值关联起来。
Key equations to memorise:
需要记忆的关键公式：
a = − ω² x
x = A cos(ω t) or x = A sin(ω t)
v = ± ω √A² − x² ; vₘₐₓ = ω A
T = 2π √m/k ; T = 2π √L/g
Etotal = ½ m ω² A² ; K = ½ m ω² (A² − x²)
Be careful with phase constants: read the question to decide whether cosine or sine fits the given initial conditions. When plotting graphs, label axes exactly: displacement–time, velocity–time, acceleration–time, or energy–displacement. Always include units on numeric values.
注意相位常数：审题时要判断余弦或正弦是否契合所给的初始条件。作图时准确标注坐标轴：位移–时间图、速度–时间图、加速度–时间图、或能量–位移图。数值结果必须带单位。
Lastly, practice deriving energy equations and the resonance explanation from first principles – examiners value clear, logical steps rather than rote recall.
最后，要多练习从基本原理出发推导能量方程和共振解释——考官看重的是清晰、逻辑性的步骤，而不是死记硬背。
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