Unit C.4: Standing Waves and Resonance单元 C.4：驻波与共振
The capstone wave unit of Theme C "Wave behaviour". Two identical waves travelling in opposite directions superpose into a standing wave: a pattern of fixed nodes and antinodes that transfers no net energy. From this single idea flow the harmonics of strings and pipes, the boundary conditions that pick out which frequencies a system will sound, and the phenomenon of resonance, where driving a system at its natural frequency produces a large amplitude response. The unit closes with damping and why bridges, buildings and instruments are tuned to it. C.4 builds directly on C.1 simple harmonic motion and C.2 the wave model.主题 C"波的行为"的收官波动单元。两列相同的波沿相反方向传播叠加成驻波：波节与波腹位置固定、不传递净能量的图样。由这一核心思想引出弦与管的谐波、决定系统会发出哪些频率的边界条件，以及共振现象——以系统固有频率驱动时产生大振幅响应。本单元以阻尼及桥梁、建筑、乐器为何要据此调校收尾。C.4 直接建立在 C.1 简谐运动与 C.2 波模型之上。
How to use this guide本指南使用说明
C.4 rewards picture-thinking. Almost every mark comes from drawing the correct standing-wave pattern (counting how many half-wavelengths fit the boundary) and from reading off the wavelength of each harmonic. The algebra is short once the picture is right: $v = f\lambda$ converts a wavelength into a frequency. Train the boundary conditions (node at a fixed end / closed end, antinode at a free end / open end) until they are automatic, then layer resonance and damping on top.C.4 奖励"画图思维"。几乎每一分都来自画对驻波图样（数有多少个半波长能装进边界）以及读出每个谐波的波长。图画对了，代数很短：用 $v = f\lambda$ 把波长换成频率。把边界条件（固定端/闭口端是波节，自由端/开口端是波腹）练到自动反应，再叠加共振与阻尼。
Memorise three harmonic formulas: string (both ends fixed) and open–open pipe both give $f_n = \tfrac{nv}{2L}$ for $n = 1, 2, 3, \dots$; the closed pipe (one end closed) gives $f_n = \tfrac{nv}{4L}$ for odd $n$ only. A standing wave has nodes (zero amplitude) and antinodes (maximum amplitude) at fixed positions. Resonance occurs when the driving frequency equals a natural frequency.
背熟三个谐波公式：弦（两端固定）与开–开管都给出 $f_n = \tfrac{nv}{2L}$（$n = 1, 2, 3, \dots$）；闭管（一端闭口）给出 $f_n = \tfrac{nv}{4L}$，且仅取奇数 $n$。驻波有位置固定的波节（振幅为零）与波腹（振幅最大）。当驱动频率等于固有频率时发生共振。
Be able to explain why a standing wave forms (superposition of two oppositely-travelling waves) and contrast it with a travelling wave (no net energy transfer; all points between nodes oscillate in phase but with position-dependent amplitude). Derive every harmonic from the boundary picture rather than memorising. Sketch a resonance curve and show how its peak shifts and broadens with damping (light vs heavy vs critical).
能解释驻波为何形成（两列反向传播波的叠加），并与行波对比（无净能量传递；两波节之间所有点同相振动但振幅随位置而变）。从边界图样推导每个谐波，而非死记。能画共振曲线并说明其峰值如何随阻尼（轻、重、临界）移动并变宽。
How a Standing Wave Forms驻波如何形成 C.4 SL+HL
Superposition. With $y_1 = A \sin(kx - \omega t)$ moving right and $y_2 = A \sin(kx + \omega t)$ moving left, the sum is $$ y = y_1 + y_2 = 2 A \sin(kx)\cos(\omega t). $$ The shape factor $\sin(kx)$ fixes the amplitude at each point $x$; the time factor $\cos(\omega t)$ makes every point oscillate together. The wave does not appear to move along — hence "standing".
Wave relation (data booklet). Use
v = f λ to convert between wavelength and frequency for the underlying waves.
叠加。右行波 $y_1 = A \sin(kx - \omega t)$ 与左行波 $y_2 = A \sin(kx + \omega t)$ 相加： $$ y = y_1 + y_2 = 2 A \sin(kx)\cos(\omega t). $$ 形状因子 $\sin(kx)$ 固定每个点 $x$ 的振幅；时间因子 $\cos(\omega t)$ 使所有点一起振动。波形看起来不沿直线移动——故称"驻"波。
波速关系（数据手册）。用
v = f λ 在底层波的波长与频率间换算。
| Feature特征 | Travelling wave行波 | Standing wave驻波 |
|---|---|---|
| Energy transfer能量传递 | Yes, along the wave有，沿传播方向 | No net transfer无净传递 |
| Amplitude振幅 | Same for all points各点相同 | Varies with position随位置变化 |
| Phase相位 | Varies with position随位置变化 | In phase between nodes波节间同相 |
| Waveform波形 | Moves along沿线移动 | Stays fixed in place位置固定 |
Two waves of equal amplitude and frequency travel in opposite directions along a stretched string, giving $y = 2A\sin(kx)\cos(\omega t)$. Show that the points where the string is permanently at rest are spaced half a wavelength apart.沿拉紧的弦反向传播的两列等幅同频波，合成 $y = 2A\sin(kx)\cos(\omega t)$。证明弦上始终静止的点间隔为半个波长。
Identify. A point is permanently at rest when its amplitude factor is zero: $2A\sin(kx) = 0$, independent of time.
识别。当某点振幅因子为零时该点始终静止：$2A\sin(kx) = 0$，与时间无关。
Set up. $\sin(kx) = 0$ requires $kx = n\pi$ for integer $n$, so $x = n\pi/k$.
列式。$\sin(kx) = 0$ 要求 $kx = n\pi$（$n$ 为整数），故 $x = n\pi/k$。
Spacing. Using $k = 2\pi/\lambda$, the gap between consecutive rest points is
间距。由 $k = 2\pi/\lambda$，相邻静止点之间的距离为
$$ \Delta x = \frac{(n+1)\pi}{k} - \frac{n\pi}{k} = \frac{\pi}{k} = \frac{\pi}{2\pi/\lambda} = \frac{\lambda}{2}. $$Evaluate. These permanently-at-rest points are the nodes. Adjacent nodes are $\lambda/2$ apart, a result used in every harmonic calculation that follows.
评估。这些始终静止的点即波节。相邻波节相距 $\lambda/2$，这一结论将用于后续所有谐波计算。
Going deeper: the trig identity behind $2A\sin(kx)\cos(\omega t)$ HL深入：$2A\sin(kx)\cos(\omega t)$ 背后的三角恒等式 HL
Add the two oppositely-travelling waves and apply the sum-to-product identity $\sin P + \sin Q = 2\sin\tfrac{P+Q}{2}\cos\tfrac{P-Q}{2}$:
把两列反向行波相加，用和差化积恒等式 $\sin P + \sin Q = 2\sin\tfrac{P+Q}{2}\cos\tfrac{P-Q}{2}$：
$$ A\sin(kx-\omega t) + A\sin(kx+\omega t) = 2A\sin(kx)\cos(\omega t). $$The $x$ and $t$ dependences have separated: spatial shape $\sin(kx)$ multiplied by a global oscillation $\cos(\omega t)$. A travelling wave keeps $x$ and $t$ locked together inside $\sin(kx - \omega t)$; that locking is what makes the pattern move. Separating them is exactly what freezes the pattern in place.
$x$ 与 $t$ 的依赖被分离：空间形状 $\sin(kx)$ 乘以整体振荡 $\cos(\omega t)$。行波把 $x$ 与 $t$ 锁在 $\sin(kx - \omega t)$ 内，正是这种锁定使波形移动；分离它们恰好使波形定格。
Nodes, Antinodes and No Net Energy Transfer波节、波腹与无净能量传递 C.4 SL+HL
Antinode (波腹). A point of maximum amplitude $2A$ (constructive interference). Occurs where $|\sin(kx)| = 1$, exactly halfway between two nodes.
Geometry. Adjacent nodes are $\tfrac{\lambda}{2}$ apart; adjacent antinodes are $\tfrac{\lambda}{2}$ apart; a node and its neighbouring antinode are $\tfrac{\lambda}{4}$ apart.
Energy. A standing wave transfers no net energy along its length: energy sloshes back and forth between kinetic (at the antinodes, max speed) and potential, but does not flow past a node.
Phase. All points between two adjacent nodes oscillate in phase (reach their extremes at the same instant). Points in neighbouring loops, on opposite sides of a node, are exactly antiphase ($\pi$ out of phase).
波腹（antinode）。振幅最大 $2A$ 的点（相长干涉）。出现在 $|\sin(kx)| = 1$ 处，恰在两波节正中间。
几何。相邻波节相距 $\tfrac{\lambda}{2}$；相邻波腹相距 $\tfrac{\lambda}{2}$；波节与相邻波腹相距 $\tfrac{\lambda}{4}$。
能量。驻波沿其长度不传递净能量：能量在动能（波腹处速度最大）与势能之间来回转化，但不会越过波节流动。
相位。相邻两波节之间的所有点同相振动（同一时刻到达各自极值）。相邻波腹环、位于波节两侧的点恰好反相（相差 $\pi$）。
A standing wave on a string has adjacent nodes located $0.30\ \mathrm{m}$ apart. The wave that produced it travels at $120\ \mathrm{m\,s^{-1}}$. Find the wavelength and frequency, and the distance from a node to the nearest antinode.弦上驻波相邻波节相距 $0.30\ \mathrm{m}$。产生它的波速为 $120\ \mathrm{m\,s^{-1}}$。求波长、频率，以及波节到最近波腹的距离。
Identify. Adjacent nodes are half a wavelength apart, so $\tfrac{\lambda}{2} = 0.30\ \mathrm{m}$.
识别。相邻波节相距半个波长，故 $\tfrac{\lambda}{2} = 0.30\ \mathrm{m}$。
Wavelength. $\lambda = 2 \times 0.30 = 0.60\ \mathrm{m}$.
波长。$\lambda = 2 \times 0.30 = 0.60\ \mathrm{m}$。
Frequency. From the data-booklet relation $v = f\lambda$:
频率。由数据手册关系 $v = f\lambda$：
$$ f = \frac{v}{\lambda} = \frac{120}{0.60} = 200\ \mathrm{Hz}. $$Node to antinode. Quarter of a wavelength: $\tfrac{\lambda}{4} = 0.15\ \mathrm{m}$.
波节到波腹。四分之一波长：$\tfrac{\lambda}{4} = 0.15\ \mathrm{m}$。
Evaluate. The frequency $200\ \mathrm{Hz}$ is that of the two underlying travelling waves; the standing pattern itself does not propagate.
评估。频率 $200\ \mathrm{Hz}$ 是两列底层行波的频率；驻波图样本身不传播。
Going deeper: where does the energy go?深入：能量去哪了？
In a travelling wave, energy is carried forward continuously. In a standing wave, the incident and reflected waves carry equal energy in opposite directions, so the net flow is zero. Energy is not lost, it is stored: at the instant the string is flat (all points passing through equilibrium) the energy is wholly kinetic, concentrated at the antinodes where the speed is greatest; a quarter-period later, when the string is at maximum displacement and momentarily at rest, the same energy is wholly potential (elastic). It oscillates between these forms twice per period but never crosses a node.
行波中能量被持续向前输运。驻波中入射波与反射波携带等量能量沿相反方向传播，故净流为零。能量没有损失，而是被储存：当弦恰好平直（所有点经过平衡位置）时能量全为动能，集中在速度最大的波腹处；四分之一周期后，弦达到最大位移并瞬时静止时，同一能量全为势能（弹性）。能量每周期在两种形式间振荡两次，但从不越过波节。
Strings Fixed at Both Ends: Harmonics两端固定的弦：谐波 C.4 SL+HL
Allowed wavelengths. Length $L$ must equal a whole number of half-wavelengths: $$ L = n\,\frac{\lambda_n}{2} \;\Rightarrow\; \lambda_n = \frac{2L}{n}, \qquad n = 1, 2, 3, \dots $$ Harmonic frequencies. With wave speed $v$ on the string and $v = f\lambda$, $$ f_n = \frac{nv}{2L}, \qquad n = 1, 2, 3, \dots $$ Naming. $n = 1$ is the fundamental (基频, first harmonic); $n = 2$ is the second harmonic (first overtone); and so on. All integer multiples of $f_1$ are present: $f_n = n f_1$.
允许的波长。弦长 $L$ 必须等于整数个半波长： $$ L = n\,\frac{\lambda_n}{2} \;\Rightarrow\; \lambda_n = \frac{2L}{n}, \qquad n = 1, 2, 3, \dots $$ 谐波频率。弦上波速为 $v$，由 $v = f\lambda$： $$ f_n = \frac{nv}{2L}, \qquad n = 1, 2, 3, \dots $$ 命名。$n = 1$ 为基频（fundamental，第一谐波）；$n = 2$ 为第二谐波（第一泛音 overtone）；以此类推。所有 $f_1$ 的整数倍都存在：$f_n = n f_1$。
A guitar string of length $0.65\ \mathrm{m}$ is fixed at both ends. Waves travel along it at $260\ \mathrm{m\,s^{-1}}$. Find the fundamental frequency and the frequency of the third harmonic. State the number of nodes (including the ends) in the third-harmonic pattern.两端固定的吉他弦长 $0.65\ \mathrm{m}$，波在弦上以 $260\ \mathrm{m\,s^{-1}}$ 传播。求基频与第三谐波频率，并说明第三谐波图样中波节数（含两端）。
Identify. Both ends fixed, so use $f_n = \tfrac{nv}{2L}$ with $L = 0.65\ \mathrm{m}$, $v = 260\ \mathrm{m\,s^{-1}}$.
识别。两端固定，用 $f_n = \tfrac{nv}{2L}$，$L = 0.65\ \mathrm{m}$、$v = 260\ \mathrm{m\,s^{-1}}$。
Fundamental ($n = 1$).
基频（$n = 1$）。
$$ f_1 = \frac{(1)(260)}{2(0.65)} = \frac{260}{1.30} = 200\ \mathrm{Hz}. $$Third harmonic ($n = 3$).
第三谐波（$n = 3$）。
$$ f_3 = 3 f_1 = 3 \times 200 = 600\ \mathrm{Hz}. $$Nodes. The $n$-th harmonic has $n$ loops and therefore $n + 1$ nodes counting both ends: the third harmonic has $4$ nodes.
波节数。第 $n$ 谐波有 $n$ 个环，故含两端共 $n + 1$ 个波节：第三谐波有 $4$ 个波节。
Evaluate. Harmonics on a fixed-fixed string are all integer multiples of the fundamental, which is what gives a plucked string its musical pitch and rich timbre.
评估。两端固定弦的谐波都是基频的整数倍，这正是拨弦产生确定音高与丰富音色的原因。
Going deeper: wave speed on a string and tuning HL深入：弦上波速与调音 HL
The wave speed on a stretched string depends on the tension $T$ and the mass per unit length $\mu$:
拉紧弦上的波速取决于张力 $T$ 与线密度 $\mu$：
$$ v = \sqrt{\frac{T}{\mu}} \quad\Rightarrow\quad f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}. $$This explains tuning: increasing tension (tightening the peg) raises $v$ and hence $f_1$; a thicker, heavier string (larger $\mu$) lowers it, which is why a guitar's bass strings are wound thick. Pressing a fret shortens $L$ and raises the pitch. The exact $v = \sqrt{T/\mu}$ formula is beyond the C.4 statement, but recognising that pitch rises with tension and falls with length and mass is expected reasoning.
这解释了调音：增大张力（拧紧弦钮）提高 $v$，从而提高 $f_1$；更粗更重的弦（$\mu$ 更大）降低它，这正是吉他低音弦缠绕较粗的原因。按品减小 $L$ 使音调升高。$v = \sqrt{T/\mu}$ 的精确公式超出 C.4 陈述，但"音调随张力升高、随长度与质量降低"是要求掌握的推理。
Open and Closed Pipes开管与闭管 C.4 SL+HL
Open–open pipe (both ends open). Antinode at each end. Like the string, $$ \lambda_n = \frac{2L}{n}, \qquad f_n = \frac{nv}{2L}, \qquad n = 1, 2, 3, \dots $$ All harmonics present.
Open–closed pipe (one end closed). Antinode at the open end, node at the closed end. The shortest fit is a quarter wavelength: $$ \lambda_n = \frac{4L}{n}, \qquad f_n = \frac{nv}{4L}, \qquad n = 1, 3, 5, \dots \ (\text{odd only}). $$ Key consequence. A closed pipe sounds only odd harmonics and its fundamental is half that of an open pipe of the same length.
开–开管（两端开口）。两端各为波腹。与弦相同， $$ \lambda_n = \frac{2L}{n}, \qquad f_n = \frac{nv}{2L}, \qquad n = 1, 2, 3, \dots $$ 所有谐波都存在。
开–闭管（一端闭口）。开口端为波腹，闭口端为波节。最短能装下的是四分之一波长： $$ \lambda_n = \frac{4L}{n}, \qquad f_n = \frac{nv}{4L}, \qquad n = 1, 3, 5, \dots\ (\text{仅奇数})。 $$ 关键结论。闭管只发出奇数谐波，且其基频为同长开管的一半。
An organ pipe closed at one end is $0.85\ \mathrm{m}$ long. The speed of sound in air is $340\ \mathrm{m\,s^{-1}}$. Find its fundamental frequency and the frequency of the next harmonic it can produce. What would the fundamental be if the pipe were open at both ends?一端闭口的风琴管长 $0.85\ \mathrm{m}$。空气中声速 $340\ \mathrm{m\,s^{-1}}$。求其基频与它能产生的下一个谐波频率。若该管两端开口，基频为何？
Identify. Closed pipe, so $f_n = \tfrac{nv}{4L}$ with odd $n$ only; $L = 0.85\ \mathrm{m}$, $v = 340\ \mathrm{m\,s^{-1}}$.
识别。闭管，故 $f_n = \tfrac{nv}{4L}$ 且 $n$ 仅取奇数；$L = 0.85\ \mathrm{m}$、$v = 340\ \mathrm{m\,s^{-1}}$。
Fundamental ($n = 1$).
基频（$n = 1$）。
$$ f_1 = \frac{(1)(340)}{4(0.85)} = \frac{340}{3.40} = 100\ \mathrm{Hz}. $$Next harmonic ($n = 3$). The closed pipe skips $n = 2$:
下一谐波（$n = 3$）。闭管跳过 $n = 2$：
$$ f_3 = 3 f_1 = 300\ \mathrm{Hz}. $$If open–open. $f_1^{\text{open}} = \tfrac{v}{2L} = \tfrac{340}{1.70} = 200\ \mathrm{Hz}$, double the closed-pipe fundamental.
若开–开管。$f_1^{\text{open}} = \tfrac{v}{2L} = \tfrac{340}{1.70} = 200\ \mathrm{Hz}$，是闭管基频的两倍。
Evaluate. Closing one end lowers the pitch by an octave and removes all even harmonics, giving the closed pipe its characteristic hollow, "woody" timbre.
评估。封闭一端使音调降低一个八度，并去除所有偶数谐波，赋予闭管特有的空、"木"音色。
Going deeper: why a closed end is a node and an open end an antinode深入：闭口端为何是波节、开口端为何是波腹
A sound wave in a pipe is a longitudinal displacement of air. At a closed end the rigid wall stops the air from moving, forcing the displacement to be zero there — a displacement node. At an open end the air is free to oscillate with maximum amplitude — a displacement antinode. (In terms of pressure the roles swap: a closed end is a pressure antinode and an open end a pressure node, because where the air cannot move the pressure varies most.) For C.4 work in displacement: node at closed, antinode at open. A real open end has a small "end correction" so the effective length is slightly longer than $L$, but the IB syllabus ignores this.
管中声波是空气的纵向位移。在闭口端刚性壁阻止空气运动，使该处位移为零——位移波节。在开口端空气可自由以最大振幅振动——位移波腹。（就压强而言角色互换：闭口端是压强波腹，开口端是压强波节，因为空气不能移动处压强变化最大。）C.4 用位移描述：闭口为波节，开口为波腹。真实开口端有微小"端修正"，有效长度略大于 $L$，但 IB 大纲忽略此项。
Natural Frequency, Driving and Resonance固有频率、受迫与共振 C.4 SL+HL
Driven (forced) oscillation (受迫振动). When a periodic external force of frequency $f_d$ is applied, the system settles into oscillating at the driving frequency $f_d$, not its own $f_0$.
Resonance (共振). When the driving frequency matches a natural frequency, $f_d = f_0$, energy is transferred most efficiently from driver to system and the amplitude grows to a sharp maximum.
Resonance curve. A plot of steady-state amplitude vs driving frequency. It peaks at $f_d = f_0$ and falls away on either side. Lighter damping gives a taller, narrower peak.
受迫（驱动）振动（forced oscillation）。施加频率为 $f_d$ 的周期性外力时，系统最终以驱动频率 $f_d$ 振动，而非自身的 $f_0$。
共振（resonance）。当驱动频率等于固有频率 $f_d = f_0$ 时，能量从驱动者向系统传递效率最高，振幅增长到尖锐的最大值。
共振曲线。稳态振幅对驱动频率的图。它在 $f_d = f_0$ 处达到峰值，两侧下降。阻尼越轻，峰越高越窄。
A length of air column closed at one end has its first resonance when driven by a tuning fork of frequency $256\ \mathrm{Hz}$ (speed of sound $340\ \mathrm{m\,s^{-1}}$). (a) Find the length of the air column at first resonance. (b) Explain what is observed if the driving frequency is increased slightly above $256\ \mathrm{Hz}$.一端闭口的空气柱在频率 $256\ \mathrm{Hz}$ 的音叉驱动下出现第一次共振（声速 $340\ \mathrm{m\,s^{-1}}$）。(a) 求第一次共振时空气柱长度。(b) 说明若驱动频率略高于 $256\ \mathrm{Hz}$ 会观察到什么。
(a) Identify. First resonance of a closed column is the fundamental: $L = \tfrac{\lambda}{4}$.
(a) 识别。闭口空气柱第一次共振即基频：$L = \tfrac{\lambda}{4}$。
Wavelength. $\lambda = \tfrac{v}{f} = \tfrac{340}{256} \approx 1.328\ \mathrm{m}$.
波长。$\lambda = \tfrac{v}{f} = \tfrac{340}{256} \approx 1.328\ \mathrm{m}$。
$$ L = \frac{\lambda}{4} = \frac{1.328}{4} \approx 0.332\ \mathrm{m}. $$(b) Off resonance. When $f_d \neq f_0$ the system is no longer driven at its natural frequency, so the loudness (amplitude) drops sharply: the air column no longer resonates strongly with the fork. To restore resonance you would shorten the column so its $f_0$ rises to match the new $f_d$.
(b) 偏离共振。当 $f_d \neq f_0$ 时系统不再以固有频率被驱动，响度（振幅）急剧下降：空气柱不再与音叉强烈共振。要恢复共振需缩短空气柱，使其 $f_0$ 升高以匹配新的 $f_d$。
Evaluate. This is exactly the resonance-tube experiment: at the resonant length the sound is loudest, confirming $f_d = f_0$.
评估。这正是共振管实验：在共振长度处声音最响，证实 $f_d = f_0$。
Going deeper: amplitude and phase across the resonance peak HL深入：共振峰两侧的振幅与相位 HL
For a lightly-damped driven oscillator the steady-state amplitude is largest near $f_d = f_0$. The phase of the oscillation relative to the driving force shifts continuously: well below resonance the system moves nearly in phase with the driver; at resonance it lags by exactly a quarter cycle ($90^{\circ}$); well above resonance it is nearly antiphase ($180^{\circ}$). The $90^{\circ}$ lag at resonance is why energy input is maximised — the driving force is always pushing in the direction the system is moving. This phase behaviour is beyond the bare C.4 statement but is exactly the reasoning that distinguishes a 7-level answer.
对轻阻尼受迫振子，稳态振幅在 $f_d = f_0$ 附近最大。振动相对驱动力的相位连续变化：远低于共振时系统几乎与驱动同相；共振时落后恰好四分之一周期（$90^{\circ}$）；远高于共振时近乎反相（$180^{\circ}$）。共振处 $90^{\circ}$ 的落后正是能量输入最大化的原因——驱动力始终沿系统运动方向推动。此相位行为超出 C.4 基本陈述，却正是区分 7 分答案的推理。
Damping and the Resonance Peak阻尼与共振峰 C.4 SL+HL
Three regimes.
- Light damping. Amplitude decays slowly; many oscillations occur. The resonance curve has a tall, narrow peak.
- Heavy (over) damping. The system returns to equilibrium slowly without oscillating. The resonance peak is low and very broad.
- Critical damping. The system returns to equilibrium in the shortest time without overshooting (no oscillation).
Real examples. Musical instruments use light damping (long, ringing notes). Car suspensions and door closers use near-critical damping. Bridges and buildings are deliberately damped so wind or footfall cannot drive a destructive resonance.
三种情形。
- 轻阻尼。振幅缓慢衰减，发生多次振荡。共振曲线峰高而窄。
- 重（过）阻尼。系统缓慢回到平衡而不振荡。共振峰低且很宽。
- 临界阻尼。系统在最短时间内回到平衡且不过冲（不振荡）。
实际例子。乐器用轻阻尼（音长、余韵悠扬）。汽车悬挂与闭门器用接近临界的阻尼。桥梁与建筑被刻意阻尼，使风或脚步无法激起破坏性共振。
Two identical pendulums are driven over a range of frequencies; pendulum X swings in air, pendulum Y has a card attached to increase air resistance. Sketch and compare their resonance curves, and state which would be preferred for a clock and which for a building's vibration absorber.两个相同的摆在一段频率范围内被驱动；摆 X 在空气中摆动，摆 Y 加一张卡片以增大空气阻力。画出并比较它们的共振曲线，并说明钟表与建筑减振器分别更适合哪一个。
Identify. Pendulum Y has greater damping than X (the card adds air resistance).
识别。摆 Y 的阻尼大于 X（卡片增加空气阻力）。
Curves. Both peak near the same natural frequency $f_0$. X (light damping) has a tall, narrow peak; Y (heavier damping) has a lower, broader peak whose maximum sits at a slightly lower frequency than X's.
曲线。两者都在大致相同的固有频率 $f_0$ 附近达峰。X（轻阻尼）峰高而窄；Y（较重阻尼）峰低而宽，其最大值所在频率略低于 X。
Choice. A clock wants a sharp, stable, long-lasting oscillation, so it uses light damping (like X). A building's vibration absorber wants to dissipate energy quickly and avoid a sharp resonant build-up, so it uses heavy / near-critical damping (like Y).
选择。钟表需要尖锐、稳定、持久的振荡，故用轻阻尼（如 X）。建筑减振器需快速耗散能量、避免尖锐的共振积累，故用重/接近临界阻尼（如 Y）。
Evaluate. Damping does not change where resonance roughly occurs, but it controls how violently the system responds at resonance — the key design lever for any structure exposed to periodic forcing.
评估。阻尼不改变共振大致发生的位置，却决定系统在共振时响应的剧烈程度——这是任何承受周期驱动的结构的关键设计杠杆。
Going deeper: why critical damping is the fast return HL深入：临界阻尼为何回归最快 HL
Imagine releasing a displaced system. With light damping it overshoots and oscillates many times before settling — slow to truly rest. With heavy damping it creeps back without overshooting, but the creep is sluggish — also slow. Critical damping is the exact boundary between these: the system returns to equilibrium in the minimum possible time without overshooting. This is why instruments such as galvanometers and analogue meter needles, and car suspensions, are designed close to critical damping: you want the needle (or the wheel) to settle fast without bouncing. A tiny bit under critical is often chosen so the response is just-barely non-oscillatory.
设想释放一个被偏移的系统。轻阻尼时它过冲并振荡多次才稳定——真正静止很慢。重阻尼时它不过冲地缓慢爬回，但爬得迟缓——也慢。临界阻尼正是二者的精确分界：系统在不过冲的前提下以最短时间回到平衡。这正是检流计、模拟仪表指针以及汽车悬挂被设计接近临界阻尼的原因：你希望指针（或车轮）快速稳定而不反弹。常选略低于临界，使响应恰好不振荡。
Exam Strategy and Common Pitfalls考试策略与常见陷阱
- Sketch the standing-wave shape before reaching for a formula. Mark the node/antinode at each boundary, then count how many half- or quarter-wavelengths fit $L$.
- 动笔套公式前先画驻波形状。在每个边界标出波节/波腹，再数有多少个半波长或四分之一波长能装进 $L$。
- Read the wavelength straight off the picture. Then convert with $v = f\lambda$. This avoids almost every harmonic-number slip.
- 直接从图样读出波长。再用 $v = f\lambda$ 换算。这能避免几乎所有谐波编号错误。
- String (both ends fixed) and open–open pipe both use $f_n = \tfrac{nv}{2L}$, all $n$.
- 弦（两端固定）与开–开管都用 $f_n = \tfrac{nv}{2L}$，所有 $n$。
- Closed pipe uses $f_n = \tfrac{nv}{4L}$ with odd $n$ only. Its "next harmonic" above the fundamental is $3f_1$, never $2f_1$.
- 闭管用 $f_n = \tfrac{nv}{4L}$ 且仅取奇数 $n$。其基频之上"下一个谐波"是 $3f_1$，绝不是 $2f_1$。
- Standing wave: no net energy transfer, position-dependent amplitude, in phase between nodes. Quote all three to secure the marks.
- 驻波：无净能量传递、振幅随位置变化、波节间同相。三点都写出以拿满分。
- Do not say points have "different phases" within a loop. They have different amplitudes but the same phase.
- 不要说一个环内各点"相位不同"。它们振幅不同但相位相同。
- State the resonance condition $f_d = f_0$ explicitly and that a driven system oscillates at $f_d$, not $f_0$.
- 明确写出共振条件 $f_d = f_0$，并说明受迫系统以 $f_d$（而非 $f_0$）振动。
- When asked about damping, describe both axes of the resonance curve: increasing damping lowers AND broadens the peak (and nudges it below $f_0$).
- 问及阻尼时，描述共振曲线的两个方面：阻尼增大使峰既降低又展宽（并略移到 $f_0$ 之下）。
Flashcards闪卡
Unit C.4 Practice Quiz单元 C.4 练习测验
Readiness Checklist备考清单
Tick each item when you can do it cold, without notes, on your first attempt.
每一条都要"裸做"做对（不看笔记、一次过）才打勾。
- Explain how a standing wave forms from two identical oppositely-travelling waves解释驻波如何由两列相同的反向行波形成
- List three ways a standing wave differs from a travelling wave (energy, amplitude, phase)列出驻波与行波的三点区别（能量、振幅、相位）
- Locate nodes and antinodes and state node-to-node ($\tfrac{\lambda}{2}$) and node-to-antinode ($\tfrac{\lambda}{4}$) spacing定位波节与波腹，并说明波节间距 ($\tfrac{\lambda}{2}$) 与波节–波腹间距 ($\tfrac{\lambda}{4}$)
- State that a standing wave transfers no net energy and that points between nodes are in phase说明驻波无净能量传递、波节间各点同相
- Use $f_n = \tfrac{nv}{2L}$ for a string fixed at both ends and identify the fundamental and overtones对两端固定弦用 $f_n = \tfrac{nv}{2L}$，并辨认基频与泛音
- Count nodes and loops for the $n$-th harmonic on a string数出弦上第 $n$ 谐波的波节数与环数
- Apply node-at-closed / antinode-at-open boundary conditions to pipes对管应用闭口波节/开口波腹的边界条件
- Use $f_n = \tfrac{nv}{4L}$ (odd $n$) for a closed pipe and explain why even harmonics are absent对闭管用 $f_n = \tfrac{nv}{4L}$（奇数 $n$），并解释偶数谐波为何缺失
- Distinguish natural frequency, driven oscillation, and resonance ($f_d = f_0$)区分固有频率、受迫振动与共振（$f_d = f_0$）
- Sketch a resonance curve and mark its peak at the natural frequency画共振曲线并标出固有频率处的峰
- Describe light, heavy and critical damping and their effect on the resonance peak描述轻、重、临界阻尼及其对共振峰的影响
- Give real examples of useful and destructive resonance (instruments, bridges, buildings)举出有益与破坏性共振的实例（乐器、桥梁、建筑）
IB Paper-Style PracticeIB 试卷风格练习
C.4 Practice and Solutions are on the roadmap, to ship under Practice Questions/Unit_C4_*.html with the bilingual built-in pattern.
C.4 配套 Practice 与 Solutions 在排期，上线后位于 Practice Questions/Unit_C4_*.html。