A-Level物理 简谐运动 弹簧振子 单摆

A-Level Physics Simple Harmonic Motion

Simple Harmonic Motion (SHM) is one of the most fundamental periodic motions in classical mechanics. It describes systems where the restoring force is directly proportional to the displacement from equilibrium and acts in the opposite direction. From the swing of a pendulum to the vibration of atoms in a crystal lattice, SHM underpins countless physical phenomena and appears in every major A-Level physics syllabus.

简谐运动（SHM）是经典力学中最基本的周期运动之一。它描述恢复力与偏离平衡位置的位移成正比且方向相反的系统。从单摆的摆动到晶格中原子的振动，简谐运动构成了无数物理现象的基础，出现在所有主要A-Level物理大纲中。

Mathematically, SHM is defined by the differential equation d²x/dt² = -(k/m)x, where x is displacement, k is the spring constant or equivalent restoring-force coefficient, and m is the mass of the oscillating object. The negative sign encodes the essential physics: the acceleration always points back toward equilibrium. This second-order linear differential equation has sinusoidal solutions, which is why SHM is intimately connected to circular motion and trigonometric functions.

数学上，简谐运动由微分方程 d²x/dt² = -(k/m)x 定义，其中 x 是位移，k 是弹簧常数或等效的恢复力系数，m 是振动物体的质量。负号编码了核心物理：加速度始终指向平衡位置。这个二阶线性微分方程有正弦解，这就是为什么简谐运动与圆周运动和三角函数紧密相连。

1. Defining Characteristics of SHM

For a system to exhibit true SHM, two conditions must be satisfied. First, the restoring force F must obey Hooke’s Law in its general form: F = -kx. Second, the acceleration a must be proportional to displacement: a = -(k/m)x. These are equivalent statements: multiply F = -kx by 1/m to get a = -(k/m)x. The proportionality constant k/m equals the square of the angular frequency ω: ω² = k/m. This relationship is the gateway to all SHM kinematics.

系统要表现出真正的简谐运动，必须满足两个条件。第一，恢复力 F 必须遵循胡克定律的普遍形式：F = -kx。第二，加速度 a 必须与位移成正比：a = -(k/m)x。这两个陈述是等价的：将 F = -kx 乘以 1/m 得到 a = -(k/m)x。比例常数 k/m 等于角频率的平方：ω² = k/m。这个关系是通往所有简谐运动运动学的门户。

Key quantities in SHM include amplitude A (maximum displacement from equilibrium), period T (time for one complete oscillation), frequency f = 1/T, angular frequency ω = 2πf, and phase constant φ (determining initial conditions). The displacement as a function of time is x(t) = A sin(ωt) if the object starts at equilibrium moving upward, or x(t) = A cos(ωt) if released from maximum positive displacement. On A-Level exam papers, identifying which trig function to use from the initial conditions is a core skill.

简谐运动中的关键量包括振幅 A（偏离平衡的最大位移）、周期 T（一次完整振动所需时间）、频率 f = 1/T、角频率 ω = 2πf 以及相常数 φ（决定初始条件）。位移作为时间的函数为 x(t) = A sin(ωt)（如果物体从平衡位置开始向上运动）或 x(t) = A cos(ωt)（如果从最大正位移释放）。在A-Level考试中，从初始条件判断使用哪个三角函数是一项核心技能。

2. The Spring-Mass Oscillator

The classic spring-mass system consists of a mass m attached to a spring of stiffness k on a frictionless horizontal surface. Pull the mass a distance A from its equilibrium length and release it, and it oscillates back and forth with angular frequency ω = √(k/m). The period, which is independent of amplitude, is T = 2π√(m/k). This amplitude-independence is called isochronism and is a unique property of linear restoring forces.

经典的弹簧振子由无摩擦水平面上连接在刚度为 k 的弹簧上的质量 m 组成。将质量拉到距离平衡长度 A 处释放，它将以角频率 ω = √(k/m) 来回振动。周期与幅值无关，为 T = 2π√(m/k)。振幅无关性称为等时性，是线性恢复力的独特性质。

A standard A-Level calculation: a 0.50 kg mass on a spring with k = 200 N/m oscillates with amplitude 4.0 cm. Find (a) the period, (b) the maximum speed, and (c) the maximum acceleration. Solution: (a) T = 2π√(0.50/200) = 2π√(0.0025) = 0.314 s. (b) v_max = ωA = (2π/T)A = (2π/0.314)(0.040) = 0.80 m/s. Alternatively, using energy: (1/2)mv²_max = (1/2)kA² gives v_max = A√(k/m) = 0.040√(200/0.50) = 0.80 m/s. (c) a_max = ω²A = (k/m)A = (200/0.50)(0.040) = 16 m/s².

一个标准A-Level计算：0.50 kg 的质量连接在 k = 200 N/m 的弹簧上，以振幅 4.0 cm 振动。求 (a) 周期，(b) 最大速度，(c) 最大加速度。解：(a) T = 2π√(0.50/200) = 2π√(0.0025) = 0.314 s。(b) v_max = ωA = (2π/T)A = (2π/0.314)(0.040) = 0.80 m/s。另一种方法，用能量：(1/2)mv²_max = (1/2)kA² 得 v_max = A√(k/m) = 0.040√(200/0.50) = 0.80 m/s。(c) a_max = ω²A = (k/m)A = (200/0.50)(0.040) = 16 m/s²。

Energy continuously transforms between kinetic and potential forms. At equilibrium, all energy is kinetic: E_k = (1/2)mv²_max = (1/2)kA². At maximum displacement, all energy is stored as elastic potential: E_p = (1/2)kA². At any intermediate position x, the total energy is conserved: E_total = (1/2)mv² + (1/2)kx² = (1/2)kA². This energy conservation is the quickest route to finding velocity at any given displacement: v = ±ω√(A² – x²).

能量在动能和势能之间不断转化。在平衡位置，所有能量为动能：E_k = (1/2)mv²_max = (1/2)kA²。在最大位移处，所有能量储存为弹性势能：E_p = (1/2)kA²。在任意中间位置 x，总能量守恒：E_total = (1/2)mv² + (1/2)kx² = (1/2)kA²。这个能量守恒是求任意位移处速度的最快捷径：v = ±ω√(A² – x²)。

3. The Simple Pendulum

A simple pendulum is a point mass m suspended from a fixed point by a light inextensible string of length L. When displaced by a small angle θ (typically less than about 10°), the restoring force is mg sin θ. For small angles, sin θ ≈ θ (in radians), giving the linear restoring force F = -(mg/L)s, where s = Lθ is the arc-length displacement. This satisfies SHM with ω = √(g/L) and period T = 2π√(L/g).

单摆是一个点质量 m，用长度为 L 的轻质不可伸长细绳悬挂在固定点上。当偏离小角度 θ（通常小于约10°）时，恢复力为 mg sin θ。对于小角度，sin θ ≈ θ（以弧度计），给出线性恢复力 F = -(mg/L)s，其中 s = Lθ 是弧长位移。这满足简谐运动，ω = √(g/L)，周期 T = 2π√(L/g)。

The small-angle approximation is critical: the pendulum is only approximately SHM. For an initial angle of 5° (0.087 rad), sin(0.087) = 0.08699, an error of only 0.01%. At 20°, the error reaches 2%, and at 90°, the pendulum is not even close to SHM. Exam questions frequently test this boundary: “State the condition under which a simple pendulum undergoes SHM.” The answer: “The angular displacement must be small (less than approximately 10°) so that sin θ ≈ θ.”

小角度近似是关键：单摆只是近似简谐运动。对于 5°（0.087 rad）的初始角度，sin(0.087) = 0.08699，误差仅为 0.01%。在 20° 时，误差达到 2%，而在 90° 时，单摆完全不是简谐运动。考试题目常测试这一边界：”说明单摆作简谐运动的条件。”答案是：”角位移必须很小（小于约10°），使得 sin θ ≈ θ。”

A classic practical exam question involves determining g using a pendulum. By measuring the period T for various lengths L and plotting T² against L, the gradient equals 4π²/g, from which g can be calculated. A well-designed experiment should include at least five different lengths, measure the time for 20 oscillations (reducing human reaction-time error), and repeat each measurement three times. The uncertainty in g can be calculated from the uncertainty in the gradient using standard error-propagation techniques.

经典的实验考题涉及用单摆测定 g。通过测量不同长度 L 的周期 T 并作 T²-L 图，斜率等于 4π²/g，由此可计算出 g。一个设计良好的实验应包括至少五个不同长度，每次测量20次振动的时间（减少人为反应时间误差），并重复每次测量三次。g 的不确定度可由斜率的不确定度用标准误差传递技术计算。

4. Velocity, Acceleration, and Phase Relationships

In SHM, displacement x, velocity v, and acceleration a are all sinusoidal functions of time with the same angular frequency ω, but they differ in phase. If x = A cos(ωt), then v = dx/dt = -Aω sin(ωt) = Aω cos(ωt + π/2), and a = d²x/dt² = -Aω² cos(ωt) = Aω² cos(ωt + π). Velocity leads displacement by π/2 (90°), and acceleration leads displacement by π (180°), meaning a is always opposite in sign to x: a = -ω²x.

在简谐运动中，位移 x、速度 v 和加速度 a 都是具有相同角频率 ω 的时间正弦函数，但它们的相位不同。若 x = A cos(ωt)，则 v = dx/dt = -Aω sin(ωt) = Aω cos(ωt + π/2)，而 a = d²x/dt² = -Aω² cos(ωt) = Aω² cos(ωt + π)。速度领先位移 π/2（90°），加速度领先位移 π（180°），意味着 a 始终与 x 异号：a = -ω²x。

This phase structure has physical meaning. At t = 0, the mass is at maximum positive displacement (x = +A). Velocity is zero because the mass has momentarily stopped before reversing direction. Acceleration is at its most negative (a = -Aω²) because the restoring force is maximal and pointing left. At t = T/4, one quarter through the cycle, the mass sweeps through equilibrium: x = 0, v = -Aω (maximum speed toward negative direction), a = 0 (no net force at equilibrium). Drawing the corresponding vector diagrams on a phasor circle is a powerful visual tool.

这种相位结构有物理意义。在 t = 0，质量处于最大正位移（x = +A）。速度为零，因为质量在反转方向前瞬间停止。加速度处于最负值（a = -Aω²），因为恢复力最大且指向左方。在 t = T/4，即周期的四分之一处，质量经过平衡位置：x = 0，v = -Aω（向负方向的最大速度），a = 0（平衡位置净力为零）。在相量圆上画出相应的矢量图是一个强大的可视化工具。

5. Damped and Forced Oscillations

In reality, no oscillating system is perfectly isolated. Damping forces, usually proportional to velocity (F_damp = -bv), remove energy from the system. Light damping (b small) produces oscillations whose amplitude decays exponentially: A(t) = A₀e^(-γt), where γ = b/(2m) is the damping coefficient. The angular frequency also shifts slightly: ω_damped = √(ω₀² – γ²). Heavy damping (γ > ω₀) prevents any oscillation: the system crawls back to equilibrium without overshooting.

现实中，没有任何振动系统是完美隔离的。阻尼力通常与速度成正比（F_damp = -bv），从系统中移除能量。轻阻尼（b 较小）产生的振动幅度呈指数衰减：A(t) = A₀e^(-γt)，其中 γ = b/(2m) 是阻尼系数。角频率也会略微偏移：ω_damped = √(ω₀² – γ²)。重阻尼（γ > ω₀）阻止任何振动：系统爬回平衡位置而不超调。

Critical damping (γ = ω₀) is the special case where the system returns to equilibrium in the shortest possible time without oscillating. This is the design goal for car shock absorbers, door-closing mechanisms, and galvanometer needles: you want rapid return with zero overshoot. Forced oscillations occur when an external periodic driving force F = F₀ cos(ω_d t) is applied. The system oscillates at the driving frequency ω_d, not its natural frequency ω₀.

临界阻尼（γ = ω₀）是系统在不振动的情况下以最短时间返回平衡位置的特例。这是汽车减震器、关门机构和检流计指针的设计目标：需要快速返回且零超调。受迫振动发生在施加外部周期性驱动力 F = F₀ cos(ω_d t) 时。系统以驱动频率 ω_d 振动，而非其固有频率 ω₀。

6. Resonance

Resonance occurs when the driving frequency ω_d approaches the natural frequency ω₀. The amplitude of forced oscillations becomes dramatically large, limited only by the damping present. For a lightly damped system, the resonance peak is narrow and tall; for a heavily damped system, it is broad and shallow. The resonant angular frequency, where amplitude is maximized, is ω_res = √(ω₀² – 2γ²), which is slightly less than ω₀ for nonzero damping.

共振现象发生在驱动频率 ω_d 接近固有频率 ω₀ 时。受迫振动的幅度变得极大，仅受现有阻尼的限制。对于轻阻尼系统，共振峰高而窄；对于重阻尼系统，共振峰低而宽。振幅最大的共振角频率为 ω_res = √(ω₀² – 2γ²)，对于非零阻尼，这略小于 ω₀。

Resonance has both destructive and constructive manifestations. The 1940 collapse of the Tacoma Narrows Bridge is the canonical cautionary tale: wind-induced vortex shedding matched the bridge’s torsional natural frequency, driving the amplitude until structural failure. Conversely, resonance is harnessed in MRI machines (nuclear magnetic resonance), quartz crystal oscillators in watches, and musical instruments where air columns resonate at specific frequencies to produce notes. A-Level exam questions frequently ask students to explain both beneficial and harmful examples of resonance.

共振有破坏性和建设性两种表现。1940年塔科马海峡大桥的坍塌是经典的警示故事：风致涡旋脱落匹配了桥梁的扭转固有频率，驱动振幅直到结构失效。相反，共振被用于MRI机器（核磁共振）、手表中的石英晶体振荡器，以及乐器中气柱在特定频率共振产生音符。A-Level考题经常要求学生解释共振的有益和有害例子。

7. Exam Tips for SHM

When solving SHM problems, always start by writing down the known quantities and identifying the unknown. Convert all units to SI before substituting into formulas. Check whether the system is starting from equilibrium or maximum displacement to choose between sine and cosine. For pendulum problems, remember that T = 2π√(L/g) applies only for small angles; the question will usually state “small amplitude” or give an angle under 10°.

解简谐运动问题时，总是从写下已知量和识别未知量开始。代入公式前将所有单位转换为SI制。根据系统是从平衡位置还是最大位移出发，选择正弦或余弦。对于单摆问题，记住 T = 2π√(L/g) 仅适用于小角度；题目通常会说明”小振幅”或给出小于10°的角度。

Energy methods are often faster than kinematic ones. If a question asks for speed at a given displacement, use v = ±ω√(A² – x²) rather than differentiating x(t). When dealing with vertical spring-mass systems, note that gravity simply shifts the equilibrium position downward by mg/k but does not change the oscillation frequency. This is a common trick: the period for a vertical spring is the same as for a horizontal one because ω = √(k/m) is independent of gravity.

能量方法通常比运动学方法更快。如果问题要求某位移处的速度，用 v = ±ω√(A² – x²) 而不用对 x(t) 求导。处理竖直弹簧振子时，注意重力只是将平衡位置向下移了 mg/k 但并不改变振动频率。这是一个常见的陷阱：竖直弹簧的周期与水平弹簧相同，因为 ω = √(k/m) 与重力无关。

For graphs, practice sketching displacement-time, velocity-time, acceleration-time, and energy-time graphs on the same time axis. Mark key moments (t = 0, T/4, T/2, 3T/4, T) clearly. The velocity graph is the gradient of the displacement graph; the acceleration graph is the gradient of the velocity graph. These gradient relationships are frequently tested in multi-choice and structured questions.

对于图像，练习在同一时间轴上画出位移-时间、速度-时间、加速度-时间和能量-时间图。清楚标记关键时刻（t = 0，T/4，T/2，3T/4，T）。速度图是位移图的斜率；加速度图是速度图的斜率。这些斜率关系经常在选择题和结构题中考查。