A-Level物理简谐运动能量转换阻尼振动

What is Simple Harmonic Motion？什么是简谐运动？

Simple harmonic motion (SHM) is a type of periodic oscillation where the restoring force is directly proportional to the displacement from equilibrium and acts in the opposite direction. 简谐运动是一种周期性振动，其回复力与偏离平衡位置的位移成正比，且方向始终指向平衡位置。The defining condition is F = -kx, where k is the spring constant and x is the displacement. 定义条件是 F = -kx，其中 k 为劲度系数，x 为位移。The negative sign indicates that the force always opposes the displacement, driving the system back towards equilibrium. 负号表示力始终与位移方向相反，将系统推向平衡位置。

SHM is the foundation for understanding many physical systems: pendulums, mass-spring systems, vibrating molecules, and even alternating current circuits. 简谐运动是理解多种物理系统的基础：单摆、弹簧振子、分子振动，乃至交流电路。The motion traces a sinusoidal waveform when plotted against time. 位移随时间的变化表现为正弦或余弦波形。Two conditions must be satisfied for SHM: the acceleration must be proportional to displacement, and the acceleration must always be directed towards the equilibrium point. 简谐运动必须满足两个条件：加速度与位移成正比，且加速度始终指向平衡位置。

Mathematical Description of SHM 简谐运动的数学描述

The displacement x of an object undergoing SHM is given by x = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase constant. 简谐运动的位移方程为 x = A cos(ωt + φ)，其中 A 为振幅，ω 为角频率，φ 为初相。Amplitude A represents the maximum displacement from equilibrium and is always a positive scalar quantity measured in metres. 振幅 A 表示离开平衡位置的最大距离，始终为正标量，单位为米。

The angular frequency ω is related to the period T and frequency f by ω = 2πf = 2π/T. 角频率 ω 与周期 T 和频率 f 的关系为 ω = 2πf = 2π/T。For a mass-spring system, ω = sqrt(k/m), meaning the oscillation frequency depends only on the physical properties of the system and not on the amplitude. 对于弹簧振子，ω = sqrt(k/m)，振动频率仅取决于系统本身的物理性质，与振幅无关，这体现了等时性。For a simple pendulum with small angular displacement (θ < 10°), ω = sqrt(g/l) where g is the gravitational field strength and l is the pendulum length. 对于小角度单摆（θ < 10°），ω = sqrt(g/l)，其中 g 为重力场强度，l 为摆长。

Velocity and acceleration are obtained by differentiating the displacement equation. 速度和加速度通过对位移方程求导得到。Velocity v = -Aω sin(ωt + φ) and the maximum speed is v_max = Aω, which occurs at the equilibrium position where x = 0. 加速度 a = -Aω^2 cos(ωt + φ) = -ω^2 x, confirming that acceleration is proportional to displacement but in the opposite direction. 加速度 a = -ω^2 x，验证了加速度与位移成正比但方向相反的关系。The maximum acceleration a_max = Aω^2 occurs at the extreme positions where x = ±A. 最大加速度 a_max = Aω^2 出现在 x = ±A 的极端位置。

The phase relationships between displacement, velocity, and acceleration are frequently examined. 位移、速度和加速度之间的相位关系是考试常见考点。Velocity leads displacement by π/2 radians (90°), meaning velocity reaches its maximum one quarter of a cycle before displacement reaches its maximum. 速度比位移超前 π/2 弧度（90°），即速度比位移提前四分之一周期达到最大值。Acceleration is π radians (180°) out of phase with displacement, meaning when displacement is maximum positive, acceleration is maximum negative. 加速度与位移相位差为 π 弧度（180°），即位移达到正最大值时，加速度达到负最大值。

Worked Calculation Example 计算示例

Consider a mass of 0.50 kg attached to a spring with spring constant k = 200 N/m, pulled 4.0 cm from equilibrium and released. 考虑一个 0.50 kg 的物体连接在劲度系数 k = 200 N/m 的弹簧上，将其拉离平衡位置 4.0 cm 后释放。Calculate the angular frequency: ω = sqrt(k/m) = sqrt(200/0.50) = sqrt(400) = 20 rad/s. 计算角频率：ω = sqrt(k/m) = sqrt(200/0.50) = sqrt(400) = 20 rad/s。The period T = 2π/ω = 2π/20 = 0.314 s. 周期 T = 2π/ω = 2π/20 = 0.314 s。The maximum speed v_max = Aω = 0.040 × 20 = 0.80 m/s. 最大速度 v_max = Aω = 0.040 × 20 = 0.80 m/s。The maximum acceleration a_max = Aω^2 = 0.040 × 400 = 16 m/s^2. 最大加速度 a_max = Aω^2 = 0.040 × 400 = 16 m/s^2。

Pendulum Worked Example 单摆计算示例

A simple pendulum of length 1.00 m is displaced by a small angle and released. 一个长度为 1.00 m 的单摆被拉开一个小角度后释放。Calculate the period: T = 2π sqrt(l/g) = 2π sqrt(1.00/9.81) = 2.01 s. 计算周期：T = 2π sqrt(l/g) = 2π sqrt(1.00/9.81) = 2.01 s。If the pendulum is taken to the Moon where g = 1.63 m/s^2, the period becomes T = 2π sqrt(1.00/1.63) = 4.92 s. 若将该单摆带到月球表面（g = 1.63 m/s^2），周期变为 T = 4.92 s。This demonstrates that the period depends on the local gravitational field strength, not on the mass of the bob. 这表明周期取决于当地重力场强度，而非摆锤的质量。

Energy Transformations in SHM 简谐运动中的能量转换

One of the most important features of SHM is the continuous interchange between kinetic and potential energy. 简谐运动最重要的特征之一就是动能与势能之间的持续转换。At the equilibrium position, velocity is maximum and kinetic energy is at its peak, while potential energy is zero. 在平衡位置，速度最大，动能达到峰值，势能为零。At the extreme positions (x = ±A), velocity is zero and all energy is stored as elastic potential energy. 在最大位移处（x = ±A），速度为零，所有能量以弹性势能形式储存。

The total mechanical energy remains constant in ideal undamped SHM: E_total = (1/2)kA^2. 理想无阻尼简谐运动中，总机械能守恒：E_total = (1/2)kA^2。This constant value depends only on the spring constant and the amplitude squared, and is proportional to the square of the amplitude. 该常数值仅取决于劲度系数和振幅的平方，与振幅的平方成正比。The kinetic energy at any displacement x is E_k = (1/2)k(A^2 – x^2) and the elastic potential energy is E_p = (1/2)kx^2. 任意位置 x 处的动能 E_k = (1/2)k(A^2 – x^2)，弹性势能 E_p = (1/2)kx^2。

Energy-time graphs for SHM show the kinetic and potential energies oscillating at twice the frequency of the displacement. 简谐运动的能量-时间图显示，动能和势能的变化频率是位移频率的两倍。This is because energy depends on velocity squared and displacement squared, which both complete two cycles per oscillation period. 这是因为能量取决于速度平方和位移平方，两者在每个振动周期内完成两个完整循环。Using the earlier worked example: E_total = (1/2) × 200 × 0.040^2 = 0.16 J. 使用前面的计算示例：E_total = (1/2) × 200 × 0.040^2 = 0.16 J。

Damped Harmonic Motion 阻尼振动

In real systems, oscillations gradually decrease in amplitude due to resistive forces such as friction or air resistance. 实际系统中，由于摩擦力或空气阻力等耗散力的存在，振幅会逐渐减小。This is called damped harmonic motion. 这称为阻尼振动。The damping force is typically proportional to velocity: F_damp = -bv, where b is the damping coefficient. 阻尼力通常与速度成正比：F_damp = -bv，其中 b 为阻尼系数。

There are three types of damping: light damping (underdamped), critical damping, and heavy damping (overdamped). 阻尼分为三种类型：欠阻尼、临界阻尼和过阻尼。In light damping, the system oscillates with a gradually decreasing amplitude, and the frequency of oscillation is slightly less than the natural frequency. 欠阻尼时，系统以逐渐减小的振幅持续振动，振动频率略低于固有频率。Critical damping brings the system to equilibrium in the shortest possible time without oscillation. 临界阻尼使系统在最短时间内回到平衡位置且不产生振动。This is essential in applications like car suspension systems, door closers, and seismometers. 这在汽车悬挂系统、闭门器和地震仪等应用中至关重要。

Heavy damping returns the system to equilibrium slowly without oscillation, taking longer than critical damping. 过阻尼使系统缓慢回到平衡位置，同样不产生振动，但耗时比临界阻尼更长。The damping ratio ζ determines which regime applies: ζ < 1 (underdamped), ζ = 1 (critically damped), ζ > 1 (overdamped). 阻尼比 ζ 决定所处的阻尼状态：ζ < 1 为欠阻尼，ζ = 1 为临界阻尼，ζ > 1 为过阻尼。

Forced Oscillations and Resonance 受迫振动与共振

When a periodic external force is applied to a damped oscillator, the system undergoes forced oscillations. 当对阻尼振子施加周期性外力时，系统进行受迫振动。Initially, the system vibrates at both its natural frequency and the driving frequency, but the natural frequency component dies away due to damping, leaving steady-state oscillation at the driving frequency. 初始时系统同时以固有频率和驱动频率振动，但固有频率分量因阻尼而衰减，最终以驱动频率进行稳态振动。

Resonance occurs when the driving frequency equals the natural frequency of the system. 当驱动频率等于系统固有频率时，发生共振。At resonance, the amplitude reaches its maximum value, and energy transfer from the driver to the oscillator is most efficient. 共振时振幅达到最大值，能量从驱动源到振子的传递效率最高。In systems with light damping, the resonance peak is tall and sharp; with heavy damping, it is broad and low. 在欠阻尼系统中，共振峰高而尖锐；在过阻尼系统中，共振峰宽而低。The phase difference between the driver and oscillator is π/2 at resonance. 共振时驱动源与振子之间的相位差为 π/2。

Resonance has critical real-world significance. 共振在现实世界中具有重要意义。The collapse of the Tacoma Narrows Bridge in 1940 was caused by wind-induced resonance. 1940年塔科马海峡大桥的坍塌就是由风致共振引起的。Soldiers break step when marching across bridges to avoid resonant frequencies. 士兵过桥时停止齐步走以避免引发共振频率。Microwave ovens use resonance at 2.45 GHz to excite water molecules and heat food. 微波炉利用 2.45 GHz 的共振频率激发水分子来加热食物。However, resonance is also useful: MRI machines use nuclear magnetic resonance, and quartz crystal oscillators in watches rely on mechanical resonance for precise timekeeping. 但共振也有正面应用：核磁共振成像利用核磁共振原理，手表中的石英晶体振荡器依赖机械共振实现精确计时。Musical instruments also depend on resonance: the body of a violin amplifies sound through resonant vibration of the wooden cavity. 乐器同样依赖共振：小提琴的琴身通过木质腔体的共振放大声音。

Experimental Determination of g 重力加速度的实验测定

A classic A-Level practical investigation uses a simple pendulum to determine the acceleration due to gravity g. 经典的 A-Level 实验探究使用单摆测定重力加速度 g。The procedure involves measuring the period T for several different pendulum lengths l, ensuring the angular amplitude is kept below 10° for the small-angle approximation to hold. 实验步骤包括测量多种不同摆长 l 下的周期 T，并确保角度振幅保持在 10° 以下以使小角度近似成立。By plotting T^2 against l, the gradient of the best-fit line equals 4π^2/g, from which g can be calculated. 通过绘制 T^2 对 l 的图像，最佳拟合线的斜率等于 4π^2/g，由此可计算出 g。Systematic errors include misalignment of the protractor when measuring angles and timing too few oscillations. 系统误差包括测量角度时量角器未对准以及计时周期数太少。

Exam Tips for A-Level Physics SHM 考试技巧

Students should memorize the four key equations: x = A cos(ωt), v = ±ω sqrt(A^2 – x^2), a = -ω^2 x, and T = 2π sqrt(m/k) for a mass-spring system. 学生应熟记四个关键方程。For the simple pendulum, use T = 2π sqrt(l/g) instead, but remember this only applies for small angles where sin θ ≈ θ. 对于单摆，使用 T = 2π sqrt(l/g)，但需注意该公式仅适用于小角度近似 sin θ ≈ θ。

Pay careful attention to the phase differences between displacement, velocity, and acceleration, as these are tested in both multiple-choice and structured questions. 仔细关注位移、速度和加速度之间的相位差，这在选择题和结构化问题中均为常考点。When sketching graphs, always label axes clearly and show correct phase relationships. 绘制图像时，务必清晰标记坐标轴并正确表示相位关系。Energy graphs should show E_k and E_p as parabolas summing to a constant total energy line. 能量图应展示动能和势能均为抛物线，两者之和为恒定总能量线。

In experimental questions, you may be asked to describe how to determine g using a simple pendulum: measure the period T for different lengths l, plot T^2 against l, and find g from the gradient since T^2 = (4π^2/g)l. 实验题可能要求描述如何使用单摆测定重力加速度 g：测量不同摆长 l 对应的周期 T，绘制 T^2 对 l 的图像，通过斜率求 g，因为 T^2 = (4π^2/g)l。

Key Bilingual Terms 核心双语术语

A-Level物理 简谐运动 能量转换 阻尼振动