📚 GCSE Maths: Simple Harmonic Motion Key Points Revision | GCSE 数学：简谐运动考点精讲
Simple harmonic motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement and acts towards the equilibrium position. In GCSE Mathematics, SHM is often explored through the lens of trigonometric functions, as the displacement of a particle in SHM can be modelled by a sine or cosine curve. Understanding SHM helps you apply sin and cos graphs to real-world oscillation problems.
简谐运动（SHM）是一种周期性运动，其回复力与位移成正比且指向平衡位置。在 GCSE 数学中，简谐运动通常通过三角函数的角度来探讨，因为简谐运动中质点的位移可以用正弦或余弦曲线来建模。理解简谐运动有助于你将正弦和余弦图像应用于实际的振动问题。
1. What is Simple Harmonic Motion? | 什么是简谐运动？
In physics and maths, simple harmonic motion describes a repeated back-and-forth movement around a central equilibrium point. The defining characteristic is that the acceleration is always directed towards the equilibrium and is proportional to the distance from it. Common examples include a mass on a spring and a simple pendulum (for small angles). The displacement-time graph of SHM is a perfect sine wave, making it a key application of trigonometric graphs.
在物理和数学中，简谐运动描述了围绕中心平衡点的往复运动。其定义特征是加速度始终指向平衡位置，且与到平衡点的距离成正比。常见的例子包括弹簧上的物块和小角度下的单摆。简谐运动的位移 – 时间图像是一条完美的正弦波，因此它是三角函数图像的重要应用。
The standard form of the SHM displacement equation is x = A sin(ωt) or x = A cos(ωt), where A, ω and t have specific meanings which we will explore.
简谐运动位移方程的标准形式为 x = A sin(ωt) 或 x = A cos(ωt)，其中 A、ω 和 t 具有特定含义，我们将在后面探讨。
2. Amplitude, Period and Frequency | 振幅、周期和频率
Amplitude (A) is the maximum displacement from the equilibrium position. It is always a positive value and represents the ‘height’ of the oscillation. Period (T) is the time taken for one complete oscillation, measured in seconds. Frequency (f) is the number of complete oscillations per second, measured in hertz (Hz). These two are related by f = 1/T.
振幅（A）是离开平衡位置的最大位移，它是一个正值，表示振动的“高度”。周期（T）是完成一次全振动所需的时间，单位是秒。频率（f）是每秒完成全振动的次数，单位是赫兹（Hz）。二者满足关系 f = 1/T。
3. The SHM Equation: x = A sin(ωt) | 简谐运动方程：x = A sin(ωt)
The displacement x of a particle undergoing SHM can be expressed as x = A sin(ωt), where A is the amplitude, ω (omega) is the angular frequency, and t is time. This form is used when the particle starts at the equilibrium position (x=0) at t=0. If it starts at maximum displacement, the equation x = A cos(ωt) is more appropriate. The argument of the sine function, ωt, is an angle in radians.
简谐运动中原点的位移 x 可以表示为 x = A sin(ωt)，其中 A 是振幅，ω（omega）是角频率，t 是时间。当质点于 t=0 时从平衡位置（x=0）开始运动，使用此形式。若质点从最大位移处开始，则方程 x = A cos(ωt) 更为合适。正弦函数的变量 ωt 是以弧度为单位的角度。
4. Angular Frequency and Its Relation to Period | 角频率及其与周期的关系
Angular frequency ω is a measure of how rapidly the oscillation cycles, expressed in radians per second. The relationship between ω, period T and frequency f is given by ω = 2πf = 2π/T. Consequently, T = 2π/ω and f = ω/(2π). In GCSE Maths, you may be asked to rearrange these formulas to find an unknown.
角频率 ω 是衡量振动快慢的物理量，单位为弧度每秒。ω 与周期 T 和频率 f 的关系为 ω = 2πf = 2π/T。因此，T = 2π/ω，f = ω/(2π)。在 GCSE 数学中，你可能会被要求运用这些公式进行变形以求解未知量。
5. Graphical Representation of SHM | 简谐运动的图形表示
When you plot displacement against time for SHM, you get a sinusoidal curve. If x = A sin(ωt), the graph starts at 0, rises to A, returns to 0, drops to -A, and completes one cycle at time T. The velocity-time graph is a cosine curve (shifted by 90° or π/2 radians) and the acceleration-time graph is an inverted sine curve, i.e., a = -ω²x. This means acceleration is always opposite in sign to displacement.
绘制简谐运动的位移 – 时间图像时，会得到正弦曲线。若 x = A sin(ωt)，图像从 0 开始，上升到 A，回到 0，下降到 -A，并在时间 T 时完成一个循环。速度 – 时间图像是一条余弦曲线（相位差 90° 或 π/2 弧度），而加速度 – 时间图像是一条倒置的正弦曲线，即 a = -ω²x。这意味着加速度的符号始终与位移相反。
6. Velocity and Acceleration in SHM | 简谐运动中的速度和加速度
In SHM, the velocity v is not constant. It is zero at the extreme positions (x = ±A) and reaches its maximum magnitude when the particle passes through equilibrium. The maximum speed v_max is given by v_max = ωA. Acceleration a is directly proportional to displacement x but with a negative sign: a = -ω²x. The constant of proportionality is -ω². Although GCSE may not require calculus proofs, recognising these relationships is essential for graph questions.
在简谐运动中，速度 v 并非恒定。它在端点位置（x = ±A）为零，当质点经过平衡位置时达到最大。最大速率 v_max 等于 ωA。加速度 a 与位移 x 成正比，但符号相反：a = -ω²x。比例常数为 -ω²。虽然 GCSE 不要求微积分证明，但识别这些关系对于处理图像类题目至关重要。
7. Energy Changes in SHM | 简谐运动中的能量变化
During SHM, energy continuously converts between kinetic and potential forms. At the equilibrium point, kinetic energy is maximum and potential energy is minimum (usually zero). At the extreme displacements, the energy is entirely potential. The total mechanical energy remains constant if no external forces do work. This concept helps explain why amplitude stays the same in ideal SHM.
在简谐运动中，能量在动能和势能之间不断转化。在平衡点处，动能最大，势能最小（通常为零）。在最大位移处，能量全部为势能。如果没有外力做功，总机械能保持不变。这一概念有助于解释理想简谐运动中振幅为何保持不变。
8. Worked Example | 典型例题
Example: A particle moves with simple harmonic motion. Its displacement x in metres from equilibrium is given by x = 0.6 sin(4π t), where t is in seconds. Find (a) the amplitude, (b) the angular frequency, (c) the period and frequency, (d) the displacement when t = 0.25 seconds.
例题：一个质点在作简谐运动，其位移 x（米）与时间 t（秒）的关系为 x = 0.6 sin(4π t)。求 (a) 振幅；(b) 角频率；(c) 周期和频率；(d) t = 0.25 秒时的位移。
Solution: (a) Comparing with x = A sin(ωt), amplitude A = 0.6 m. (b) Angular frequency ω = 4π rad/s. (c) Period T = 2π/ω = 2π/(4π) = 0.5 s. Frequency f = 1/T = 2 Hz. (d) Substitute t=0.25: x = 0.6 sin(4π × 0.25) = 0.6 sin(π) = 0.6 × 0 = 0 m. The particle is at the equilibrium position.
解答： (a) 与 x = A sin(ωt) 对比，振幅 A = 0.6 m。(b) 角频率 ω = 4π rad/s。(c) 周期 T = 2π/ω = 2π/(4π) = 0.5 s。频率 f = 1/T = 2 Hz。(d) 代入 t=0.25: x = 0.6 sin(4π × 0.25) = 0.6 sin(π) = 0.6 × 0 = 0 m。质点处于平衡位置。
9. Exam Tips for SHM Questions | 简谐运动考点提示
Always check whether the equation uses sine or cosine – this tells you the starting position. Use radian mode on your calculator when evaluating trig functions. Be comfortable rearranging T = 2π/ω and ω = 2πf. If a question asks for maximum speed, recall v_max = ωA. For multistep problems, write down given information clearly and identify which parameters correspond to A, ω, and initial phase.
务必检查方程使用的是正弦还是余弦——这决定了初始位置。计算三角函数值时，将计算器设置为弧度模式。熟练掌握 T = 2π/ω 和 ω = 2πf 等公式的变形。如果题目问最大速率，记住 v_max = ωA。对于多步骤问题，清晰地列出已知信息，并确定哪些参数对应 A、ω 和初相。
10. Common Pitfalls and Misconceptions | 常见错误与误区
Many students confuse angular frequency ω with frequency f. Remember ω has units rad/s while f is in Hz. Another common error is using degrees instead of radians inside the sine function, which will produce incorrect results. Also, be careful with the negative sign in a = -ω²x; it indicates direction, not a negative magnitude. Finally, do not assume amplitude is always half the total range – it is the maximum displacement from equilibrium, not peak-to-peak.
许多学生将角频率 ω 与频率 f 混淆。记住 ω 的单位是 rad/s，而 f 的单位是 Hz。另一个常见错误是在正弦函数内使用角度制而非弧度制，这会导致错误的结果。此外，注意 a = -ω²x 中的负号，它表示方向，而非大小上的负值。最后，不要误以为振幅总是总范围的一半——它是相对于平衡位置的最大位移，而不是峰峰值。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导