ALevel物理 简谐运动SHM 位移 加速度 能量
简谐运动 Simple Harmonic Motion（SHM）是 A-Level 物理力学部分最重要、也最常考的周期性运动模型。无论是 AQA、Edexcel、OCR 还是 CAIE 剑桥考局，SHM 都以大题形式出现在试卷中，分值通常在 8 到 15 分之间，而且常常与能量守恒、图形分析以及实验设计题结合考查。很多同学觉得 SHM 难，是因为它同时涉及微积分思想、三角函数图像和能量图像三类工具。本文用中英双语把 SHM 的定义、方程、能量转化和典型计算完整拆解一遍，帮助你一次理清全部核心考点。
Simple Harmonic Motion is one of the most heavily examined topics in A-Level Physics. Whether you sit AQA, Edexcel, OCR or Cambridge CAIE, SHM usually appears as a full structured question worth 8 to 15 marks, often combined with energy conservation, graph analysis and practical design work. Students find it hard because it mixes calculus-style reasoning, trigonometric graphs and energy curves in one topic. This guide explains the definition, key equations, energy changes and a fully worked calculation, so you can master every core idea in one pass.
1. SHM定义 加速度与位移成正比且反向
要判断一个运动是不是简谐运动，考试要求你使用严格的定义：物体的加速度 a 与它离开平衡位置的位移 x 成正比，并且方向始终指向平衡位置。写成公式就是 a = -ω²x，其中 ω 称为角频率，单位是 rad/s。负号是 SHM 的灵魂，它表示加速度总是与位移方向相反：物体在平衡位置右侧时，加速度指向左侧；物体在左侧时，加速度又指向右侧。正因为加速度永远指向平衡位置，物体才会在平衡位置附近来回振动而不是飞走。考试中判断物体是否做 SHM，只需检查 a 与 x 是否满足线性关系 a = -ω²x。
The formal definition of SHM is what examiners ask you to state and use: the acceleration a of the object is proportional to its displacement x from the equilibrium position, and is always directed towards that equilibrium position. In symbols, a = -ω²x, where ω is the angular frequency in rad/s. The minus sign is the heart of SHM: it shows that acceleration always opposes displacement. When the object sits to the right of equilibrium, acceleration points left; when it is on the left, acceleration points right. Because acceleration constantly pulls the object back towards equilibrium, the object oscillates about that point instead of flying away. To test whether any motion is SHM, check whether a and x satisfy the linear relation a = -ω²x.
2. 位移时间方程 余弦函数与相位
求解微分方程 a = -ω²x 可以得到位移随时间的变化规律。若从最大位移处开始计时，位移方程为 x = A cos(ωt)；若从平衡位置开始计时，则为 x = A sin(ωt)。这里的 A 是振幅，单位是 m，表示离开平衡位置的最大距离。角频率 ω 与周期 T、频率 f 的关系是 ω = 2π/T = 2πf。周期 T 是完成一次完整振动所需的时间，单位是秒；频率 f 是每秒完成的振动次数，单位是 Hz。注意 ωt 的单位是弧度，所以在计算时要先把时间 t 换算成与 ω 匹配的单位，例如 ω = π rad/s 时，t = 0.5 s 对应相位 ωt = π/2。
Solving the differential equation a = -ω²x gives displacement as a function of time. If timing starts at maximum displacement, x = A cos(ωt); if it starts at the equilibrium position, x = A sin(ωt). Here A is the amplitude in metres, the maximum distance from equilibrium. The angular frequency ω links to the period T and frequency f through ω = 2π/T = 2πf. The period is the time for one complete oscillation, measured in seconds, while the frequency counts oscillations per second in hertz. Remember that ωt is measured in radians, so always match the time unit to ω; for example with ω = π rad/s, t = 0.5 s gives the phase ωt = π/2.
3. 速度与加速度图像 极值位置辨析
对位移方程求导可得速度 v = -Aω sin(ωt)，再求导得加速度 a = -Aω² cos(ωt) = -ω²x。更常用的等价形式是 v = ±ω√(A² - x²)：这个公式直接告诉你速度与位置的关系。当物体经过平衡位置 x = 0 时速度最大，v_max = Aω；当物体到达振幅端点 x = ±A 时速度为零，因为此时物体刚好反向。加速度则完全相反：在平衡位置时加速度为零，在端点处加速度最大，a_max = Aω²。许多选择题故意把速度和加速度的极值位置互换来迷惑考生，记住口诀：平衡位置速度最大加速度为零，端点位置速度为零加速度最大。
Differentiating the displacement equation gives velocity v = -Aω sin(ωt), and differentiating again gives acceleration a = -Aω² cos(ωt) = -ω²x. An equivalent and very useful form is v = ±ω√(A² - x²), which links speed directly to position. At the equilibrium position x = 0 the speed is greatest, v_max = Aω; at the amplitude extremes x = ±A the speed falls to zero because the object is about to reverse. Acceleration behaves in the opposite way: it is zero at equilibrium and reaches its maximum Aω² at the extremes. Multiple choice questions often swap these extreme positions to trap you. Remember the rule: maximum speed and zero acceleration at equilibrium, zero speed and maximum acceleration at the ends.
4. 能量转化 动能与弹性势能守恒
SHM 系统在做无阻尼振动时机械能守恒。以水平弹簧振子为例，系统总能量 E = ½mω²A²，它由动能和弹性势能组成。在平衡位置，速度最大，动能达到最大值 ½mω²A²，弹性势能为零；在端点处，物体瞬间静止，动能为零，全部能量转化为弹性势能 ½mω²A²。在任意位置 x，动能表达式为 ½mω²(A² - x²)，势能表达式为 ½mω²x²。两个能量图像都是抛物线形状，动能图像开口向下，势能图像开口向上，两者之和恒为水平直线。考试常让你比较不同位置的动能势能大小，或者根据能量图像判断振幅与总能量。
For undamped SHM the total mechanical energy is conserved. Taking a horizontal mass-spring system, the total energy is E = ½mω²A², shared between kinetic energy and elastic potential energy. At equilibrium the speed is greatest, so kinetic energy reaches its maximum ½mω²A² and potential energy is zero. At the extremes the object stops momentarily, kinetic energy drops to zero and all energy sits in the elastic store as ½mω²A². At any general position x the kinetic energy is ½mω²(A² - x²) and the potential energy is ½mω²x². Both energy graphs are parabolas, one opening downwards and one upwards, and their sum is always the same horizontal line. Exam questions frequently ask you to compare energy at different points or to read amplitude and total energy from the curves.
5. 弹簧振子与单摆 周期公式对比
两种最常考的 SHM 系统是弹簧振子和单摆。弹簧振子的角频率 ω = √(k/m)，周期 T = 2π√(m/k)，其中 k 是弹簧劲度系数，单位 N/m，m 是振子质量。注意弹簧振子的周期只取决于质量和劲度系数，与振幅大小无关，这就是所谓的等时性。单摆做小角度摆动时近似为 SHM，角频率 ω = √(g/l)，周期 T = 2π√(l/g)，其中 l 是摆长，g 是重力加速度。单摆周期与摆球质量无关，也与振幅无关（小角度时）。实验题常考：如何用 T² 对 l 作图测量 g，斜率等于 4π²/g。记住两个公式不要混淆，弹簧振子看 k 和 m，单摆看 l 和 g。
The two SHM systems you must know are the mass-spring oscillator and the simple pendulum. For a mass on a spring, ω = √(k/m) and the period is T = 2π√(m/k), where k is the spring constant in N/m and m is the oscillating mass. Crucially, this period depends only on mass and spring constant, never on amplitude, which is the property called isochronism. A simple pendulum approximates SHM for small angles, with ω = √(g/l) and period T = 2π√(l/g), where l is the pendulum length and g the gravitational field strength. The pendulum period is independent of the bob mass and, for small swings, of amplitude. Practical questions often ask you to plot T² against l and find g from the gradient 4π²/g. Do not mix the two formulas: springs use k and m, pendulums use l and g.
6. 阻尼与共振 振幅变化的定性分析
实际振动系统总存在阻尼（damping），例如空气阻力或摩擦力。阻尼使系统能量逐渐减少，振幅随时间指数衰减，但周期几乎不变（欠阻尼时）。根据阻尼强弱，系统可分为欠阻尼、临界阻尼和过阻尼三类。临界阻尼使系统最快回到平衡位置且不振荡，所以汽车减震器和门窗闭合器都设计成接近临界阻尼。当驱动频率接近系统固有频率 f₀ 时，振幅急剧增大，这种现象称为共振（resonance）。驱动频率等于固有频率时振幅最大。共振既有危害也有应用：桥梁需要避免共振，而微波炉、收音机调谐都利用共振原理。考试常让你画振幅对驱动频率的图像，并解释共振曲线的峰值。
Real oscillating systems always experience damping from air resistance or friction. Damping removes energy from the system, so the amplitude decays exponentially with time while the period stays almost constant for light damping. Depending on its strength, damping is classified as light, critical or heavy. Critical damping returns the system to equilibrium in the shortest time without any oscillation, which is why car shock absorbers and door closers are tuned close to critical damping. When the driving frequency approaches the natural frequency f₀ of the system, the amplitude rises sharply; this is resonance, and the amplitude peaks when the driving frequency equals f₀. Resonance can be dangerous, as with bridges, but it is also useful in microwave ovens and radio tuning. Expect questions where you sketch amplitude against driving frequency and explain the shape of the resonance curve.
7. 典型计算题 完整三步解答
下面通过一道标准题目演示完整解题过程。题目：一个质量 0.50 kg 的物体做 SHM，振幅 0.080 m，周期 1.6 s。求：(a) 角频率 ω；(b) 最大速度；(c) 物体在 x = 0.040 m 处的速度大小；(d) 系统的总能量。第一步，求角频率：ω = 2π/T = 2π/1.6 = 3.93 rad/s。第二步，求最大速度：v_max = Aω = 0.080 × 3.93 = 0.31 m/s。第三步，用 v = ω√(A² - x²) 求任意位置速度：v = 3.93 × √(0.080² - 0.040²) = 3.93 × √0.0048 = 3.93 × 0.0693 = 0.27 m/s。第四步，求总能量 E = ½mω²A² = 0.5 × 0.50 × 3.93² × 0.080² = 0.5 × 0.50 × 15.4 × 0.0064 = 0.0247 J。注意有效数字：题目数据都是两位有效数字，答案应保留两位。
Here is a standard question solved step by step. A 0.50 kg object performs SHM with amplitude 0.080 m and period 1.6 s. Find (a) the angular frequency, (b) the maximum speed, (c) the speed at x = 0.040 m, and (d) the total energy. Step one: ω = 2π/T = 2π/1.6 = 3.93 rad/s. Step two: v_max = Aω = 0.080 × 3.93 = 0.31 m/s. Step three: use v = ω√(A² - x²), giving v = 3.93 × √(0.080² - 0.040²) = 3.93 × √0.0048 = 3.93 × 0.0693 = 0.27 m/s. Step four: total energy E = ½mω²A² = 0.5 × 0.50 × 15.4 × 0.0064 = 0.0247 J. Watch significant figures: the data carry two significant figures, so answers should be quoted to two.
8. 高频失分点 相位起点与单位换算
最后总结 SHM 大题的五个高频失分点。第一，相位起点选错：题目说从平衡位置开始计时就用正弦形式，从最大位移开始就用余弦形式，混用会丢掉全部图像分。第二，忘记角度单位：ωt 必须用弧度计算，计算器要调到弧度模式 RAD，很多同学用度数模式导致答案全错。第三，周期与频率混淆：T = 1/f，ω = 2πf，换算时不要把 2π 放错位置。第四，能量表达式记混：总能量 ½mω²A² 中 A 必须平方，且 ω 要换成 rad/s 再平方。第五，忽略方向描述：SHM 是矢量运动，速度方向改变时要在答案里说明何时反向。逐条核对这五点，SHM 大题基本可以拿满。
Here are the five most common marks lost in SHH questions. First, choosing the wrong phase start: start timing at equilibrium and use sine, start at maximum displacement and use cosine; mixing them costs all graph marks. Second, angle units: ωt is always in radians, so your calculator must be in RAD mode; degree mode silently destroys every answer. Third, confusing period and frequency: T = 1/f and ω = 2πf, so place the 2π correctly. Fourth, errors in the energy formula: in E = ½mω²A² the amplitude must be squared, and ω must be converted to rad/s before squaring. Fifth, ignoring direction: SHM is a vector motion, so when velocity reverses you must say so explicitly. Check these five points one by one and you can secure almost full marks on any SHM question.
掌握简谐运动，核心就是一条定义式 a = -ω²x、两个周期公式和一个能量守恒框架。把位移、速度、加速度在平衡位置与端点处的取值对比记牢，再把今天这道计算题独立重做一遍，SHM 就不再是难点。如果你在 A-Level 物理的其他章节也有疑问，欢迎随时联系我们，专业老师会帮你做系统的考点梳理与真题训练。
Mastering SHM comes down to one defining equation a = -ω²x, two period formulas and one energy conservation framework. Once you memorise how displacement, velocity and acceleration behave at equilibrium and at the extremes, and rework today's calculation by yourself, SHM stops being difficult. If you need help with other A-Level Physics topics, contact us any time and our tutors will build a structured revision plan with real past-paper practice.
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