📚 IB Physics: Core Concepts of Simple Harmonic Motion & Period Calculation | IB物理：简谐运动核心概念与周期计算
Simple Harmonic Motion (SHM) is one of the most fundamental topics in IB Physics, appearing in both Standard Level (Topic 4) and Higher Level (Topic 9) syllabi. It bridges mechanics and wave phenomena, requiring students to understand the interplay between displacement, velocity, acceleration, and restoring forces. This guide breaks down the essential definitions, conditions, energy transformations, and — most importantly — the period calculations that appear repeatedly in IB examinations.
简谐运动（SHM）是IB物理中最基础的主题之一，同时出现在标准水平（Topic 4）和高级水平（Topic 9）的教学大纲中。它连接了力学与波动现象，要求学生理解位移、速度、加速度与回复力之间的相互作用。本指南将系统梳理核心定义、成立条件、能量转化，以及最重要的——在IB考试中反复出现的周期计算。
1. Definition & Conditions | 定义与条件
Simple harmonic motion is defined as the oscillatory motion in which the acceleration of the object is directly proportional to its displacement from equilibrium and always directed toward the equilibrium position. This means the acceleration is always opposite in direction to the displacement.
简谐运动定义为：物体的加速度与其相对于平衡位置的位移成正比，且方向始终指向平衡位置的振动运动。也就是说，加速度的方向始终与位移方向相反。
The mathematical condition for SHM can be written as:
a = −ω²x
where a is acceleration, x is displacement from equilibrium, and ω is the angular frequency (measured in rad s⁻¹). The negative sign indicates that acceleration always points toward equilibrium — a restoring effect.
其中 a 为加速度，x 为相对于平衡位置的位移，ω 为角频率（单位 rad s⁻¹）。负号表示加速度始终指向平衡位置——即回复效应。
For an oscillating system to exhibit SHM, two key conditions must be satisfied:
- Restoring force: : The net force must be proportional to displacement and directed toward equilibrium (F = −kx, Hooke's Law).
- 回复力：合力必须与位移成正比且指向平衡位置（F = −kx，胡克定律）。
- Linear range: The system must obey this linear relationship within its range of oscillation; for a pendulum this requires small angles (typically θ < 10°).
- 线性范围：系统在振动范围内必须满足这一线性关系；对于单摆而言，需要小角度条件（通常 θ < 10°）。
- No damping: In ideal SHM, energy is conserved and oscillations continue indefinitely (energy losses ignored).
- 无阻尼：理想简谐运动中能量守恒，振动无限持续（忽略能量损失）。
2. Key Kinematic Quantities | 关键运动学量
Three primary quantities describe SHM: displacement, velocity, and acceleration. Each varies sinusoidally with time, and their phases are offset relative to one another.
描述简谐运动的三个主要物理量为：位移、速度和加速度。它们都随时间按正弦规律变化，且彼此之间存在相位差。
For an object starting at maximum displacement (amplitude A) at t = 0, the displacement equation is:
x = A cos(ωt)
If the object starts at equilibrium and moves in the positive direction, use the sine form: x = A sin(ωt). Both forms are acceptable in IB answers; just be consistent with initial conditions.
若物体在 t = 0 时从最大位移（振幅 A）处开始，位移方程为：x = A cos(ωt)。若物体从平衡位置出发并向正方向运动，则使用正弦形式：x = A sin(ωt)。IB答题中两种形式均可，但需与初始条件保持一致。
Differentiating the displacement equation yields velocity:
v = −Aω sin(ωt)
The maximum speed occurs as the object passes through equilibrium and is given by vmax = Aω. At this point the velocity is nonzero while the restoring force and acceleration are zero.
对位移方程求导可得速度：v = −Aω sin(ωt)。最大速度出现在物体经过平衡位置时，大小为 vmax = Aω。在该点速度不为零，而回复力和加速度均为零。
Differentiating again gives acceleration:
a = −ω²A cos(ωt) = −ω²x
This confirms that acceleration is proportional to displacement but in the opposite direction. The maximum acceleration, which occurs at the turning points, is amax = ω²A.
再次求导得加速度：a = −ω²A cos(ωt) = −ω²x。这证实了加速度与位移成正比但方向相反。最大加速度出现在转向点，大小为 amax = ω²A。
| Quantity | 物理量 | Maximum | 最大值 | Position of Maximum | 最大值位置 |
| Displacement x | 位移 x | A | Turning points | 转向点 |
| Velocity v | 速度 v | Aω | Equilibrium | 平衡位置 |
| Acceleration a | 加速度 a | ω²A | Turning points | 转向点 |
3. Angular Frequency & Period | 角频率与周期
The angular frequency ω connects the period of oscillation to the physical properties of the system. For any SHM system, the period T is the time for one complete oscillation and is related to ω by:
角频率 ω 将振动周期与系统的物理性质联系起来。对于任何简谐运动系统，周期 T 是一次完整振动所需的时间，与 ω 的关系为：
T = 2π / ω
Note that in ideal SHM, the period is independent of the amplitude. This is a hallmark property of SHM — whether you pull a pendulum back 2° or 8°, the period remains essentially the same (within the small-angle approximation).
注意：在理想简谐运动中，周期与振幅无关。这是简谐运动的重要特性——无论你将单摆拉开2°还是8°，周期基本相同（在小角度近似范围内）。
One of the most common IB exam traps is confusing ω with frequency f. They are related by ω = 2πf, where f is the ordinary frequency measured in hertz (Hz). Thus:
IB考试中最常见的陷阱之一是混淆 ω 与频率 f。它们的关系为 ω = 2πf，其中 f 是普通频率，单位为赫兹（Hz）。因此：
T = 1/f = 2π/ω
4. Period of a Mass–Spring System | 弹簧振子的周期
For a mass m attached to a spring with spring constant k, the restoring force is given by Hooke's Law: F = −kx. Combining this with Newton's Second Law (F = ma), we obtain ma = −kx, which simplifies to a = −(k/m)x.
对于质量为 m、连接在劲度系数为 k 的弹簧上的物体，回复力由胡克定律给出：F = −kx。结合牛顿第二定律（F = ma），可得 ma = −kx，化简为 a = −(k/m)x。
Comparing this with the general SHM condition a = −ω²x, we identify ω² = k/m, giving:
将此与一般简谐运动条件 a = −ω²x 比较，可得 ω² = k/m，从而：
T = 2π√m/k
Key observations:
- Heavier mass → longer period: More inertia means the system oscillates more slowly.
- 质量越大 → 周期越长：惯性越大，系统振动越慢。
- Stiffer spring → shorter period: A larger k value means a stronger restoring force and faster oscillation.
- 弹簧越硬 → 周期越短：k 值越大意味着回复力越强，振动越快。
- Period is independent of amplitude and gravitational field: A mass–spring system on the Moon has the same period as on Earth (assuming ideal conditions).
- 周期与振幅和重力场无关：在月球上的弹簧振子与在地球上的周期相同（理想条件下）。
This formula is essential for both SL and HL examinations. In practical labs, this relationship is often verified by plotting T² against m, which yields a straight line with slope 4π²/k.
该公式在SL和HL考试中都至关重要。在实验课中，通常通过绘制 T² 对 m 的图线来验证此关系，所得直线的斜率为 4π²/k。
5. Period of a Simple Pendulum | 单摆的周期
For a simple pendulum of length L F = −mg sin θ. For small angles, sin θ ≈ θ, and using arc length s = Lθ, we can show that:
对于摆长为 L、小角度振动的单摆，回复力是重力沿运动切向方向的分量：F = −mg sin θ。在小角度下，sin θ ≈ θ，利用弧长 s = Lθ，可以证明：
a = −(g/L)x
Therefore ω² = g/L, and the period is:
因此 ω² = g/L，周期为：
T = 2π√L/g
Critical points to remember for IB exams:
IB考试中需记住的关键要点：
- Period depends only on L and g — not on mass. A heavier bob does not change the period.
- 周期只取决于 L 和 g —— 与质量无关。更重的摆锤不会改变周期。
- Small-angle approximation is required. For angles beyond approximately 10°, sin θ ≠ θ and the motion deviates from true SHM; the period becomes slightly longer.
- 必须满足小角度近似。当角度超过约10°时，sin θ ≠ θ，运动偏离真正的简谐运动；周期会略有增加。
- g can be determined experimentally by measuring T and L: g = 4π²L/T². This is a standard IB Physics practical investigation.
- 可通过实验测定 g：通过测量 T 和 L 计算：g = 4π²L/T²。这是IB物理标准实验探究课题。
misconception is that the pendulum period changes when the mass is increased. It does not — the relation T = 2π√L/g contains no mass term. However, changing the length of the string or moving the pendulum to a location with different gravitational field strength absolutely will change the period.
一个常见误解是增加摆锤质量会改变单摆周期。事实并非如此——T = 2π√L/g 中不含质量项。然而，改变摆长或将单摆移到重力场强度不同的位置，确实会改变周期。
6. Energy in SHM | 简谐运动中的能量
In ideal SHM, mechanical energy is conserved. Energy continuously transforms between kinetic energy (KE) and potential energy (PE), with the total remaining constant.
在理想简谐运动中，机械能守恒。能量在动能（KE）和势能（PE）之间持续转化，总能量保持恒定。
At the turning points (x = ±A): all energy is potential.
在转向点（x = ±A）：所有能量均为势能。
Etotal = ½kA²
At the equilibrium position (x = 0): all energy is kinetic, and the speed is maximum.
在平衡位置（x = 0）：所有能量均为动能，速度达到最大值。
Etotal = ½mv_max² = ½kA²
At any intermediate displacement x, the energies are:
在任意中间位移 x 处，能量分别为：
KE = ½k(A² − x²)
PE = ½kx²
For a pendulum, the same principle applies but with gravitational potential energy replacing elastic potential energy. At maximum height (amplitude), the bob has maximum gravitational PE; at the lowest point, all energy is kinetic.
对于单摆，同样的原理适用，只是弹性势能换成了重力势能。在最大高度（振幅处），摆锤具有最大的重力势能；在最低点，所有能量均为动能。
Exam tip: When asked to sketch energy–displacement graphs, remember that PE is a parabola (∝ x²), KE is an inverted parabola, and total energy is a horizontal line — all of which remain positive throughout the motion.
考试提示：当要求绘制能量–位移图时，记住势能是抛物线（∝ x²），动能是倒抛物线，总能量是水平直线——在整个运动过程中所有量均为正值。
7. Phase and Phase Difference | 相位与相位差
Phase describes the position within the oscillation cycle at a given time. Two oscillating systems are said to be in phase when they reach their maximum displacements in the same direction at the same time.
相位描述的是在给定时刻物体在振动周期中的位置。当两个振动系统在同一时刻沿同一方向达到最大位移时，称它们同相。
In SHM, displacement, velocity, and acceleration have specific phase relationships:
在简谐运动中，位移、速度和加速度之间存在特定的相位关系：
- Velocity leads displacement by π/2 (90°): when displacement is maximum, velocity is zero; when displacement is zero, velocity is maximum.
- 速度超前位移 π/2（90°）：位移最大时速度为零；位移为零时速度最大。
- Acceleration is π (180°) out of phase with displacement: they are in opposite directions at all times.
- 加速度与位移反相 π（180°）：它们始终方向相反。
- Acceleration is π/2 (90°) behind velocity: maximum acceleration occurs at the turning points where velocity is zero.
- 加速度落后速度 π/2（90°）：最大加速度出现在转向点，此时速度为零。
Graphically, these phase relationships appear as shifted sine/cosine curves. Being able to read these graphs and connect them to physical motion is a core IB skill.
从图形上看，这些相位关系表现为平移的正弦/余弦曲线。能够解读这些图像并将它们与物理运动联系起来，是IB的核心技能。
8. SHM and Circular Motion | 简谐运动与圆周运动
SHM is the one-dimensional projection of uniform circular motion. Consider an object moving in a circle of radius A with constant angular speed ω. If you project its position onto a diameter, the projection executes SHM.
简谐运动是匀速圆周运动的一维投影。考虑一个物体在半径为 A 的圆上以恒定角速度 ω 运动。如果将其位置投影到直径上，投影点做简谐运动。
This geometric interpretation explains several important relationships:
这种几何解释阐明了几个重要的关系：
- The amplitude of SHM equals the radius of the reference circle: A.
- 简谐运动的振幅等于参考圆的半径：A。
- The angular frequency of SHM equals the angular speed of the circular motion: ω.
- 简谐运动的角频率等于圆周运动的角速度：ω。
- The period of SHM equals the time for one full revolution: T = 2π/ω.
- 简谐运动的周期等于旋转一周所需的时间：T = 2π/ω。
r cosine and how to adjust the phase constant φ in the general equation x = A cos(ωt + φ).
参考圆在解决涉及初始条件的问题时特别有用。如果粒子从某个任意位置和相位开始运动，画出参考圆有助于判断应使用正弦还是余弦函数，以及如何调整一般方程 x = A cos(ωt + φ) 中的相位常数 φ。
For IB Paper 2 problems that combine SHM with circular motion (e.g., a particle on a rotating turntable viewed from the side), this conceptual link is often the key to finding the correct answer.
对于结合简谐运动与圆周运动的IB Paper 2题（例如从侧面观察转盘上粒子的运动），这种概念联系往往是找到正确答案的关键。
9. Damping and Resonance | 阻尼与共振
Real-world oscillating systems lose energy over time due to friction, air resistance, or other dissipative forces. This gradual loss of amplitude is called damping. While ideal SHM assumes no energy loss, IB Physics requires an understanding of how damping affects real systems.
现实中的振动系统会因摩擦、空气阻力或其他耗散力而随时间损失能量。这种振幅的逐渐减小称为阻尼。虽然理想简谐运动假设无能量损失，但IB物理要求理解阻尼对实际系统的影响。
Three levels of damping are recognized:
阻尼分为三个级别：
- Light damping: Amplitude decreases gradually over many cycles; the period remains approximately constant.
- 欠阻尼：振幅在多个周期内逐渐减小；周期近似保持不变。
- Critical damping: The system returns to equilibrium in the shortest possible time without oscillating.
- 临界阻尼：系统在不振动的情况下以最短时间回到平衡位置。
- Heavy (over) damping: The system returns to equilibrium very slowly, without oscillating.
- 过阻尼：系统缓慢回到平衡位置，不发生振动。
plitude. The natural frequency f₀ is related to the natural period by f₀ = 1/T. For a mass–spring system, f₀ = (1/2π)√k/m; for a pendulum, f₀ = (1/2π)√g/L.
当驱动频率与系统的固有频率相等时，会发生共振，导致振幅急剧增大。固有频率 f₀ 与固有周期的关系为 f₀ = 1/T。对于弹簧振子，f₀ = (1/2π)√k/m；对于单摆，f₀ = (1/2π)√g/L。
IB exam questions often ask you to interpret a resonance curve (amplitude vs. driving frequency). Key features include: the peak occurs at f₀, and increased damping lowers the peak amplitude while slightly broadening the curve.
IB考题通常要求解读共振曲线（振幅 vs. 驱动频率）。关键特征包括：峰值出现在 f₀ 处；阻尼增大时峰值振幅降低，曲线稍微变宽。
10. Worked Examples | 典型例题精解
Example 1: Mass–Spring System | 例1：弹簧振子系统
ttached to a spring with k = 100 N m⁻¹. Calculate the period and frequency of oscillation.
一个 0.25 kg 的物体连接在劲度系数 k = 100 N m⁻¹ 的弹簧上。计算振动周期和频率。
Solution | 解答：
T = 2π√m/k = 2π√0.25 / 100 = 2π√0.0025 = 2π × 0.05 = 0.314 s
f = 1/T = 1/0.314 = 3.18 Hz
Example 2: Pendulum on the Moon | 例2：月球上的单摆
od of 2.0 s on Earth (g = 9.81 m s⁻²). What is its period on the Moon where g = 1.62 m s⁻²?
一个单摆在地球上（g = 9.81 m s⁻²）的周期为 2.0 s。在月球上（g = 1.62 m s⁻²）其周期为多少？
Solution | 解答：
First, find the length L from the Earth period:
首先由地球上的周期求摆长 L：
L = gT²/(4π²) = 9.81 × 4 / (4π²) = 0.994 m
Then calculate the Moon period:
然后计算月球上的周期：
Tmoon = 2π√L/gmoon = 2π√0.994/1.62 = 2π√0.6136 = 2π × 0.783 = 4.92 s
Note how a weaker gravitational field produces a longer period — the pendulum swings more slowly on the Moon. This example illustrates why gravitational field strength is a crucial parameter for pendulum systems but completely irrelevant for ideal mass–spring systems.
注意较弱的引力场会产生更长的周期——单摆在月球上摆动得更慢。这个例子说明重力场强度对单摆系统至关重要，但对理想弹簧振子系统完全无关。
Example 3: Maximum Speed | 例3：最大速度
An object undergoes SHM with amplitude 0.05 m and period 0.40 s. Calculate its maximum speed and maximum acceleration.
一个物体做简谐运动，振幅 0.05 m，周期 0.40 s。计算其最大速度和最大加速度。
Solution | 解答：
ω = 2π/T = 2π/0.40 = 15.7 rad s⁻¹
vmax = Aω = 0.05 × 15.7 = 0.785 m s⁻¹
amax = Aω² = 0.05 × (15.7)² = 0.05 × 246.5 = 12.3 m s⁻²
This example reinforces the relationship between period, angular frequency, and the kinematic quantities. Always convert period to angular frequency first before attempting to find velocities or accelerations.
此例强化了周期、角频率与运动学量之间的关系。在求速度或加速度之前，务必先将周期转换为角频率。
11. Common IB Exam Mistakes | 常见IB考试错误
Over years of marking IB Physics papers, certain errors appear repeatedly. Being aware of these can significantly boost your exam performance.
在多年批改IB物理试卷的过程中，某些错误反复出现。了解这些错误可以显著提升你的考试成绩。
- Forgetting the small-angle approximation for pendulums: The formula T = 2π√L/g is ONLY valid for θ < 10°. IB questions may specify "small oscillations" — if not stated, note this assumption in your answer.
- 忘记单摆的小角度近似：T = 2π√L/g 仅在 θ < 10° 时有效。IB题目可能会注明"小角度摆动"——若未注明，应在答案中说明这一假设。
- Confusing frequency f and angular frequency ω: f is in Hz, ω is in rad s⁻¹; they differ by a factor of 2π. When using T = 2π√m/k, do not insert k/m directly into f.
- 混淆频率 f 与角频率 ω：f 单位为 Hz，ω 单位为 rad s⁻¹；两者相差 2π 倍。使用 T = 2π√m/k 时，不要将 k/m 直接代入 f。
- Sign errors with acceleration direction: The restoring force always points toward equilibrium. In vector problems, carefully define your positive direction first.
- 加速度方向符号错误：回复力始终指向平衡位置。在矢量问题中，先明确定义正方向。
- Using x = A sin(ωt) without checking initial conditions: Verify where the object starts at t = 0 before selecting the sine or cosine form.
- 未检查初始条件就使用 x = A sin(ωt)：在选用正弦或余弦形式前，需先确认 t = 0 时物体的位置。
- Omitting units: Period must be in seconds, angular frequency in rad s⁻¹, and amplitude in the same length unit throughout the calculation.
- 遗漏单位：周期必须以秒为单位，角频率以 rad s⁻¹ 为单位，振幅在整个计算过程中保持相同的长度单位。
12. Exam Strategy Checklist | 备考策略清单
To succeed in SHM questions on both Paper 1 and Paper 2, build a systematic approach:
要在Paper 1和Paper 2的简谐运动题目中取得成功，需要建立系统化的解题方法：
- Identify the system type: Is it a mass–spring (T = 2π√m/k) or a pendulum (T = 2π√L/g)? This determines which formula set applies.
- 判断系统类型：是弹簧振子（T = 2π√m/k）还是单摆（T = 2π√L/g）？这决定了适用哪组公式。
- Extract all given quantities: Write down m, k, L, g, A, T, or ω as given in the problem. Convert to SI units early.
- 提取所有已知量：写下题目中给出的 m、k、L、g、A、T 或 ω。尽早转换为SI单位。
- Check assumptions: For pendulums, verify small angles; for ideal SHM, confirm no damping is assumed.
- 检查假设条件：对于单摆，确认小角度；对于理想简谐运动，确认假设无阻尼。
- Draw a diagram or reference circle: Visualizing the motion helps determine phase and initial conditions.
- 画图或参考圆：可视化运动有助于判断相位和初始条件。
- For graphical questions: Identify which quantity is plotted on each axis, then relate slopes and intercepts to ω, k, or g.
- 对于图像题：确认每个坐标轴上的物理量，然后将斜率和截距与 ω、k 或 g 联系起来。
- Interpret the problem physically: Before calculating, predict the answer qualitatively — will the period increase or decrease based on changing parameters?
- 从物理角度理解问题：在计算之前，定性地预测答案——参数变化后周期会增大还是减小？
Remember: the period equations for SHM are among the most tested formulas in the IB Physics syllabus. Mastery of when and how to apply them, accompanied by a solid grasp of the sinusoidal relationships between x, v, and a, will position you strongly for both multiple-choice and extended-response questions.
请记住：周期公式是IB物理教学大纲中考查频率最高的公式之一。熟练掌握何时使用以及如何使用它们，同时扎实理解 x、v、a 之间的正弦关系，将使你在选择题和扩展回答题中均占据优势。
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