📚 Simple Harmonic Motion | 简谐运动
Simple harmonic motion (SHM) lies at the heart of many oscillations in physics, from pendulums and mass-spring systems to the vibrations of atoms. In SHM, the restoring force always pulls the object back toward equilibrium with a strength proportional to the displacement. Understanding SHM gives you a powerful framework for analysing wave motion, alternating currents and even quantum mechanics.
简谐运动是物理中许多振荡的核心，从单摆和弹簧质量系统到原子的振动。在简谐运动中，回复力总是将物体拉回平衡位置，且力的大小与位移成正比。理解简谐运动为您提供了一个强大的分析框架，可用于研究波动、交流电甚至量子力学。
1. Defining Simple Harmonic Motion | 定义简谐运动
An oscillation is classified as simple harmonic motion only when two conditions are satisfied. First, the magnitude of the restoring force, or of the acceleration, is directly proportional to the displacement from the equilibrium position. Second, the direction of the force is always opposite to the direction of the displacement. If either condition fails, the motion is periodic but not simple harmonic.
一个振动只有在满足两个条件时才被归类为简谐运动。第一，回复力（或加速度）的大小与偏离平衡位置的位移成正比。第二，力的方向始终与位移的方向相反。只要其中任一条件不成立，该运动就是周期运动而不是简谐运动。
e force constant (also called the stiffness), measured in N m⁻¹, and x is the displacement from equilibrium, measured in metres. The minus sign is not decoration: it carries the physical meaning that the force always points back toward the centre of oscillation.
数学上我们把这个定义关系写作 F = -kx，其中 k 是力常数（也称劲度系数），单位为 N m⁻¹；x 是偏离平衡位置的位移，单位为米。负号不是装饰：它承载了"力始终指向振荡中心"这一物理含义。
F = -kx
The equilibrium position is the point where the net force on the object is zero. Displace the object by x and release it, and the restoring force accelerates it back toward equilibrium. It overshoots, is slowed, stops, and returns, so the motion repeats indefinitely in the absence of friction.
平衡位置是物体所受合力为零的点。把物体拉开位移 x 后释放，回复力使它加速返回平衡位置；它会冲过平衡点，被减速、停下、再返回，因此在没有摩擦的理想情况下，这种运动将永远重复下去。
Because the force grows in proportion to the displacement, the motion is perfectly regular: for an ideal oscillator the period is independent of amplitude. A larger amplitude means a longer path but also a larger average speed, and the two effects cancel exactly.
由于力与位移成正比，这种运动十分规整：对理想振子而言，周期与振幅无关。振幅越大，路径越长，但平均速率也越大，两者恰好相互抵消。
2. From Newton's Second Law to a = -ω²x | 从牛顿第二定律到 a = -ω²x
Taking the mass m into account, Newton' )x. The acceleration is proportional to the displacement and always directed toward equilibrium, exactly as the definition of SHM requires.
把质量 m 考虑进来，将牛顿第二定律 F = ma 与 F = -kx 结合，可得 ma = -kx，因此 a = -(k/m)x。加速度与位移成正比，并且始终指向平衡位置，这正好符合简谐运动的定义。
It is conventional to define the angular frequency ω so that ω² = k/m, that is ω = √k/m. The equation of motion then takes its standard, universal form, in which the constant k/m has been absorbed into a single quantity that is the same for every simple harmonic oscillator in the universe.
习惯上定义角频率 ω，使 ω² = k/m，即 ω = √k/m。运动方程于是取得了标准而普适的形式，常数 k/m 被吸收进一个对宇宙中所有简谐振子都相同的量中。
a = -ω²x
(rad s⁻¹) and is related to the period T and the frequency f by ω = 2πf and T = 2π/ω = 2π√m/k. For a spring-mass system this shows that a stiffer spring or a smaller mass gives a higher frequency, which matches everyday experience.
角频率的单位是弧度每秒（rad s⁻¹），它与周期 T 和频率 f 的关系为 ω = 2πf 以及 T = 2π/ω = 2π√m/k。对弹簧振子系统来说，这意味着弹簧越硬、质量越小，频率越高，这与日常经验一致。
on does satisfy a = -ω²x. The same is true of x = A sin(ωt) and of any combination of the two, which is why the general solution is written x = A cos(ωt + φ), where φ is the phase constant.
你可以用代入法验证标准解。若 x = A cos(ωt)，求导两次得 a = -ω²A cos(ωt) = -ω²x，可见余弦函数确实满足 a = -ω²x。x = A sin(ωt) 以及两者的任意组合同样满足，因此通解写作 x = A cos(ωt + φ)，其中 φ 是初相位。
3. Displacement, Velocity and Acceleration Graphs | 位移、速度与加速度图像
t), we obtain v = -Aω sin(ωt) and a = -Aω² cos(ωt). All three quantities vary sinusoidally with the same period T, but they are not in step with one another.
把位移方程对时间求导，便得到速度与加速度随时间变化的函数。从 x = A cos(ωt) 出发，可得 v = -Aω sin(ωt) 与 a = -Aω² cos(ωt)。这三个量都以相同的周期 T 作正弦式变化，但彼此并不同步。
The velocity curve is shifted a quarter of a cycle (π/2, or 90°) ahead of the displacement curve, and the acceleration curve is exactly half a cycle (π, or 180°) out of phase with the displacement. When the displacement is at a maximum the acceleration is also at a maximum but in the opposite direction, and the velocity is momentarily zero at the turning point.
速度曲线比位移曲线提前四分之一周期（π/2，即 90°），而加速度曲线与位移曲线恰好相差半周期（π，即 180°）。当位移达到最大值时，加速度也达到最大值但方向相反，而速度在该折返点瞬间为零。
x = A cos(ωt) v = -Aω sin(ωt) a = -Aω² cos(ωt)
The table below summarises what happens at the three key positions of an oscillation. Examiners frequently ask you to identify these, so learn them as a pattern rather than deriving them each time.
下表总结了振动中三个关键位置的情况。考官经常要求你指出这些结论，因此建议把它们当作固定模式记住，而不必每次重新推导。
| Position | 位置 | Displacement x | 位移 x | Velocity v | 速度 v | Acceleration a | 加速度 a | Energy | 能量 |
|---|---|---|---|---|
| Equilibrium | 平衡位置 | 0 | Maximum, Aω | 最大，Aω | 0 | All kinetic | 全为动能 |
| Midway | 中点位置 | ±A/2 | ±0.87Aω | ±0.5Aω² | Mixed | 混合 |
| Extreme | 振幅端点 | ±A | 0 | Maximum, Aω² | 最大，Aω² | All potential | 全为势能 |
Note the elegant relationship between the two maxima: amax = ω vmax = ω²A. This single line frequently saves time in multiple-choice questions.
注意两个最大值之间的优雅关系：amax = ω vmax = ω²A。这一行式子常常能在选择题中为你节省时间。
4. The Reference Circle and Phase | 参考圆与相位
A helpful way to picture SHM is the reference circle. Imagine a particle moving at constant angular speed ω around a circle of radius A. The projection of that particle onto a fixed diameter performs exactly simple harmonic motion with amplitude A and angular frequency ω. This construction explains where the sine and cosine functions come from.
理解简谐运动的一个直观方法是参考圆。设想一个质点以恒定角速度 ω 沿半径为 A 的圆运动，它在某条固定直径上的投影正好作振幅为 A、角频率为 ω 的简谐运动。这一构造解释了正弦、余弦函数为何会出现。
The angle swept out by the radius since the start of timing is called the phase, written as (ωt + φ), where φ is the phase constant determined by the initial conditions. Two oscillators with the same frequency but different phase constants reach their maximum displacements at different times.
半径自计时起转过的角度称为相位，写作 (ωt + φ)，其中 φ 是由初始条件决定的初相位。频率相同但初相位不同的两个振子，会在不同时刻到达最大位移。
Phase differences matter in exam questions. A phase difference of π/2 rad between two vibrating systems means one reaches maximum displacement a quarter of a period before the other, while a phase difference of π rad means they move in exact opposition. Always work in radians when using ω, because ω is defined in rad s⁻¹ and mixing degrees with radians is a classic source of error.
相位差在考试题中很重要。两个振动系统之间相位差为 π/2 rad，意味着一个比另一个提前四分之一周期到达最大位移；而相位差为 π rad 则意味着它们完全反向运动。使用 ω 时务必以弧度计算，因为 ω 以 rad s⁻¹ 定义，把角度制与弧度制混用是典型的错误来源。
5. Velocity and Acceleration at a Given Displacement | 已知位移处的速度与加速度
iminating time between the displacement and velocity equations, using sin²θ + cos²θ = 1, leads to a very useful result.
虽然 v = -Aω sin(ωt) 给出某一时刻的速度，但考题常常要求你求出物体处于某一特定位移时的速率。利用 sin²θ + cos²θ = 1 消去位移方程与速度方程中的时间，可得到一个非常有用的结果。
v = ±ω√A² - x²
vel, one on the way out and one on the way back. The maximum speed occurs when x = 0, so vmax = ωA, and the speed falls to zero at x = ±A.
正负号提醒你：除端点之外的任何位置都存在两个可能的运动方向，一是向外，二是返回。当 x = 0 时速率最大，即 vmax = ωA；当 x = ±A 时速率降为零。
is simpler because it depends only on position, not on direction of travel: a = -ω²x, so amax cceleration-time graph is a mirror image (inverted) of the displacement-time graph, while the velocity-time graph is shifted by a quarter cycle.
加速度则更简单，因为它只取决于位置而与运动方向无关：a = -ω²x，因此在端点处 amax = ω²A。这正是加速度—时间图是位移—时间图上下翻转的镜像、而速度—时间图整体平移四分之一周期的原因。
6. Energy in Simple Harmonic Motion | 简谐运动中的能量
In an ideal, undamped oscillator the total energy is constant, but it is continually exchanged between kinetic and potential forms. The kinetic energy is ½mv², which after substitution becomes ½mω²(A² - x²). The potential energy stored in the restoring-force system is ½mω²x².
在理想无阻尼振子中，总能量保持不变，但它在动能与势能之间不断转换。动能为 ½mv²，代入后变为 ½mω²(A² - x²)；储存在回复力系统中的势能为 ½mω²x²。
Etotal = KE + PE = ½mω²A² = ½kA²
At the centre of the oscillation all the energy is kinetic and the potential energy is zero; at the extremes the reverse is true. The total energy is proportional to the square of the amplitude, which is why doubling the amplitude of a driven system quadruples the energy it must store and release.
在振荡中心，全部能量都是动能，势能为零；在端点处则相反。总能量与振幅的平方成正比，这就是为什么把受迫系统的振幅加倍，它需要储存和释放的能量会变为四倍。
Two further results are worth remembering. The kinetic and potential energies both vary with period T/2, twice as fast as the displacement, because they depend on the square of a sinusoidal quantity. Averaged over a full cycle, the mean kinetic energy equals the mean potential energy, each being ¼kA².
还有两个值得记住的结论。动能与势能的变化周期都是 T/2，比位移快一倍，因为它们依赖于正弦量的平方。对一个完整周期取平均，平均动能等于平均势能，各为 ¼kA²。
7. The Mass-Spring Oscillator | 弹簧振子系统
a frictionless surface, the restoring force is provided entirely by the spring. Since ω = √k/m, the period follows directly and contains no dependence on g or on the amplitude.
对于连接在劲度系数为 k 的弹簧上的质量 m，在水平或竖直的无摩擦情况下振动时，回复力完全由弹簧提供。由于 ω = √k/m，周期可直接写出，且式中不包含 g，也不包含振幅。
T = 2π√m/k
This is an important difference from the simple pendulum, whose period does depend on the gravitational field strength. It also means that a mass-spring oscillator taken to the Moon would keep the same period, while a pendulum clock would run slow there.
这是它与单摆的一个重要区别，因为单摆的周期确实依赖于重力场强度。这也意味着把弹簧振子带到月球上，其周期不变，而摆钟在月球上会走慢。
. When displaced from that new equilibrium position, the additional force is still proportional to the extra displacement, so the motion is simple harmonic about the stretched equilibrium point with the same period. The weight simply shifts the centre of oscillation.
竖直悬挂的弹簧在平衡时会伸长 e，且满足 ke = mg。若从这个新的平衡位置再拉开一段距离，增加的力仍与额外的位移成正比，因此运动是围绕这个已伸长的平衡位置进行的简谐运动，周期不变。重力只是把振荡中心整体平移了。
In practice the mass of the spring itself is neglected, and the oscillation is assumed to be undamped. For real springs, only a fraction of the spring's mass (about one third) contributes effectively to the inertia.
实际处理时，弹簧自身质量被忽略，并假设振动无阻尼。对真实弹簧而言，大约只有其质量的三分之一会有效地贡献到惯性中。
8. The Simple Pendulum | 单摆
A simple pendulum consists of a small bob of mass m suspended from a fixed point by a light, inextensible string of length l. When the bob is displaced through a small angle θ, the component of weight perpendicular to the string provides the restoring force, of magnitude mg sinθ.
单摆由一个质量 m 的小球用长为 l 的轻质不可伸长细线悬挂于固定点构成。当小球被拉开一个小角度 θ 时，重力的切向分量提供回复力，大小为 mg sinθ。
th k = mg/l, which confirms that the small-angle pendulum is a simple harmonic oscillator.
对小角度而言，弧长 x = lθ，且在弧度制下 sinθ ≈ θ，于是回复力近似为 -mgx/l。这正是 F = -kx 的形式，其中 k = mg/l，从而证明小角度单摆是简谐振子。
T = 2π√l/g
Notice what is absent from this result: the mass of the bob and, for small amplitudes, the amplitude itself. A pendulum is therefore an excellent way to measure g, since plotting T² against l gives a straight
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