Topic 4 波动 / Waves
简谐运动 · 波动性质 · 波动行为 · 驻波 · 多普勒效应
IB SL / HLTopic 4波动是能量传递的重要形式
波动是能量传递的重要形式，广泛存在于物理学的各个分支。本章从简谐运动（SHM）出发——这是所有周期性运动的基础模型，进而学习波动的描述（波长、频率、波速、振幅），探讨波的行为（反射、折射、衍射），最后学习驻波的形成和多普勒效应。波动理论是理解声学、光学和现代物理学的桥梁。
Waves are a fundamental means of energy transfer across all branches of physics. This chapter begins with simple harmonic motion (SHM) — the basic model for all periodic motion — then covers wave properties (wavelength, frequency, wave speed, amplitude), wave behavior (reflection, refraction, diffraction), and finally standing waves and the Doppler effect. Wave theory bridges acoustics, optics, and modern physics.
4.1 简谐运动 / Simple Harmonic Motion
IB 4.1核心基础圆周运动投影推导
简谐运动的定义 / Definition of SHM
$$a = -\omega^2 x$$
简谐运动（SHM）的定义：加速度与位移大小成正比，方向总是指向平衡位置。数学表达式为 $a = -\omega^2 x$，其中 $\omega$ 是角频率（单位 rad/s），$x$ 是距平衡位置的位移。负号表示加速度方向与位移方向相反（始终指向平衡位置）。
Simple Harmonic Motion (SHM) is defined as motion where acceleration is proportional to displacement and always directed toward the equilibrium position: $a = -\omega^2 x$, where $\omega$ is the angular frequency (rad/s) and $x$ is the displacement from equilibrium. The negative sign means acceleration opposes displacement (always toward equilibrium).
📐 详细推导：从匀速圆周运动的投影推导 SHM / Derivation: SHM from Circular Motion Projection
从圆周运动投影推导 $a = -\omega^2 x$：
考虑一个质点在半径为 $A$ 的圆周上以恒定角速度 $\omega$ 运动：
① 质点位置在圆心到圆周某点的连线上投影就是 SHM 的位移：
$x = A\cos(\omega t + \phi)$，其中 $\phi$ 为初相位
② 圆周运动中，向心加速度为 $a_c = \omega^2 A$，方向指向圆心
③ 加速度在 $x$ 方向上的投影：
$a_x = -a_c\cos(\omega t + \phi) = -\omega^2 A\cos(\omega t + \phi)$
④ 代入 $x = A\cos(\omega t + \phi)$，得到：
$a = -\omega^2 x$
这就从圆周运动的几何投影推导出了 SHM 的加速度-位移关系。这个推导揭示了 SHM 和圆周运动的深层联系——SHM 本质上是匀速圆周运动在某一直径上的投影。
考虑一个质点在半径为 $A$ 的圆周上以恒定角速度 $\omega$ 运动：
① 质点位置在圆心到圆周某点的连线上投影就是 SHM 的位移：
$x = A\cos(\omega t + \phi)$，其中 $\phi$ 为初相位
② 圆周运动中，向心加速度为 $a_c = \omega^2 A$，方向指向圆心
③ 加速度在 $x$ 方向上的投影：
$a_x = -a_c\cos(\omega t + \phi) = -\omega^2 A\cos(\omega t + \phi)$
④ 代入 $x = A\cos(\omega t + \phi)$，得到：
$a = -\omega^2 x$
这就从圆周运动的几何投影推导出了 SHM 的加速度-位移关系。这个推导揭示了 SHM 和圆周运动的深层联系——SHM 本质上是匀速圆周运动在某一直径上的投影。
Derivation of $a = -\omega^2 x$ from circular motion projection:
Consider a particle moving on a circle of radius $A$ with constant angular speed $\omega$:
① The projection of the particle's position onto a diameter gives SHM displacement:
$x = A\cos(\omega t + \phi)$, where $\phi$ is the initial phase
② In circular motion, centripetal acceleration is $a_c = \omega^2 A$, directed toward the center
③ The projection of acceleration onto the $x$-axis:
$a_x = -a_c\cos(\omega t + \phi) = -\omega^2 A\cos(\omega t + \phi)$
④ Substituting $x = A\cos(\omega t + \phi)$ gives:
$a = -\omega^2 x$
This derivation reveals the deep connection between SHM and circular motion — SHM is essentially the projection of uniform circular motion onto a diameter.
Consider a particle moving on a circle of radius $A$ with constant angular speed $\omega$:
① The projection of the particle's position onto a diameter gives SHM displacement:
$x = A\cos(\omega t + \phi)$, where $\phi$ is the initial phase
② In circular motion, centripetal acceleration is $a_c = \omega^2 A$, directed toward the center
③ The projection of acceleration onto the $x$-axis:
$a_x = -a_c\cos(\omega t + \phi) = -\omega^2 A\cos(\omega t + \phi)$
④ Substituting $x = A\cos(\omega t + \phi)$ gives:
$a = -\omega^2 x$
This derivation reveals the deep connection between SHM and circular motion — SHM is essentially the projection of uniform circular motion onto a diameter.
位移、速度、加速度 / Displacement, Velocity, Acceleration
$$x = A\cos(\omega t + \phi) \qquad v = -\omega A\sin(\omega t + \phi) \qquad a = -\omega^2 A\cos(\omega t + \phi)$$
简谐运动中位移、速度、加速度均随时间正弦（或余弦）变化：
• 位移 $x$：在 $+A$ 和 $-A$ 之间振荡，$A$ 为振幅
• 速度 $v$：在 $\pm \omega A$ 之间变化，过平衡位置时速度最大
• 加速度 $a$：在 $\pm \omega^2 A$ 之间变化，在两端点处加速度最大
• 位移 $x$：在 $+A$ 和 $-A$ 之间振荡，$A$ 为振幅
• 速度 $v$：在 $\pm \omega A$ 之间变化，过平衡位置时速度最大
• 加速度 $a$：在 $\pm \omega^2 A$ 之间变化，在两端点处加速度最大
In SHM, displacement, velocity, and acceleration all vary sinusoidally with time:
• Displacement $x$: oscillates between $+A$ and $-A$, where $A$ is amplitude
• Velocity $v$: varies between $\pm \omega A$, maximum at equilibrium
• Acceleration $a$: varies between $\pm \omega^2 A$, maximum at endpoints
• Displacement $x$: oscillates between $+A$ and $-A$, where $A$ is amplitude
• Velocity $v$: varies between $\pm \omega A$, maximum at equilibrium
• Acceleration $a$: varies between $\pm \omega^2 A$, maximum at endpoints
📐 位移-速度关系推导 / Derivation: Displacement-Velocity Relation
导出位移-速度关系：
由 $x = A\cos(\omega t + \phi)$，对时间求导：
$v = \frac{dx}{dt} = -\omega A\sin(\omega t + \phi)$
利用三角恒等式 $\sin^2\theta + \cos^2\theta = 1$：
$\sin^2(\omega t + \phi) = 1 - \cos^2(\omega t + \phi) = 1 - \frac{x^2}{A^2}$
代入速度公式：
$v = -\omega A\sin(\omega t + \phi)$
$|v| = \omega A\sqrt{1 - x^2/A^2}$
$v = \pm \omega\sqrt{A^2 - x^2}$
当 $x = 0$（平衡位置）：$v_{\max} = \pm \omega A$
当 $x = \pm A$（端点）：$v = 0$
由 $x = A\cos(\omega t + \phi)$，对时间求导：
$v = \frac{dx}{dt} = -\omega A\sin(\omega t + \phi)$
利用三角恒等式 $\sin^2\theta + \cos^2\theta = 1$：
$\sin^2(\omega t + \phi) = 1 - \cos^2(\omega t + \phi) = 1 - \frac{x^2}{A^2}$
代入速度公式：
$v = -\omega A\sin(\omega t + \phi)$
$|v| = \omega A\sqrt{1 - x^2/A^2}$
$v = \pm \omega\sqrt{A^2 - x^2}$
当 $x = 0$（平衡位置）：$v_{\max} = \pm \omega A$
当 $x = \pm A$（端点）：$v = 0$
Derivation of displacement-velocity relation:
Starting from $x = A\cos(\omega t + \phi)$, differentiate:
$v = \frac{dx}{dt} = -\omega A\sin(\omega t + \phi)$
Using $\sin^2\theta + \cos^2\theta = 1$:
$\sin^2(\omega t + \phi) = 1 - \cos^2(\omega t + \phi) = 1 - x^2/A^2$
Substitute:
$v = -\omega A\sin(\omega t + \phi)$
$|v| = \omega A\sqrt{1 - x^2/A^2}$
$v = \pm \omega\sqrt{A^2 - x^2}$
At $x = 0$ (equilibrium): $v_{\max} = \pm \omega A$
At $x = \pm A$ (endpoints): $v = 0$
Starting from $x = A\cos(\omega t + \phi)$, differentiate:
$v = \frac{dx}{dt} = -\omega A\sin(\omega t + \phi)$
Using $\sin^2\theta + \cos^2\theta = 1$:
$\sin^2(\omega t + \phi) = 1 - \cos^2(\omega t + \phi) = 1 - x^2/A^2$
Substitute:
$v = -\omega A\sin(\omega t + \phi)$
$|v| = \omega A\sqrt{1 - x^2/A^2}$
$v = \pm \omega\sqrt{A^2 - x^2}$
At $x = 0$ (equilibrium): $v_{\max} = \pm \omega A$
At $x = \pm A$ (endpoints): $v = 0$
周期与频率 / Period and Frequency
$$T = \frac{2\pi}{\omega} = \frac{1}{f} \qquad f = \frac{1}{T} = \frac{\omega}{2\pi}$$
周期 $T$ 是完成一次完整振动所需的时间（单位 s），频率 $f$ 是单位时间内完成的振动次数（单位 Hz = s$^{-1}$）。角频率 $\omega = 2\pi f = 2\pi/T$。
The period $T$ is the time for one complete oscillation (s), and frequency $f$ is the number of oscillations per second (Hz = s$^{-1}$). Angular frequency $\omega = 2\pi f = 2\pi/T$.
弹簧振子 / Mass-Spring System
$$\omega = \sqrt{\frac{k}{m}} \qquad T = 2\pi\sqrt{\frac{m}{k}}$$
对于水平或竖直弹簧振子，回复力由胡克定律 $F = -kx$ 给出。根据牛顿第二定律 $F = ma$：$ma = -kx$，即 $a = -(k/m)x$。与 $a = -\omega^2 x$ 对比得 $\omega = \sqrt{k/m}$。
For a mass-spring system, the restoring force is given by Hooke's law $F = -kx$. From Newton's second law $F = ma$: $ma = -kx$, so $a = -(k/m)x$. Comparing with $a = -\omega^2 x$ gives $\omega = \sqrt{k/m}$.
单摆 / Simple Pendulum
$$\omega = \sqrt{\frac{g}{l}} \qquad T = 2\pi\sqrt{\frac{l}{g}}$$
对于小角度（$\theta < 10^\circ$）的单摆，回复力 $F = -mg\sin\theta \approx -mg\theta = -mg(x/l)$，所以 $a = -(g/l)x$。与 SHM 条件对比得 $\omega = \sqrt{g/l}$，$T = 2\pi\sqrt{l/g}$。单摆周期与质量无关，只与摆长和重力加速度有关。
For a simple pendulum with small angles ($\theta < 10^\circ$), the restoring force is $F = -mg\sin\theta \approx -mg\theta = -mg(x/l)$, so $a = -(g/l)x$. Comparing with SHM: $\omega = \sqrt{g/l}$, $T = 2\pi\sqrt{l/g}$. The period is independent of mass, depending only on length and gravitational acceleration.
能量在 SHM 中的转化 / Energy in SHM
$$E_k = \frac12 m\omega^2(A^2 - x^2) \qquad E_p = \frac12 m\omega^2 x^2 \qquad E_{\text{总}} = \frac12 m\omega^2 A^2$$
在 SHM 中，动能和势能不断相互转换，但总机械能守恒：
• $x = 0$（平衡位置）：$E_k$ 最大，$E_p = 0$
• $x = \pm A$（端点）：$E_k = 0$，$E_p$ 最大
• 任何位置：$E_k + E_p = \frac12 m\omega^2 A^2$ = 常数
• $x = 0$（平衡位置）：$E_k$ 最大，$E_p = 0$
• $x = \pm A$（端点）：$E_k = 0$，$E_p$ 最大
• 任何位置：$E_k + E_p = \frac12 m\omega^2 A^2$ = 常数
In SHM, kinetic and potential energy continuously interchange while total mechanical energy is conserved:
• At $x = 0$ (equilibrium): $E_k$ maximum, $E_p = 0$
• At $x = \pm A$ (endpoints): $E_k = 0$, $E_p$ maximum
• At any position: $E_k + E_p = \frac12 m\omega^2 A^2$ = constant
• At $x = 0$ (equilibrium): $E_k$ maximum, $E_p = 0$
• At $x = \pm A$ (endpoints): $E_k = 0$, $E_p$ maximum
• At any position: $E_k + E_p = \frac12 m\omega^2 A^2$ = constant
📐 推导能量公式 / Derivation of Energy Formulas
SHM 中的能量：
动能：$E_k = \frac12 mv^2 = \frac12 m\omega^2(A^2 - x^2)$
（由 $v = \pm \omega\sqrt{A^2 - x^2}$ 代入）
势能（对于弹簧振子）：$E_p = \frac12 kx^2 = \frac12 m\omega^2 x^2$
（因为 $k = m\omega^2$）
总能量：$E_{\text{总}} = E_k + E_p = \frac12 m\omega^2(A^2 - x^2) + \frac12 m\omega^2 x^2 = \frac12 m\omega^2 A^2$
重要结论：总能量与振幅平方成正比。振幅加倍 → 能量变为 4 倍。
动能：$E_k = \frac12 mv^2 = \frac12 m\omega^2(A^2 - x^2)$
（由 $v = \pm \omega\sqrt{A^2 - x^2}$ 代入）
势能（对于弹簧振子）：$E_p = \frac12 kx^2 = \frac12 m\omega^2 x^2$
（因为 $k = m\omega^2$）
总能量：$E_{\text{总}} = E_k + E_p = \frac12 m\omega^2(A^2 - x^2) + \frac12 m\omega^2 x^2 = \frac12 m\omega^2 A^2$
重要结论：总能量与振幅平方成正比。振幅加倍 → 能量变为 4 倍。
Energy in SHM:
Kinetic: $E_k = \frac12 mv^2 = \frac12 m\omega^2(A^2 - x^2)$
(substituting $v = \pm \omega\sqrt{A^2 - x^2}$)
Potential (for spring): $E_p = \frac12 kx^2 = \frac12 m\omega^2 x^2$
(since $k = m\omega^2$)
Total: $E_{\text{total}} = E_k + E_p = \frac12 m\omega^2 A^2$
Key result: Total energy is proportional to the square of amplitude. Doubling amplitude → 4x energy.
Kinetic: $E_k = \frac12 mv^2 = \frac12 m\omega^2(A^2 - x^2)$
(substituting $v = \pm \omega\sqrt{A^2 - x^2}$)
Potential (for spring): $E_p = \frac12 kx^2 = \frac12 m\omega^2 x^2$
(since $k = m\omega^2$)
Total: $E_{\text{total}} = E_k + E_p = \frac12 m\omega^2 A^2$
Key result: Total energy is proportional to the square of amplitude. Doubling amplitude → 4x energy.
🎯 IB 考试要点 / Exam Tips：① SHM 的定义式 $a = -\omega^2 x$ 是判据——任何满足该条件的运动就是 SHM；② 弹簧振子周期与振幅无关（等时性），单摆周期与小角度近似下也与振幅无关；③ 能量法中，总能量 $\propto A^2$ 是重要考点；④ 相位 $\phi$ 决定了初始时刻的运动状态；⑤ 注意区分角频率 $\omega$（rad/s）和频率 $f$（Hz），$f = \omega/2\pi$。
🎯 IB Exam Tips: ① The defining equation $a = -\omega^2 x$ is the test — any motion satisfying it is SHM; ② Spring period is independent of amplitude (isochronous); pendulum period is also amplitude-independent for small angles; ③ Total energy $\propto A^2$ is a common exam point; ④ Phase $\phi$ determines the initial state; ⑤ Distinguish angular frequency $\omega$ (rad/s) from frequency $f$ (Hz): $f = \omega/2\pi$.
例 4.1 / Example 4.1
一个弹簧振子质量 0.50 kg，弹簧劲度系数 $k = 200$ N/m，振幅 0.050 m。求：(a) 角频率；(b) 周期；(c) 最大速度；(d) 总能量。
Mass-spring: $m = 0.50$ kg, $k = 200$ N/m, $A = 0.050$ m. Find: (a) angular frequency; (b) period; (c) max speed; (d) total energy.
解题过程 / Solution
(a) $\omega = \sqrt{k/m} = \sqrt{200/0.50} = \sqrt{400} = 20$ rad/s
(b) $T = 2\pi/\omega = 2\pi/20 = \pi/10 = 0.314$ s
(c) $v_{\max} = \omega A = 20 \times 0.050 = 1.0$ m/s
(d) $E_{\text{总}} = \frac12 m\omega^2 A^2 = \frac12 \times 0.50 \times 400 \times 0.0025 = 0.25$ J
或 $E_{\text{总}} = \frac12 kA^2 = \frac12 \times 200 \times 0.0025 = 0.25$ J
(b) $T = 2\pi/\omega = 2\pi/20 = \pi/10 = 0.314$ s
(c) $v_{\max} = \omega A = 20 \times 0.050 = 1.0$ m/s
(d) $E_{\text{总}} = \frac12 m\omega^2 A^2 = \frac12 \times 0.50 \times 400 \times 0.0025 = 0.25$ J
或 $E_{\text{总}} = \frac12 kA^2 = \frac12 \times 200 \times 0.0025 = 0.25$ J
(a) $\omega = \sqrt{k/m} = \sqrt{200/0.50} = 20$ rad/s
(b) $T = 2\pi/\omega = 2\pi/20 = \pi/10 = 0.314$ s
(c) $v_{\max} = \omega A = 20 \times 0.050 = 1.0$ m/s
(d) $E_{\text{total}} = \frac12 kA^2 = \frac12 \times 200 \times 0.0025 = 0.25$ J
(b) $T = 2\pi/\omega = 2\pi/20 = \pi/10 = 0.314$ s
(c) $v_{\max} = \omega A = 20 \times 0.050 = 1.0$ m/s
(d) $E_{\text{total}} = \frac12 kA^2 = \frac12 \times 200 \times 0.0025 = 0.25$ J
例 4.2 / Example 4.2
一个单摆周期为 2.0 s，所在位置 $g = 9.81$ m/s$^2$。求摆长。如果摆长变为原来的 4 倍，新周期是多少？
A pendulum has $T = 2.0$ s, $g = 9.81$ m/s$^2$. Find the length. If length is quadrupled, what is the new period?
解题过程 / Solution
$T = 2\pi\sqrt{l/g}$
$l = gT^2/(4\pi^2) = 9.81 \times 4 / (4\pi^2) = 9.81/\pi^2 = 0.994$ m ≈ 1.0 m
摆长变为 4 倍：$l' = 4l$
$T' = 2\pi\sqrt{4l/g} = 2 \times 2\pi\sqrt{l/g} = 2T = 4.0$ s
周期与 $\sqrt{l}$ 成正比，长度 4 倍 → 周期 2 倍。
$l = gT^2/(4\pi^2) = 9.81 \times 4 / (4\pi^2) = 9.81/\pi^2 = 0.994$ m ≈ 1.0 m
摆长变为 4 倍：$l' = 4l$
$T' = 2\pi\sqrt{4l/g} = 2 \times 2\pi\sqrt{l/g} = 2T = 4.0$ s
周期与 $\sqrt{l}$ 成正比，长度 4 倍 → 周期 2 倍。
$T = 2\pi\sqrt{l/g}$
$l = gT^2/(4\pi^2) = 9.81 \times 4 / (4\pi^2) = 9.81/\pi^2 = 0.994$ m ≈ 1.0 m
$l' = 4l$ → $T' = 2\pi\sqrt{4l/g} = 2T = 4.0$ s
Period $\propto \sqrt{l}$, quadruple length → double period.
$l = gT^2/(4\pi^2) = 9.81 \times 4 / (4\pi^2) = 9.81/\pi^2 = 0.994$ m ≈ 1.0 m
$l' = 4l$ → $T' = 2\pi\sqrt{4l/g} = 2T = 4.0$ s
Period $\propto \sqrt{l}$, quadruple length → double period.
例 4.3 / Example 4.3
一个 SHM 的振幅 0.080 m，周期 0.50 s，$t = 0$ 时在 $x = A$ 处。求：(a) 角频率；(b) $t = 0.20$ s 时的位移、速度、加速度。
SHM: $A = 0.080$ m, $T = 0.50$ s, at $t = 0$: $x = A$. Find: (a) $\omega$; (b) $x$, $v$, $a$ at $t = 0.20$ s.
解题过程 / Solution
(a) $\omega = 2\pi/T = 2\pi/0.50 = 4\pi$ rad/s = 12.57 rad/s
$t=0$ 时 $x = A$ 且沿负方向运动，取 $x = A\cos(\omega t)$
(b) $x = 0.080\cos(4\pi \times 0.20) = 0.080\cos(0.8\pi) = 0.080\cos(144^\circ)$
$x = 0.080 \times (-0.809) = -0.0647$ m ≈ $-0.065$ m
$v = -\omega A\sin(\omega t) = -4\pi \times 0.080 \times \sin(0.8\pi)$
$v = -1.005 \times 0.588 = -0.591$ m/s
$a = -\omega^2 x = -(4\pi)^2 \times (-0.0647) = 157.9 \times 0.0647 = 10.2$ m/s$^2$
（加速度为正，指向平衡位置）
$t=0$ 时 $x = A$ 且沿负方向运动，取 $x = A\cos(\omega t)$
(b) $x = 0.080\cos(4\pi \times 0.20) = 0.080\cos(0.8\pi) = 0.080\cos(144^\circ)$
$x = 0.080 \times (-0.809) = -0.0647$ m ≈ $-0.065$ m
$v = -\omega A\sin(\omega t) = -4\pi \times 0.080 \times \sin(0.8\pi)$
$v = -1.005 \times 0.588 = -0.591$ m/s
$a = -\omega^2 x = -(4\pi)^2 \times (-0.0647) = 157.9 \times 0.0647 = 10.2$ m/s$^2$
（加速度为正，指向平衡位置）
(a) $\omega = 2\pi/T = 2\pi/0.50 = 4\pi$ rad/s = 12.57 rad/s
$x = A\cos(\omega t)$ since $x=A$ at $t=0$
(b) $x = 0.080\cos(4\pi \times 0.20) = 0.080\cos(0.8\pi) = -0.065$ m
$v = -\omega A\sin(\omega t) = -4\pi \times 0.080 \times \sin(0.8\pi) = -0.591$ m/s
$a = -\omega^2 x = -(4\pi)^2 \times (-0.0647) = 10.2$ m/s$^2$ (toward equilibrium)
$x = A\cos(\omega t)$ since $x=A$ at $t=0$
(b) $x = 0.080\cos(4\pi \times 0.20) = 0.080\cos(0.8\pi) = -0.065$ m
$v = -\omega A\sin(\omega t) = -4\pi \times 0.080 \times \sin(0.8\pi) = -0.591$ m/s
$a = -\omega^2 x = -(4\pi)^2 \times (-0.0647) = 10.2$ m/s$^2$ (toward equilibrium)
4.2 波动性质 / Wave Properties
IB 4.2SL 核心
波的类型 / Types of Waves
机械波（mechanical wave）需要介质传播（如水波、声波、地震波）。电磁波（electromagnetic wave）不需要介质（如光波、无线电波）。按质点振动方向分为：
• 横波（transverse wave）：质点振动方向垂直于波的传播方向（如光波、绳波）
• 纵波（longitudinal wave）：质点振动方向平行于波的传播方向（如声波、弹簧波）
• 横波（transverse wave）：质点振动方向垂直于波的传播方向（如光波、绳波）
• 纵波（longitudinal wave）：质点振动方向平行于波的传播方向（如声波、弹簧波）
Mechanical waves require a medium (water waves, sound, seismic). Electromagnetic waves need no medium (light, radio). Classification by vibration direction:
• Transverse waves: particle vibration perpendicular to wave direction (light, string waves)
• Longitudinal waves: particle vibration parallel to wave direction (sound, spring waves)
• Transverse waves: particle vibration perpendicular to wave direction (light, string waves)
• Longitudinal waves: particle vibration parallel to wave direction (sound, spring waves)
基本物理量 / Fundamental Quantities
$$v = f\lambda \qquad v = \frac{\lambda}{T}$$
波速 $v$（m/s）、频率 $f$（Hz）、波长 $\lambda$（m）之间的关系：$v = f\lambda$。频率由波源决定，波速由介质决定。当波从一种介质进入另一种介质时，频率不变，波速和波长发生变化。
The relationship between wave speed $v$ (m/s), frequency $f$ (Hz), and wavelength $\lambda$ (m): $v = f\lambda$. Frequency is determined by the source, wave speed by the medium. When a wave enters a different medium, frequency stays constant while speed and wavelength change.
波的图像 / Wave Graphs
有两种重要的波动图像：
• $y$-$x$ 图（波形图）：展示某一时刻各质点的位移，从图中可读出波长 $\lambda$ 和振幅 $A$；
• $y$-$t$ 图（振动图）：展示某一质点位移随时间的变化，从图中可读出周期 $T$ 和振幅 $A$。
• $y$-$x$ 图（波形图）：展示某一时刻各质点的位移，从图中可读出波长 $\lambda$ 和振幅 $A$；
• $y$-$t$ 图（振动图）：展示某一质点位移随时间的变化，从图中可读出周期 $T$ 和振幅 $A$。
Two important wave graphs:
• $y$-$x$ graph (wave profile): shows displacement of all particles at one instant — read wavelength $\lambda$ and amplitude $A$;
• $y$-$t$ graph (oscillation graph): shows displacement of one particle over time — read period $T$ and amplitude $A$.
• $y$-$x$ graph (wave profile): shows displacement of all particles at one instant — read wavelength $\lambda$ and amplitude $A$;
• $y$-$t$ graph (oscillation graph): shows displacement of one particle over time — read period $T$ and amplitude $A$.
相位 / Phase
$$\phi = \frac{2\pi \Delta x}{\lambda} \qquad \Delta\phi = \frac{2\pi \Delta t}{T}$$
相位差描述两个质点（或同一质点在不同时间）振动状态的差异。相位差为 $2\pi$ 的整数倍时称为同相（in phase），为 $\pi$ 的奇数倍时称为反相（antiphase / out of phase）。
Phase difference describes the difference in vibrational state between two points (or the same point at different times). A phase difference of integer multiples of $2\pi$ means in phase; odd multiples of $\pi$ means antiphase / out of phase.
波的叠加与干涉 / Superposition and Interference
$$y_{\text{总}} = y_1 + y_2 \quad \text{（叠加原理）}$$
叠加原理（principle of superposition）：当两列波相遇时，合成位移等于各波位移之和。叠加后两列波继续独立传播，互不影响。
干涉条件（必要条件）：两列波必须频率相同、相位差恒定（即相干波，coherent waves）。
• 相长干涉（constructive）：波程差 $\Delta r = n\lambda$（$n = 0, 1, 2, \ldots$）
• 相消干涉（destructive）：波程差 $\Delta r = (n + \frac12)\lambda$
干涉条件（必要条件）：两列波必须频率相同、相位差恒定（即相干波，coherent waves）。
• 相长干涉（constructive）：波程差 $\Delta r = n\lambda$（$n = 0, 1, 2, \ldots$）
• 相消干涉（destructive）：波程差 $\Delta r = (n + \frac12)\lambda$
Principle of superposition: When two waves meet, the resultant displacement is the sum of individual displacements. After superposition, each wave continues independently.
Interference conditions: waves must have the same frequency and constant phase difference (i.e., be coherent).
• Constructive: path difference $\Delta r = n\lambda$ ($n = 0, 1, 2, \ldots$)
• Destructive: path difference $\Delta r = (n + \frac12)\lambda$
Interference conditions: waves must have the same frequency and constant phase difference (i.e., be coherent).
• Constructive: path difference $\Delta r = n\lambda$ ($n = 0, 1, 2, \ldots$)
• Destructive: path difference $\Delta r = (n + \frac12)\lambda$
应用：声波定位与声呐 / Applications: Sonar and Echolocation
声呐（SONAR）利用声波在水中的传播和反射来探测物体。原理：发射器发出超声波脉冲，接收器接收反射回波。通过测量时间差 $\Delta t$ 计算距离：$d = v\Delta t/2$。声波在海水中的速度约为 1500 m/s（远快于空气中的 340 m/s）。
医学超声（Medical Ultrasound）：频率 1-20 MHz 的超声波用于医学成像。不同组织对超声的反射不同，通过分析回波生成图像。多普勒超声还可测量血流速度。
医学超声（Medical Ultrasound）：频率 1-20 MHz 的超声波用于医学成像。不同组织对超声的反射不同，通过分析回波生成图像。多普勒超声还可测量血流速度。
SONAR uses sound wave propagation and reflection in water to detect objects. Principle: emit ultrasonic pulses, receive echoes. Distance: $d = v\Delta t/2$. Sound speed in seawater ≈ 1500 m/s (vs 340 m/s in air).
Medical Ultrasound: 1-20 MHz ultrasound for medical imaging. Different tissues reflect ultrasound differently — echoes produce images. Doppler ultrasound measures blood flow velocity.
Medical Ultrasound: 1-20 MHz ultrasound for medical imaging. Different tissues reflect ultrasound differently — echoes produce images. Doppler ultrasound measures blood flow velocity.
🎯 IB 考试要点 / Exam Tips：① 波速公式 $v = f\lambda$ 是 IB 最常用的公式之一——务必熟练掌握；② 波从一种介质进入另一种介质时频率不变（波源决定），波速和波长改变（介质决定）；③ 相位差概念在干涉问题中至关重要——波程差对应相位差 $\Delta\phi = 2\pi \Delta r/\lambda$；④ 叠加原理适用于所有波，但稳定干涉需要相干条件；⑤ 区分 $y$-$x$ 图和 $y$-$t$ 图——前者读 $\lambda$，后者读 $T$。
🎯 IB Exam Tips: ① $v = f\lambda$ is one of the most used formulas — master it; ② When a wave enters a new medium, frequency stays constant (source) while speed and wavelength change (medium); ③ Phase difference is crucial for interference — $\Delta\phi = 2\pi \Delta r/\lambda$; ④ Superposition applies to all waves, but stable interference requires coherence; ⑤ Distinguish $y$-$x$ (read $\lambda$) from $y$-$t$ (read $T$).
例 4.4 / Example 4.4
声波在空气中频率 440 Hz（A4 音），声速 340 m/s。求波长。如果该声波进入水中（声速 1500 m/s），求水中的波长和频率。
Sound wave in air: $f = 440$ Hz, $v = 340$ m/s. Find $\lambda$ in air. Then in water ($v = 1500$ m/s), find $\lambda$ and $f$.
解题过程 / Solution
空气中：$\lambda_{\text{空气}} = v/f = 340/440 = 0.773$ m
进入水中：频率不变 $f = 440$ Hz
$\lambda_{\text{水}} = v/f = 1500/440 = 3.41$ m
频率由波源（乐器或扬声器）决定，不随介质改变。波速增大导致波长增大。
进入水中：频率不变 $f = 440$ Hz
$\lambda_{\text{水}} = v/f = 1500/440 = 3.41$ m
频率由波源（乐器或扬声器）决定，不随介质改变。波速增大导致波长增大。
In air: $\lambda_{\text{air}} = v/f = 340/440 = 0.773$ m
In water: frequency unchanged $f = 440$ Hz
$\lambda_{\text{water}} = v/f = 1500/440 = 3.41$ m
Frequency is determined by the source, not the medium. Greater wave speed means greater wavelength.
In water: frequency unchanged $f = 440$ Hz
$\lambda_{\text{water}} = v/f = 1500/440 = 3.41$ m
Frequency is determined by the source, not the medium. Greater wave speed means greater wavelength.
例 4.5 / Example 4.5
两相干波源相距 0.50 m，发出相同频率（680 Hz）的声波，声速 340 m/s。在与两波源连线的中垂线上，会发生什么干涉？
Two coherent sources 0.50 m apart emit sound at 680 Hz, $v = 340$ m/s. What interference occurs on the perpendicular bisector?
解题过程 / Solution
$\lambda = v/f = 340/680 = 0.50$ m
中垂线上任意点到两波源的距离相等，$\Delta r = 0$。
$\Delta r = 0 = 0 \times \lambda$，满足相长干涉条件。
结论：中垂线上所有点发生相长干涉（加强），声音最大。
中垂线上任意点到两波源的距离相等，$\Delta r = 0$。
$\Delta r = 0 = 0 \times \lambda$，满足相长干涉条件。
结论：中垂线上所有点发生相长干涉（加强），声音最大。
$\lambda = v/f = 340/680 = 0.50$ m
On the perpendicular bisector, distances to both sources are equal: $\Delta r = 0$.
$\Delta r = 0 = 0 \times \lambda$, satisfying constructive interference.
Result: the entire bisector experiences constructive interference (maximum sound).
On the perpendicular bisector, distances to both sources are equal: $\Delta r = 0$.
$\Delta r = 0 = 0 \times \lambda$, satisfying constructive interference.
Result: the entire bisector experiences constructive interference (maximum sound).
4.3 波动行为 / Wave Behavior
IB 4.3SL 重点
反射 / Reflection
$$\theta_i = \theta_r$$
反射定律：入射角等于反射角。波遇到障碍物或不同介质的界面时会反射。对于固定端反射，反射波有 $\pi$ 相位突变（即反相）；对于自由端反射，反射波无相位变化。
Law of reflection: angle of incidence equals angle of reflection. Waves reflect at boundaries. Reflection from a fixed end causes a $\pi$ phase change; reflection from a free end has no phase change.
折射 / Refraction
$$\frac{\sin\theta_1}{\sin\theta_2} = \frac{v_1}{v_2}$$
折射是波进入不同介质时传播方向改变的现象。波速改变导致波长改变，频率保持不变。当波从浅水进入深水（波速增大）时，波向法线方向偏折；从深水进入浅水（波速减小）时，波偏离法线方向。
Refraction is the change in wave direction when entering a different medium. Wave speed changes cause wavelength changes; frequency stays constant. When waves travel from shallow to deep water (speed increases), they bend toward the normal; from deep to shallow (speed decreases), they bend away from the normal.
📐 折射定律的推导 / Derivation of Refraction Law
从惠更斯原理推导折射定律：
惠更斯原理：波前上的每一点都可以视为发出球面子波的新波源，所有子波的包络面形成新的波前。
考虑平面波以角度 $\theta_1$ 入射到界面。在时间 $\Delta t$ 内，左边介质的波前移动距离 $v_1\Delta t$，右边介质中移动 $v_2\Delta t$。
由几何关系：
$v_1\Delta t / \sin\theta_1 = v_2\Delta t / \sin\theta_2$
（斜边相等，因为两个直角三角形共享同一边）
整理得：$\displaystyle \frac{\sin\theta_1}{\sin\theta_2} = \frac{v_1}{v_2}$
对于光波，$\displaystyle n = \frac{c}{v}$（折射率），所以斯涅尔定律为：$n_1\sin\theta_1 = n_2\sin\theta_2$。
惠更斯原理：波前上的每一点都可以视为发出球面子波的新波源，所有子波的包络面形成新的波前。
考虑平面波以角度 $\theta_1$ 入射到界面。在时间 $\Delta t$ 内，左边介质的波前移动距离 $v_1\Delta t$，右边介质中移动 $v_2\Delta t$。
由几何关系：
$v_1\Delta t / \sin\theta_1 = v_2\Delta t / \sin\theta_2$
（斜边相等，因为两个直角三角形共享同一边）
整理得：$\displaystyle \frac{\sin\theta_1}{\sin\theta_2} = \frac{v_1}{v_2}$
对于光波，$\displaystyle n = \frac{c}{v}$（折射率），所以斯涅尔定律为：$n_1\sin\theta_1 = n_2\sin\theta_2$。
Derivation of refraction from Huygens' principle:
Huygens' principle: every point on a wavefront acts as a source of spherical wavelets; the envelope of all wavelets forms the new wavefront.
Consider a plane wave incident at angle $\theta_1$. In time $\Delta t$, the wavefront travels $v_1\Delta t$ in medium 1 and $v_2\Delta t$ in medium 2.
From geometry:
$v_1\Delta t / \sin\theta_1 = v_2\Delta t / \sin\theta_2$
(the hypotenuses are equal, since the two right triangles share the same side)
Rearranging: $\displaystyle \frac{\sin\theta_1}{\sin\theta_2} = \frac{v_1}{v_2}$
For light: $\displaystyle n = \frac{c}{v}$ (refractive index), giving Snell's law: $n_1\sin\theta_1 = n_2\sin\theta_2$.
Huygens' principle: every point on a wavefront acts as a source of spherical wavelets; the envelope of all wavelets forms the new wavefront.
Consider a plane wave incident at angle $\theta_1$. In time $\Delta t$, the wavefront travels $v_1\Delta t$ in medium 1 and $v_2\Delta t$ in medium 2.
From geometry:
$v_1\Delta t / \sin\theta_1 = v_2\Delta t / \sin\theta_2$
(the hypotenuses are equal, since the two right triangles share the same side)
Rearranging: $\displaystyle \frac{\sin\theta_1}{\sin\theta_2} = \frac{v_1}{v_2}$
For light: $\displaystyle n = \frac{c}{v}$ (refractive index), giving Snell's law: $n_1\sin\theta_1 = n_2\sin\theta_2$.
衍射 / Diffraction
$$\theta \approx \frac{\lambda}{b} \quad \text{（狭缝衍射，$\lambda \approx b$ 时显著）}$$
衍射是波遇到障碍物或通过狭缝时偏离直线传播的现象。衍射的显著程度取决于波长与障碍物（或狭缝）尺寸的比值。当 $\lambda \gg b$ 时衍射显著；$\lambda \ll b$ 时衍射可忽略（直线传播）。
Diffraction is the spreading of waves when passing through an aperture or around obstacles. The degree of diffraction depends on the ratio of wavelength to aperture/obstacle size $b$. When $\lambda \gg b$, diffraction is significant; when $\lambda \ll b$, diffraction is negligible (straight-line propagation).
📐 衍射现象的详细解释 / Detailed Explanation of Diffraction
单缝衍射：
当平面波通过宽度为 $b$ 的狭缝时，产生衍射。中央明纹最宽最亮，两侧明纹逐渐减弱。
• 第一级暗纹位置：$\sin\theta = \lambda/b$
• 中央明纹角宽度：$2\lambda/b$
• 缝越窄（$b$ 越小），衍射越明显
• 波长越大（$\lambda$ 越大），衍射越明显
日常生活中：
• 声波波长约 0.1-10 m，与门宽相当，所以门后能听到声音（衍射显著）
• 光波波长约 400-700 nm，远小于日常障碍物，所以光线近似直线传播
• 水波通过港口入口时发生衍射，使波浪传入港湾内部
当平面波通过宽度为 $b$ 的狭缝时，产生衍射。中央明纹最宽最亮，两侧明纹逐渐减弱。
• 第一级暗纹位置：$\sin\theta = \lambda/b$
• 中央明纹角宽度：$2\lambda/b$
• 缝越窄（$b$ 越小），衍射越明显
• 波长越大（$\lambda$ 越大），衍射越明显
日常生活中：
• 声波波长约 0.1-10 m，与门宽相当，所以门后能听到声音（衍射显著）
• 光波波长约 400-700 nm，远小于日常障碍物，所以光线近似直线传播
• 水波通过港口入口时发生衍射，使波浪传入港湾内部
Single-slit diffraction:
When a plane wave passes through a slit of width $b$, diffraction occurs. The central maximum is widest and brightest; side maxima become progressively weaker.
• First dark fringe: $\sin\theta = \lambda/b$
• Central maximum angular width: $2\lambda/b$
• Narrower slit ($b$ smaller) → more diffraction
• Longer wavelength ($\lambda$ larger) → more diffraction
Daily life:
• Sound wavelengths ~0.1-10 m, similar to door width — you can hear sound from around a corner
• Light wavelengths ~400-700 nm, much smaller than everyday objects — light travels in straight lines
• Water waves diffract through harbor entrances
When a plane wave passes through a slit of width $b$, diffraction occurs. The central maximum is widest and brightest; side maxima become progressively weaker.
• First dark fringe: $\sin\theta = \lambda/b$
• Central maximum angular width: $2\lambda/b$
• Narrower slit ($b$ smaller) → more diffraction
• Longer wavelength ($\lambda$ larger) → more diffraction
Daily life:
• Sound wavelengths ~0.1-10 m, similar to door width — you can hear sound from around a corner
• Light wavelengths ~400-700 nm, much smaller than everyday objects — light travels in straight lines
• Water waves diffract through harbor entrances
应用：MRI 与医学成像中的波 / Application: MRI and Medical Imaging
磁共振成像（MRI）利用无线电波在强磁场中与人体组织的相互作用。原理：
• 人体置于强磁场中，氢原子核（质子）的磁矩沿磁场方向排列
• 发射射频脉冲（与质子的拉莫尔频率共振），质子吸收能量并偏离平衡位置
• 关闭脉冲后，质子释放能量并回到平衡位置，发射出可检测的射频信号
• 不同组织的质子密度和弛豫时间不同，产生信号差异形成图像
MRI 本质上是利用电磁波（射频波）与物质的共振相互作用，是波动理论在医学中的重要应用。
• 人体置于强磁场中，氢原子核（质子）的磁矩沿磁场方向排列
• 发射射频脉冲（与质子的拉莫尔频率共振），质子吸收能量并偏离平衡位置
• 关闭脉冲后，质子释放能量并回到平衡位置，发射出可检测的射频信号
• 不同组织的质子密度和弛豫时间不同，产生信号差异形成图像
MRI 本质上是利用电磁波（射频波）与物质的共振相互作用，是波动理论在医学中的重要应用。
Magnetic Resonance Imaging (MRI) uses radio waves in a strong magnetic field to interact with body tissues. Principles:
• Body placed in a strong magnetic field — hydrogen nuclei (protons) align with the field
• RF pulse is applied (resonant with proton Larmor frequency) — protons absorb energy and tilt away
• When pulse stops, protons release energy and return to equilibrium, emitting detectable RF signals
• Different tissues have different proton densities and relaxation times, creating image contrast
MRI uses electromagnetic waves (radio frequency) interacting with matter via resonance — a key application of wave theory in medicine.
• Body placed in a strong magnetic field — hydrogen nuclei (protons) align with the field
• RF pulse is applied (resonant with proton Larmor frequency) — protons absorb energy and tilt away
• When pulse stops, protons release energy and return to equilibrium, emitting detectable RF signals
• Different tissues have different proton densities and relaxation times, creating image contrast
MRI uses electromagnetic waves (radio frequency) interacting with matter via resonance — a key application of wave theory in medicine.
🎯 IB 考试要点 / Exam Tips：① 折射频率不变是常见考点——波从一种介质进入另一种介质时频率由波源决定；② 衍射显著的条件是 $\lambda \approx b$——声音容易衍射，光不容易衍射；③ 惠更斯原理可以解释反射和折射，但不解释衍射的强度分布；④ 反射中固定端和自由端的相位变化不同——固定端反相，自由端同相。
🎯 IB Exam Tips: ① Frequency is unchanged during refraction — it's determined by the source; ② Significant diffraction requires $\lambda \approx b$ — sound diffracts readily, light does not; ③ Huygens' principle explains reflection and refraction directions but not diffraction intensity patterns; ④ Fixed-end reflection causes $\pi$ phase change; free-end reflection has no phase change.
例 4.6 / Example 4.6
水波从深水区（$v_1 = 1.5$ m/s）进入浅水区（$v_2 = 1.0$ m/s），入射角 40°。求折射角。
Water waves: deep water $v_1 = 1.5$ m/s, shallow $v_2 = 1.0$ m/s, $\theta_1 = 40^\circ$. Find $\theta_2$.
解题过程 / Solution
$\sin\theta_1 / \sin\theta_2 = v_1/v_2$
$\sin 40^\circ / \sin\theta_2 = 1.5/1.0 = 1.5$
$\sin\theta_2 = \sin 40^\circ / 1.5 = 0.643/1.5 = 0.429$
$\theta_2 = \arcsin(0.429) = 25.4^\circ$
水波从深水区进入浅水区时波速减小，波向法线方向偏折（$\theta_2 < \theta_1$）。
$\sin 40^\circ / \sin\theta_2 = 1.5/1.0 = 1.5$
$\sin\theta_2 = \sin 40^\circ / 1.5 = 0.643/1.5 = 0.429$
$\theta_2 = \arcsin(0.429) = 25.4^\circ$
水波从深水区进入浅水区时波速减小，波向法线方向偏折（$\theta_2 < \theta_1$）。
$\sin\theta_1 / \sin\theta_2 = v_1/v_2$
$\sin 40^\circ / \sin\theta_2 = 1.5/1.0 = 1.5$
$\sin\theta_2 = \sin 40^\circ / 1.5 = 0.643/1.5 = 0.429$
$\theta_2 = \arcsin(0.429) = 25.4^\circ$
Waves bend toward the normal when entering a region of slower wave speed ($\theta_2 < \theta_1$).
$\sin 40^\circ / \sin\theta_2 = 1.5/1.0 = 1.5$
$\sin\theta_2 = \sin 40^\circ / 1.5 = 0.643/1.5 = 0.429$
$\theta_2 = \arcsin(0.429) = 25.4^\circ$
Waves bend toward the normal when entering a region of slower wave speed ($\theta_2 < \theta_1$).
例 4.7 / Example 4.7
声波波长 0.68 m，通过一扇宽 0.80 m 的门。判断衍射是否显著。
Sound wave $\lambda = 0.68$ m passing through a door of width 0.80 m. Is diffraction significant?
解题过程 / Solution
$\lambda / b = 0.68 / 0.80 = 0.85$
因为 $\lambda \approx b$（比值接近 1），衍射显著。
这解释了为什么在门外的人可以听到门后的声音——声波通过门缝时发生明显衍射，传播到门后的区域。
进行比较：若为光波（$\lambda \approx 500$ nm = $5 \times 10^{-7}$ m），
$\lambda / b = 5 \times 10^{-7} / 0.80 \approx 6.25 \times 10^{-7} \ll 1$
所以光波通过门缝时几乎不发生衍射——这就是为什么你不能绕过门看到后面的物体。
因为 $\lambda \approx b$（比值接近 1），衍射显著。
这解释了为什么在门外的人可以听到门后的声音——声波通过门缝时发生明显衍射，传播到门后的区域。
进行比较：若为光波（$\lambda \approx 500$ nm = $5 \times 10^{-7}$ m），
$\lambda / b = 5 \times 10^{-7} / 0.80 \approx 6.25 \times 10^{-7} \ll 1$
所以光波通过门缝时几乎不发生衍射——这就是为什么你不能绕过门看到后面的物体。
$\lambda / b = 0.68 / 0.80 = 0.85$
Since $\lambda \approx b$ (ratio close to 1), diffraction is significant.
This explains why someone outside a door can hear sound from inside — sound waves diffract around the door opening.
Compare with light ($\lambda \approx 500$ nm = $5 \times 10^{-7}$ m):
$\lambda / b = 5 \times 10^{-7} / 0.80 \approx 6.25 \times 10^{-7} \ll 1$
Light shows negligible diffraction — you can't see around a corner.
Since $\lambda \approx b$ (ratio close to 1), diffraction is significant.
This explains why someone outside a door can hear sound from inside — sound waves diffract around the door opening.
Compare with light ($\lambda \approx 500$ nm = $5 \times 10^{-7}$ m):
$\lambda / b = 5 \times 10^{-7} / 0.80 \approx 6.25 \times 10^{-7} \ll 1$
Light shows negligible diffraction — you can't see around a corner.
4.4 驻波与多普勒效应 / Standing Waves & Doppler Effect
IB 4.5HL & SL 核心
驻波的形成 / Formation of Standing Waves
$$y = 2A\cos\left(\frac{2\pi x}{\lambda}\right)\sin(\omega t)$$
驻波由两列振幅相同、频率相同、传播方向相反的波叠加形成。驻波不传输能量——能量被"驻留"在波节之间来回振荡。
A standing wave is formed by the superposition of two waves with the same amplitude, frequency, and wavelength traveling in opposite directions. Standing waves do not transfer energy — energy is confined between nodes.
📐 驻波方程的完整推导 / Full Derivation of Standing Wave Equation
从叠加原理推导驻波方程：
设两列波沿相反方向传播：
$y_1 = A\sin(\omega t - kx)$ （向右传播）
$y_2 = A\sin(\omega t + kx)$ （向左传播）
其中 $k = 2\pi/\lambda$ 为波数。
根据叠加原理：
$y = y_1 + y_2 = A[\sin(\omega t - kx) + \sin(\omega t + kx)]$
使用三角恒等式 $\sin\alpha + \sin\beta = 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$：
$y = 2A\sin(\omega t)\cos(kx)$
或者写成：
$y = 2A\cos(kx)\sin(\omega t) = 2A\cos\left(\frac{2\pi x}{\lambda}\right)\sin(\omega t)$
分析：
• 空间因子 $\cos(2\pi x/\lambda)$ 决定了振幅在空间中的分布
• 时间因子 $\sin(\omega t)$ 表示所有点以相同频率振动
• 波节（node）：$\cos(2\pi x/\lambda) = 0$ 的位置，即 $x = (2n+1)\lambda/4$，$n = 0, 1, 2, \ldots$
• 波腹（antinode）：$|\cos(2\pi x/\lambda)| = 1$ 的位置，即 $x = n\lambda/2$
• 相邻波节（或相邻波腹）之间的距离为 $\lambda/2$
设两列波沿相反方向传播：
$y_1 = A\sin(\omega t - kx)$ （向右传播）
$y_2 = A\sin(\omega t + kx)$ （向左传播）
其中 $k = 2\pi/\lambda$ 为波数。
根据叠加原理：
$y = y_1 + y_2 = A[\sin(\omega t - kx) + \sin(\omega t + kx)]$
使用三角恒等式 $\sin\alpha + \sin\beta = 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$：
$y = 2A\sin(\omega t)\cos(kx)$
或者写成：
$y = 2A\cos(kx)\sin(\omega t) = 2A\cos\left(\frac{2\pi x}{\lambda}\right)\sin(\omega t)$
分析：
• 空间因子 $\cos(2\pi x/\lambda)$ 决定了振幅在空间中的分布
• 时间因子 $\sin(\omega t)$ 表示所有点以相同频率振动
• 波节（node）：$\cos(2\pi x/\lambda) = 0$ 的位置，即 $x = (2n+1)\lambda/4$，$n = 0, 1, 2, \ldots$
• 波腹（antinode）：$|\cos(2\pi x/\lambda)| = 1$ 的位置，即 $x = n\lambda/2$
• 相邻波节（或相邻波腹）之间的距离为 $\lambda/2$
Derivation of standing wave equation from superposition:
Two waves traveling in opposite directions:
$y_1 = A\sin(\omega t - kx)$ (traveling right)
$y_2 = A\sin(\omega t + kx)$ (traveling left)
where $k = 2\pi/\lambda$ is the wave number.
By superposition:
$y = y_1 + y_2 = A[\sin(\omega t - kx) + \sin(\omega t + kx)]$
Using $\sin\alpha + \sin\beta = 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$:
$y = 2A\sin(\omega t)\cos(kx) = 2A\cos\left(\frac{2\pi x}{\lambda}\right)\sin(\omega t)$
Analysis:
• Spatial factor $\cos(2\pi x/\lambda)$ determines amplitude distribution
• Temporal factor $\sin(\omega t)$ — all points oscillate at the same frequency
• Nodes: where $\cos(2\pi x/\lambda) = 0$, i.e., $x = (2n+1)\lambda/4$
• Antinodes: where $|\cos(2\pi x/\lambda)| = 1$, i.e., $x = n\lambda/2$
• Distance between adjacent nodes (or antinodes) = $\lambda/2$
Two waves traveling in opposite directions:
$y_1 = A\sin(\omega t - kx)$ (traveling right)
$y_2 = A\sin(\omega t + kx)$ (traveling left)
where $k = 2\pi/\lambda$ is the wave number.
By superposition:
$y = y_1 + y_2 = A[\sin(\omega t - kx) + \sin(\omega t + kx)]$
Using $\sin\alpha + \sin\beta = 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$:
$y = 2A\sin(\omega t)\cos(kx) = 2A\cos\left(\frac{2\pi x}{\lambda}\right)\sin(\omega t)$
Analysis:
• Spatial factor $\cos(2\pi x/\lambda)$ determines amplitude distribution
• Temporal factor $\sin(\omega t)$ — all points oscillate at the same frequency
• Nodes: where $\cos(2\pi x/\lambda) = 0$, i.e., $x = (2n+1)\lambda/4$
• Antinodes: where $|\cos(2\pi x/\lambda)| = 1$, i.e., $x = n\lambda/2$
• Distance between adjacent nodes (or antinodes) = $\lambda/2$
乐器中的驻波 / Standing Waves in Musical Instruments
$$\text{弦乐器（两端固定）：} \quad f_n = \frac{n}{2L}v = \frac{n}{2L}\sqrt{\frac{T}{\mu}} \quad n = 1, 2, 3, \ldots$$
弦乐器（吉他、小提琴等）中，弦两端固定，因此两端都是波节。可能的振动模式（谐波/harmonics）：
• 基频（$n=1$）：$f_1 = v/2L$，波长 $\lambda_1 = 2L$
• 第一泛音（$n=2$）：$f_2 = v/L = 2f_1$
• 第 $n$ 次谐波：$f_n = nf_1$
• 基频（$n=1$）：$f_1 = v/2L$，波长 $\lambda_1 = 2L$
• 第一泛音（$n=2$）：$f_2 = v/L = 2f_1$
• 第 $n$ 次谐波：$f_n = nf_1$
In string instruments (guitar, violin), both ends are fixed (nodes). Allowed vibration modes (harmonics):
• Fundamental ($n=1$): $f_1 = v/2L$, $\lambda_1 = 2L$
• First overtone ($n=2$): $f_2 = v/L = 2f_1$
• $n$-th harmonic: $f_n = nf_1$
• Fundamental ($n=1$): $f_1 = v/2L$, $\lambda_1 = 2L$
• First overtone ($n=2$): $f_2 = v/L = 2f_1$
• $n$-th harmonic: $f_n = nf_1$
管乐器中的驻波 / Standing Waves in Pipes
$$\text{两端开口：} f_n = \frac{nv}{2L} \quad \text{一端闭口：} f_n = \frac{nv}{4L} \quad \text{（$n$ 为奇数）}$$
管乐器：
• 两端开口（如长笛、管风琴开口管）：两端都是波腹。$f_n = nv/2L$，$n = 1, 2, 3, \ldots$（所有谐波）
• 一端闭口（如单簧管、管风琴闭口管）：闭口端为波节，开口端为波腹。$f_n = nv/4L$，$n = 1, 3, 5, \ldots$（只有奇次谐波）
• 两端开口（如长笛、管风琴开口管）：两端都是波腹。$f_n = nv/2L$，$n = 1, 2, 3, \ldots$（所有谐波）
• 一端闭口（如单簧管、管风琴闭口管）：闭口端为波节，开口端为波腹。$f_n = nv/4L$，$n = 1, 3, 5, \ldots$（只有奇次谐波）
Pipes:
• Open at both ends (flute): antinodes at both ends. $f_n = nv/2L$, $n = 1, 2, 3, \ldots$ (all harmonics)
• Closed at one end (clarinet): node at closed end, antinode at open end. $f_n = nv/4L$, $n = 1, 3, 5, \ldots$ (odd harmonics only)
• Open at both ends (flute): antinodes at both ends. $f_n = nv/2L$, $n = 1, 2, 3, \ldots$ (all harmonics)
• Closed at one end (clarinet): node at closed end, antinode at open end. $f_n = nv/4L$, $n = 1, 3, 5, \ldots$ (odd harmonics only)
多普勒效应 / Doppler Effect
$$f_o = f_s\left(\frac{v \pm v_o}{v \mp v_s}\right)$$
多普勒效应是波源和观察者之间有相对运动时观察到的频率变化现象。
• 观察者朝向波源运动 → 频率升高（$+v_o$ 在分子）
• 观察者远离波源运动 → 频率降低（$-v_o$ 在分子）
• 波源朝向观察者运动 → 频率升高（$-v_s$ 在分母）
• 波源远离观察者运动 → 频率降低（$+v_s$ 在分母）
其中 $f_o$ 是观察者接收的频率，$f_s$ 是波源发出的频率，$v$ 是波速，$v_o$ 是观察者速度，$v_s$ 是波源速度。
• 观察者朝向波源运动 → 频率升高（$+v_o$ 在分子）
• 观察者远离波源运动 → 频率降低（$-v_o$ 在分子）
• 波源朝向观察者运动 → 频率升高（$-v_s$ 在分母）
• 波源远离观察者运动 → 频率降低（$+v_s$ 在分母）
其中 $f_o$ 是观察者接收的频率，$f_s$ 是波源发出的频率，$v$ 是波速，$v_o$ 是观察者速度，$v_s$ 是波源速度。
The Doppler effect is the change in observed frequency when there is relative motion between the source and observer.
• Observer moving toward source → frequency increases ($+v_o$ in numerator)
• Observer moving away from source → frequency decreases ($-v_o$ in numerator)
• Source moving toward observer → frequency increases ($-v_s$ in denominator)
• Source moving away from observer → frequency decreases ($+v_s$ in denominator)
Where $f_o$ is observed frequency, $f_s$ is source frequency, $v$ is wave speed, $v_o$ is observer speed, $v_s$ is source speed.
• Observer moving toward source → frequency increases ($+v_o$ in numerator)
• Observer moving away from source → frequency decreases ($-v_o$ in numerator)
• Source moving toward observer → frequency increases ($-v_s$ in denominator)
• Source moving away from observer → frequency decreases ($+v_s$ in denominator)
Where $f_o$ is observed frequency, $f_s$ is source frequency, $v$ is wave speed, $v_o$ is observer speed, $v_s$ is source speed.
📐 多普勒公式推导 / Derivation of Doppler Formula
波源运动的情况：
波源以速度 $v_s$ 朝向静止观察者运动。波源在 $T$ 时间内发出的波在空间中的长度：
$\lambda_o = \lambda - v_s T = v/f_s - v_s/f_s = (v - v_s)/f_s$
观察者接收到的频率：
$f_o = v/\lambda_o = vf_s/(v - v_s)$
观察者运动的情况：
观察者以速度 $v_o$ 朝向静止波源运动。波相对于观察者的速度：
$v_{\text{相对}} = v + v_o$
观察者接收到的频率：
$f_o = (v + v_o)/\lambda = (v + v_o)f_s/v$
一般情况：
将两种情况合并，得到完整公式：
$f_o = f_s \displaystyle \frac{v \pm v_o}{v \mp v_s}$
分子：观察者运动，朝向用 $+$，远离用 $-$
分母：波源运动，朝向用 $-$，远离用 $+$
波源以速度 $v_s$ 朝向静止观察者运动。波源在 $T$ 时间内发出的波在空间中的长度：
$\lambda_o = \lambda - v_s T = v/f_s - v_s/f_s = (v - v_s)/f_s$
观察者接收到的频率：
$f_o = v/\lambda_o = vf_s/(v - v_s)$
观察者运动的情况：
观察者以速度 $v_o$ 朝向静止波源运动。波相对于观察者的速度：
$v_{\text{相对}} = v + v_o$
观察者接收到的频率：
$f_o = (v + v_o)/\lambda = (v + v_o)f_s/v$
一般情况：
将两种情况合并，得到完整公式：
$f_o = f_s \displaystyle \frac{v \pm v_o}{v \mp v_s}$
分子：观察者运动，朝向用 $+$，远离用 $-$
分母：波源运动，朝向用 $-$，远离用 $+$
Source moving:
Source moving at $v_s$ toward stationary observer. Wavelength compressed:
$\lambda_o = (v - v_s)/f_s$
$f_o = v/\lambda_o = vf_s/(v - v_s)$
Observer moving:
Observer moving at $v_o$ toward stationary source. Relative wave speed:
$v_{\text{rel}} = v + v_o$
$f_o = (v + v_o)f_s/v$
General case:
$f_o = f_s \displaystyle \frac{v \pm v_o}{v \mp v_s}$
Numerator: observer motion, + toward, - away
Denominator: source motion, - toward, + away
Source moving at $v_s$ toward stationary observer. Wavelength compressed:
$\lambda_o = (v - v_s)/f_s$
$f_o = v/\lambda_o = vf_s/(v - v_s)$
Observer moving:
Observer moving at $v_o$ toward stationary source. Relative wave speed:
$v_{\text{rel}} = v + v_o$
$f_o = (v + v_o)f_s/v$
General case:
$f_o = f_s \displaystyle \frac{v \pm v_o}{v \mp v_s}$
Numerator: observer motion, + toward, - away
Denominator: source motion, - toward, + away
应用：多普勒测速 / Application: Doppler Speed Measurement
雷达测速（如警用雷达枪）：向车辆发射无线电波，接收反射波。反射波的频率因车辆运动而发生多普勒频移。测量频移 $\Delta f$ 即可计算车速：$\Delta f \approx 2f_s v_{\text{车}}/c$。
医学多普勒超声：利用超声波的多普勒频移测量血流速度。探头向血管发射超声，红细胞反射的回波发生频移——血液朝向探头运动时频率升高，远离时降低。通过频移推算血流速度。
天文多普勒：恒星光谱线的多普勒频移可以告诉我们恒星是朝向地球运动（蓝移/blue shift）还是远离地球运动（红移/red shift），从而测量宇宙膨胀。
医学多普勒超声：利用超声波的多普勒频移测量血流速度。探头向血管发射超声，红细胞反射的回波发生频移——血液朝向探头运动时频率升高，远离时降低。通过频移推算血流速度。
天文多普勒：恒星光谱线的多普勒频移可以告诉我们恒星是朝向地球运动（蓝移/blue shift）还是远离地球运动（红移/red shift），从而测量宇宙膨胀。
Radar speed detection: radio waves reflected from a moving vehicle experience a Doppler shift: $\Delta f \approx 2f_s v_{\text{car}}/c$.
Medical Doppler ultrasound: measures blood flow velocity. Ultrasound reflected from moving red blood cells shifts in frequency — toward the probe increases frequency, away decreases it.
Astronomical Doppler: spectral line shifts reveal whether stars move toward Earth (blue shift) or away (red shift), enabling measurement of cosmic expansion.
Medical Doppler ultrasound: measures blood flow velocity. Ultrasound reflected from moving red blood cells shifts in frequency — toward the probe increases frequency, away decreases it.
Astronomical Doppler: spectral line shifts reveal whether stars move toward Earth (blue shift) or away (red shift), enabling measurement of cosmic expansion.
🎯 IB 考试要点 / Exam Tips：① 驻波中相邻波节（或波腹）间距为 $\lambda/2$，不是 $\lambda$；② 乐器中弦的基频为 $f_1 = v/2L$，$v = \sqrt{T/\mu}$，张力 $T$ 越大、线密度 $\mu$ 越小、弦长 $L$ 越短 → 频率越高；③ 管乐器中注意开口端是波腹（位移最大处），闭口端是波节（位移为零处）；④ 多普勒效应公式中的符号要特别注意——建议记作"分子：观朝加（频率升高），分母：源朝减（频率升高）"；⑤ 多普勒效应适用于所有波，但对光波有相对论公式。
🎯 IB Exam Tips: ① In standing waves, adjacent nodes (or antinodes) are $\lambda/2$ apart, not $\lambda$; ② String fundamental $f_1 = v/2L$, $v = \sqrt{T/\mu}$ — higher tension $T$, lower linear density $\mu$, shorter length $L$ → higher frequency; ③ In pipes, open end is an antinode, closed end is a node; ④ Doppler sign convention: numerator + toward observer raises $f$, denominator - toward source raises $f$; ⑤ Doppler effect applies to all waves, but light has a relativistic formula.
例 4.8 / Example 4.8
一根吉他弦长 0.65 m，线密度 $5.0 \times 10^{-4}$ kg/m，张力 120 N。求基频和一阶泛音频率。
Guitar string: $L = 0.65$ m, $\mu = 5.0 \times 10^{-4}$ kg/m, $T = 120$ N. Find fundamental and first overtone frequencies.
解题过程 / Solution
波速：$v = \sqrt{T/\mu} = \sqrt{120/5.0 \times 10^{-4}} = \sqrt{240000} = 490$ m/s
基频（$n=1$）：$f_1 = v/2L = 490 / (2 \times 0.65) = 490/1.30 = 377$ Hz
一阶泛音（$n=2$）：$f_2 = 2f_1 = 2 \times 377 = 754$ Hz
在吉他上，弦的音高可以通过调节张力（调音旋钮）或改变有效弦长（按压品柱）来改变。
基频（$n=1$）：$f_1 = v/2L = 490 / (2 \times 0.65) = 490/1.30 = 377$ Hz
一阶泛音（$n=2$）：$f_2 = 2f_1 = 2 \times 377 = 754$ Hz
在吉他上，弦的音高可以通过调节张力（调音旋钮）或改变有效弦长（按压品柱）来改变。
Wave speed: $v = \sqrt{T/\mu} = \sqrt{120/5.0 \times 10^{-4}} = 490$ m/s
Fundamental ($n=1$): $f_1 = v/2L = 490 / (2 \times 0.65) = 377$ Hz
First overtone ($n=2$): $f_2 = 2f_1 = 754$ Hz
On a guitar, pitch is changed by adjusting tension (tuning pegs) or changing effective length (pressing frets).
Fundamental ($n=1$): $f_1 = v/2L = 490 / (2 \times 0.65) = 377$ Hz
First overtone ($n=2$): $f_2 = 2f_1 = 754$ Hz
On a guitar, pitch is changed by adjusting tension (tuning pegs) or changing effective length (pressing frets).
例 4.9 / Example 4.9
一根一端闭口的管长 0.85 m，声速 340 m/s。求基频和第一个可产生的泛音频率。
A pipe closed at one end: $L = 0.85$ m, $v = 340$ m/s. Find fundamental and first possible overtone.
解题过程 / Solution
一端闭口管：$f_n = nv/4L$，$n = 1, 3, 5, \ldots$（只有奇数）
基频（$n=1$）：$f_1 = 340 / (4 \times 0.85) = 340/3.40 = 100$ Hz
下一个可产生的泛音（$n=3$）：$f_3 = 3 \times 340 / (4 \times 0.85) = 1020/3.40 = 300$ Hz
注意：$n=2$ 不存在，因此一端闭口管的泛音是基频的 3 倍、5 倍……而不是 2 倍、3 倍……这就是单簧管（一端闭口）音色与长笛（两端开口）不同的原因。
基频（$n=1$）：$f_1 = 340 / (4 \times 0.85) = 340/3.40 = 100$ Hz
下一个可产生的泛音（$n=3$）：$f_3 = 3 \times 340 / (4 \times 0.85) = 1020/3.40 = 300$ Hz
注意：$n=2$ 不存在，因此一端闭口管的泛音是基频的 3 倍、5 倍……而不是 2 倍、3 倍……这就是单簧管（一端闭口）音色与长笛（两端开口）不同的原因。
Pipe closed at one end: $f_n = nv/4L$, $n = 1, 3, 5, \ldots$ (odd only)
Fundamental ($n=1$): $f_1 = 340 / (4 \times 0.85) = 100$ Hz
Next possible overtone ($n=3$): $f_3 = 3 \times 340 / (4 \times 0.85) = 300$ Hz
Note: $n=2$ is not allowed — overtones are 3x, 5x the fundamental. This is why a clarinet (closed pipe) sounds different from a flute (open pipe).
Fundamental ($n=1$): $f_1 = 340 / (4 \times 0.85) = 100$ Hz
Next possible overtone ($n=3$): $f_3 = 3 \times 340 / (4 \times 0.85) = 300$ Hz
Note: $n=2$ is not allowed — overtones are 3x, 5x the fundamental. This is why a clarinet (closed pipe) sounds different from a flute (open pipe).
例 4.10 / Example 4.10
警车鸣笛（频率 1000 Hz）以 30 m/s 驶向静止观察者。声速 340 m/s。求观察者听到的频率。警车远离时呢？
Police siren: $f_s = 1000$ Hz, $v_s = 30$ m/s toward stationary observer. $v = 340$ m/s. Find $f_o$ when approaching and when receding.
解题过程 / Solution
波源朝向观察者运动（$-$ 在分母）：
$f_o = f_s \cdot v/(v - v_s) = 1000 \times 340/(340 - 30) = 1000 \times 340/310 = 1097$ Hz
波源远离观察者运动（$+$ 在分母）：
$f_o = f_s \cdot v/(v + v_s) = 1000 \times 340/(340 + 30) = 1000 \times 340/370 = 919$ Hz
警车经过观察者的瞬间，频率从 1097 Hz 突然下降到 919 Hz——这就是警笛声从高变低的"呜哇"效果。
$f_o = f_s \cdot v/(v - v_s) = 1000 \times 340/(340 - 30) = 1000 \times 340/310 = 1097$ Hz
波源远离观察者运动（$+$ 在分母）：
$f_o = f_s \cdot v/(v + v_s) = 1000 \times 340/(340 + 30) = 1000 \times 340/370 = 919$ Hz
警车经过观察者的瞬间，频率从 1097 Hz 突然下降到 919 Hz——这就是警笛声从高变低的"呜哇"效果。
Source approaching ($-$ in denominator):
$f_o = 1000 \times 340/(340 - 30) = 1097$ Hz
Source receding ($+$ in denominator):
$f_o = 1000 \times 340/(340 + 30) = 919$ Hz
When the police car passes the observer, the frequency drops abruptly from 1097 Hz to 919 Hz — the familiar "wow-wow" effect.
$f_o = 1000 \times 340/(340 - 30) = 1097$ Hz
Source receding ($+$ in denominator):
$f_o = 1000 \times 340/(340 + 30) = 919$ Hz
When the police car passes the observer, the frequency drops abruptly from 1097 Hz to 919 Hz — the familiar "wow-wow" effect.
🏆 挑战题 / Challenge Problems
多概念综合 / Multi-concept难度 / Hard
以下题目需要综合运用本章的多个知识点。建议先复习所有内容后再尝试。
These problems require integrating multiple concepts from this chapter. Review all sections before attempting.
挑战 1 / Challenge 1 SHM 能量与弹簧系统 / SHM Energy and Spring System
一个质量为 0.20 kg 的物体连接在劲度系数 $k = 80$ N/m 的弹簧上，在光滑水平面上做 SHM，振幅 0.040 m。求：(a) 总能量；(b) 物体在 $x = 0.025$ m 时的动能和势能；(c) 动能等于势能时的位移。
A 0.20 kg mass on a spring ($k = 80$ N/m) oscillates with $A = 0.040$ m on a frictionless surface. Find: (a) total energy; (b) $E_k$ and $E_p$ at $x = 0.025$ m; (c) displacement when $E_k = E_p$.
解题过程 / Solution
(a) $E_{\text{总}} = \frac12 kA^2 = \frac12 \times 80 \times (0.040)^2 = \frac12 \times 80 \times 0.0016 = 0.064$ J
(b) 在 $x = 0.025$ m：
$E_p = \frac12 kx^2 = \frac12 \times 80 \times (0.025)^2 = 0.5 \times 80 \times 0.000625 = 0.025$ J
$E_k = E_{\text{总}} - E_p = 0.064 - 0.025 = 0.039$ J
(c) $E_k = E_p$ 时，$E_p = \frac12 E_{\text{总}}$：
$\frac12 kx^2 = \frac12 \times \frac12 kA^2$
$x^2 = \frac12 A^2$
$x = \pm A/\sqrt{2} = \pm 0.040/\sqrt{2} = \pm 0.0283$ m
这个结果 $\pm A/\sqrt{2}$ 对任何 SHM 系统都成立——动能和势能相等发生在位移为 $A/\sqrt{2}$ 处。
(b) 在 $x = 0.025$ m：
$E_p = \frac12 kx^2 = \frac12 \times 80 \times (0.025)^2 = 0.5 \times 80 \times 0.000625 = 0.025$ J
$E_k = E_{\text{总}} - E_p = 0.064 - 0.025 = 0.039$ J
(c) $E_k = E_p$ 时，$E_p = \frac12 E_{\text{总}}$：
$\frac12 kx^2 = \frac12 \times \frac12 kA^2$
$x^2 = \frac12 A^2$
$x = \pm A/\sqrt{2} = \pm 0.040/\sqrt{2} = \pm 0.0283$ m
这个结果 $\pm A/\sqrt{2}$ 对任何 SHM 系统都成立——动能和势能相等发生在位移为 $A/\sqrt{2}$ 处。
(a) $E_{\text{total}} = \frac12 kA^2 = 0.064$ J
(b) At $x = 0.025$ m: $E_p = \frac12 kx^2 = 0.025$ J, $E_k = E_{\text{total}} - E_p = 0.039$ J
(c) When $E_k = E_p$: $\frac12 kx^2 = \frac14 kA^2$
$x = \pm A/\sqrt{2} = \pm 0.0283$ m
This result $\pm A/\sqrt{2}$ holds for any SHM system — kinetic and potential energy are equal at $x = A/\sqrt{2}$.
(b) At $x = 0.025$ m: $E_p = \frac12 kx^2 = 0.025$ J, $E_k = E_{\text{total}} - E_p = 0.039$ J
(c) When $E_k = E_p$: $\frac12 kx^2 = \frac14 kA^2$
$x = \pm A/\sqrt{2} = \pm 0.0283$ m
This result $\pm A/\sqrt{2}$ holds for any SHM system — kinetic and potential energy are equal at $x = A/\sqrt{2}$.
挑战 2 / Challenge 2 多普勒效应与相对运动 / Doppler Effect and Relative Motion
一辆救护车以 25 m/s 的速度驶向一座山壁并鸣笛（频率 800 Hz）。声速 340 m/s。求：(a) 山壁反射的回波频率（即救护车司机听到的回声频率）；(b) 若救护车静止并以 20 m/s 驶向山壁的观察者听到的频率。
An ambulance moves at 25 m/s toward a cliff, siren at 800 Hz. $v_{\text{sound}} = 340$ m/s. Find: (a) frequency of echo reflected from cliff (what the driver hears); (b) if ambulance is stationary, observer moving at 20 m/s toward cliff hears what frequency?
解题过程 / Solution
(a) 回波需分两步计算：
第一步：山壁作为"观察者"，接收救护车发出的声音。
波源运动（25 m/s 朝向山壁）：$f_{\text{壁}} = f_s \cdot v/(v - v_s) = 800 \times 340/(340 - 25) = 800 \times 340/315 = 863.5$ Hz
第二步：山壁反射声音，相当于"波源"（山壁）以频率 $f_{\text{壁}}$ 发出声音，救护车作为观察者向波源运动。
观察者运动（25 m/s 朝向山壁）：$f_{\text{echo}} = f_{\text{壁}} \times (v + v_o)/v = 863.5 \times (340 + 25)/340 = 863.5 \times 365/340 \approx 927$ Hz
所以救护车司机听到的回声频率为约 927 Hz（比原频率 800 Hz 高很多）。
(b) 救护车静止（$f_s = 800$ Hz），观察者以 20 m/s 朝向山壁运动。山壁反射的声音到达观察者：
山壁反射频率 = 800 Hz（波源静止）
观察者运动朝向波源：$f_o = 800 \times (340 + 20)/340 = 800 \times 360/340 = 847$ Hz
第一步：山壁作为"观察者"，接收救护车发出的声音。
波源运动（25 m/s 朝向山壁）：$f_{\text{壁}} = f_s \cdot v/(v - v_s) = 800 \times 340/(340 - 25) = 800 \times 340/315 = 863.5$ Hz
第二步：山壁反射声音，相当于"波源"（山壁）以频率 $f_{\text{壁}}$ 发出声音，救护车作为观察者向波源运动。
观察者运动（25 m/s 朝向山壁）：$f_{\text{echo}} = f_{\text{壁}} \times (v + v_o)/v = 863.5 \times (340 + 25)/340 = 863.5 \times 365/340 \approx 927$ Hz
所以救护车司机听到的回声频率为约 927 Hz（比原频率 800 Hz 高很多）。
(b) 救护车静止（$f_s = 800$ Hz），观察者以 20 m/s 朝向山壁运动。山壁反射的声音到达观察者：
山壁反射频率 = 800 Hz（波源静止）
观察者运动朝向波源：$f_o = 800 \times (340 + 20)/340 = 800 \times 360/340 = 847$ Hz
(a) Two-step calculation:
Step 1: Cliff as "observer" — source moving toward cliff:
$f_{\text{cliff}} = 800 \times 340/(340 - 25) = 863.5$ Hz
Step 2: Cliff reflects — now cliff is "source" at 863.5 Hz, ambulance moving toward it as observer:
$f_{\text{echo}} = 863.5 \times (340 + 25)/340 \approx 927$ Hz
The ambulance driver hears the echo at ~927 Hz.
(b) Stationary source (800 Hz), observer moving at 20 m/s toward cliff:
$f_o = 800 \times (340 + 20)/340 = 847$ Hz
Step 1: Cliff as "observer" — source moving toward cliff:
$f_{\text{cliff}} = 800 \times 340/(340 - 25) = 863.5$ Hz
Step 2: Cliff reflects — now cliff is "source" at 863.5 Hz, ambulance moving toward it as observer:
$f_{\text{echo}} = 863.5 \times (340 + 25)/340 \approx 927$ Hz
The ambulance driver hears the echo at ~927 Hz.
(b) Stationary source (800 Hz), observer moving at 20 m/s toward cliff:
$f_o = 800 \times (340 + 20)/340 = 847$ Hz
挑战 3 / Challenge 3 驻波与弦的边界校正 / Standing Waves with End Correction
一根长 0.60 m 的弦两端固定，线密度 $4.0 \times 10^{-4}$ kg/m，张力 90 N。求：(a) 基频和前三阶谐波频率；(b) 如果弦的中间 1/3 被手指轻轻接触（成为波节），可以产生的最高可能频率是多少？
A 0.60 m string, $\mu = 4.0 \times 10^{-4}$ kg/m, $T = 90$ N. Find: (a) fundamental and first three harmonics; (b) if the middle third is lightly touched (creating a node), what's the highest possible frequency?
解题过程 / Solution
(a) 波速：$v = \sqrt{T/\mu} = \sqrt{90/(4.0 \times 10^{-4})} = \sqrt{225000} = 474$ m/s
$f_n = nv/(2L)$，$n = 1, 2, 3, \ldots$
$f_1 = 474/(2 \times 0.60) = 474/1.20 = 395$ Hz
$f_2 = 2 \times 395 = 790$ Hz
$f_3 = 3 \times 395 = 1185$ Hz
$f_4 = 4 \times 395 = 1580$ Hz
(b) 在弦的中点轻触产生波节，相当于将弦分成两段独立的振动部分，每段长 $L/2 = 0.30$ m。
每段的基频：$f_1' = v/(2 \times 0.30) = 474/0.60 = 790$ Hz
实际上，轻触中点使得弦只能产生以该点为波节的模式——即偶次谐波（$n = 2, 4, 6, \ldots$）。其中 $n=2$ 对应 790 Hz，$n=4$ 对应 1580 Hz。如果手指轻轻接触不吸收能量，可以产生的频率为 $n \times 395$ Hz（$n$ 为偶数）。在 2000 Hz 以下的最高频率为 $n=4$ 时的 1580 Hz。
$f_n = nv/(2L)$，$n = 1, 2, 3, \ldots$
$f_1 = 474/(2 \times 0.60) = 474/1.20 = 395$ Hz
$f_2 = 2 \times 395 = 790$ Hz
$f_3 = 3 \times 395 = 1185$ Hz
$f_4 = 4 \times 395 = 1580$ Hz
(b) 在弦的中点轻触产生波节，相当于将弦分成两段独立的振动部分，每段长 $L/2 = 0.30$ m。
每段的基频：$f_1' = v/(2 \times 0.30) = 474/0.60 = 790$ Hz
实际上，轻触中点使得弦只能产生以该点为波节的模式——即偶次谐波（$n = 2, 4, 6, \ldots$）。其中 $n=2$ 对应 790 Hz，$n=4$ 对应 1580 Hz。如果手指轻轻接触不吸收能量，可以产生的频率为 $n \times 395$ Hz（$n$ 为偶数）。在 2000 Hz 以下的最高频率为 $n=4$ 时的 1580 Hz。
(a) $v = \sqrt{T/\mu} = 474$ m/s
$f_1 = v/2L = 474/1.20 = 395$ Hz
$f_2 = 790$ Hz, $f_3 = 1185$ Hz, $f_4 = 1580$ Hz
(b) Touching the midpoint creates a node there — the string effectively becomes two half-length segments ($L/2 = 0.30$ m).
Each segment's fundamental: $f_1' = v/(2 \times 0.30) = 790$ Hz
This only allows even harmonics ($n = 2, 4, 6, \ldots$) of the original string. The 4th harmonic (1580 Hz) is the highest below 2000 Hz.
$f_1 = v/2L = 474/1.20 = 395$ Hz
$f_2 = 790$ Hz, $f_3 = 1185$ Hz, $f_4 = 1580$ Hz
(b) Touching the midpoint creates a node there — the string effectively becomes two half-length segments ($L/2 = 0.30$ m).
Each segment's fundamental: $f_1' = v/(2 \times 0.30) = 790$ Hz
This only allows even harmonics ($n = 2, 4, 6, \ldots$) of the original string. The 4th harmonic (1580 Hz) is the highest below 2000 Hz.
挑战 4 / Challenge 4 SHM + 多普勒综合 / SHM + Doppler Combined
一个音叉（频率 440 Hz）安装在弹簧振子上做 SHM，振幅 0.050 m，频率 2.0 Hz。音叉面向静止观察者振动。声速 340 m/s。求观察者听到的最高和最低频率。
A tuning fork (440 Hz) mounted on a spring oscillates in SHM: $A = 0.050$ m, $f_{\text{SHM}} = 2.0$ Hz. The fork faces a stationary observer. $v_{\text{sound}} = 340$ m/s. Find the highest and lowest frequencies heard.
解题过程 / Solution
音叉在 SHM 中的速度：$v(t) = \pm \omega A\sin(\omega t)$，最大速度 $v_{\max} = \omega A$。
$\omega_{\text{SHM}} = 2\pi f_{\text{SHM}} = 2\pi \times 2.0 = 4\pi$ rad/s
$v_{\max} = \omega A = 4\pi \times 0.050 = 0.628$ m/s
音叉朝向观察者运动时频率最高：
$f_{\max} = f_s \cdot v/(v - v_s) = 440 \times 340/(340 - 0.628) = 440 \times 340/339.372 \approx 440 \times 1.00185 \approx 440.8$ Hz
音叉远离观察者运动时频率最低：
$f_{\min} = f_s \cdot v/(v + v_s) = 440 \times 340/(340 + 0.628) = 440 \times 340/340.628 \approx 440 \times 0.99816 \approx 439.2$ Hz
频率变化范围约为 $\pm 0.8$ Hz（约 0.18% 的变化）。因为音叉速度远小于声速，所以多普勒频移很小。
一般地，对于 $v_s \ll v$：$\Delta f \approx f_s \cdot v_s/v$。本题中 $\Delta f \approx 440 \times 0.628/340 = 0.81$ Hz。
$\omega_{\text{SHM}} = 2\pi f_{\text{SHM}} = 2\pi \times 2.0 = 4\pi$ rad/s
$v_{\max} = \omega A = 4\pi \times 0.050 = 0.628$ m/s
音叉朝向观察者运动时频率最高：
$f_{\max} = f_s \cdot v/(v - v_s) = 440 \times 340/(340 - 0.628) = 440 \times 340/339.372 \approx 440 \times 1.00185 \approx 440.8$ Hz
音叉远离观察者运动时频率最低：
$f_{\min} = f_s \cdot v/(v + v_s) = 440 \times 340/(340 + 0.628) = 440 \times 340/340.628 \approx 440 \times 0.99816 \approx 439.2$ Hz
频率变化范围约为 $\pm 0.8$ Hz（约 0.18% 的变化）。因为音叉速度远小于声速，所以多普勒频移很小。
一般地，对于 $v_s \ll v$：$\Delta f \approx f_s \cdot v_s/v$。本题中 $\Delta f \approx 440 \times 0.628/340 = 0.81$ Hz。
The fork's SHM speed: $v_{\max} = \omega A = 4\pi \times 0.050 = 0.628$ m/s
Moving toward observer (max frequency):
$f_{\max} = 440 \times 340/(340 - 0.628) \approx 440.8$ Hz
Moving away (min frequency):
$f_{\min} = 440 \times 340/(340 + 0.628) \approx 439.2$ Hz
Frequency variation is $\pm 0.8$ Hz (~0.18% change). The fork speed is much less than sound speed, so the Doppler shift is small.
General approximation for $v_s \ll v$: $\Delta f \approx f_s \cdot v_s/v = 440 \times 0.628/340 = 0.81$ Hz.
Moving toward observer (max frequency):
$f_{\max} = 440 \times 340/(340 - 0.628) \approx 440.8$ Hz
Moving away (min frequency):
$f_{\min} = 440 \times 340/(340 + 0.628) \approx 439.2$ Hz
Frequency variation is $\pm 0.8$ Hz (~0.18% change). The fork speed is much less than sound speed, so the Doppler shift is small.
General approximation for $v_s \ll v$: $\Delta f \approx f_s \cdot v_s/v = 440 \times 0.628/340 = 0.81$ Hz.
挑战 5 / Challenge 5 声波干涉与驻波 / Sound Interference and Standing Waves
两个扬声器相距 2.0 m，发出同频率（340 Hz）同相位的声波。声速 340 m/s。一个听众沿两个扬声器的中垂线从远处走近。求：(a) 波长；(b) 听众听到的第一个声音最大处距两扬声器连线的距离（即中垂线上距中心多远的距离处第一次出现强度最大？提示：考虑两列波到达听众处的相位差）。
Two speakers 2.0 m apart emit in-phase sound at 340 Hz. $v = 340$ m/s. A listener walks along the perpendicular bisector from far away. Find: (a) $\lambda$; (b) distance from the center line where the first maximum occurs.
解题过程 / Solution
(a) $\lambda = v/f = 340/340 = 1.0$ m
(b) 中垂线上所有点到两扬声器的距离相等，$\Delta r = 0$，此为 $n=0$ 级相长干涉（最大声强）。
题目实际想问的是：听众沿平行于扬声器连线的方向移动时，第一次出现最大声强的位置。
设听众距离中垂线中心的距离为 $y$，两扬声器间距为 $d = 2.0$ m。听众到两个扬声器的波程差：$\Delta r = \sqrt{(d/2)^2 + y^2} - \sqrt{(d/2)^2 + y^2}$，在中垂线上此值为 0。
更合理的题目：听众在距离扬声器连线 $D = 10$ m 的平行线上行走（即扬声器连线方向）。两列声波到达听众的波程差为 $\Delta r = d\sin\theta$，其中 $\theta$ 为偏离中垂线的角度。
第一级最大（$n=1$）：$\Delta r = \lambda = 1.0$ m
$d\sin\theta = \lambda$，$\sin\theta = \lambda/d = 1.0/2.0 = 0.5$
$\theta = 30^\circ$
听众到中垂线的横向距离：$y = D\tan\theta = 10 \times \tan 30^\circ = 10 \times 0.577 = 5.77$ m
(b) 中垂线上所有点到两扬声器的距离相等，$\Delta r = 0$，此为 $n=0$ 级相长干涉（最大声强）。
题目实际想问的是：听众沿平行于扬声器连线的方向移动时，第一次出现最大声强的位置。
设听众距离中垂线中心的距离为 $y$，两扬声器间距为 $d = 2.0$ m。听众到两个扬声器的波程差：$\Delta r = \sqrt{(d/2)^2 + y^2} - \sqrt{(d/2)^2 + y^2}$，在中垂线上此值为 0。
更合理的题目：听众在距离扬声器连线 $D = 10$ m 的平行线上行走（即扬声器连线方向）。两列声波到达听众的波程差为 $\Delta r = d\sin\theta$，其中 $\theta$ 为偏离中垂线的角度。
第一级最大（$n=1$）：$\Delta r = \lambda = 1.0$ m
$d\sin\theta = \lambda$，$\sin\theta = \lambda/d = 1.0/2.0 = 0.5$
$\theta = 30^\circ$
听众到中垂线的横向距离：$y = D\tan\theta = 10 \times \tan 30^\circ = 10 \times 0.577 = 5.77$ m
(a) $\lambda = v/f = 340/340 = 1.0$ m
(b) On the perpendicular bisector, path difference is always 0 ($n=0$ constructive interference).
For a listener walking parallel to the speaker line at distance $D = 10$ m:
First maximum ($n=1$): $\Delta r = d\sin\theta = \lambda$
$\sin\theta = \lambda/d = 1.0/2.0 = 0.5$, $\theta = 30^\circ$
Lateral distance: $y = D\tan\theta = 10 \times \tan 30^\circ = 5.77$ m
(b) On the perpendicular bisector, path difference is always 0 ($n=0$ constructive interference).
For a listener walking parallel to the speaker line at distance $D = 10$ m:
First maximum ($n=1$): $\Delta r = d\sin\theta = \lambda$
$\sin\theta = \lambda/d = 1.0/2.0 = 0.5$, $\theta = 30^\circ$
Lateral distance: $y = D\tan\theta = 10 \times \tan 30^\circ = 5.77$ m
附录：核心公式速查 / Formula Reference
| 知识点 / Topic | 公式 / Formula | 说明 / Notes |
|---|---|---|
| SHM 定义 / SHM Condition | $a = -\omega^2 x$ | 加速度与位移成正比，方向相反 |
| SHM 位移 / Displacement | $x = A\cos(\omega t + \phi)$ | $A$ 为振幅，$\phi$ 为初相位 |
| SHM 速度 / Velocity | $v = \pm \omega\sqrt{A^2 - x^2}$ | $v_{\max} = \omega A$ 在 $x = 0$ 处 |
| SHM 加速度 / Acceleration | $a = -\omega^2 x$ | $a_{\max} = \omega^2 A$ 在 $x = \pm A$ 处 |
| 角频率与周期 | $\omega = 2\pi/T = 2\pi f$ | $T = 1/f$ |
| 弹簧振子 / Spring | $\omega = \sqrt{k/m}$, $T = 2\pi\sqrt{m/k}$ | $k$ 为劲度系数 |
| 单摆 / Pendulum | $\omega = \sqrt{g/l}$, $T = 2\pi\sqrt{l/g}$ | 小角度近似（$\theta < 10^\circ$） |
| SHM 总能量 / Total Energy | $E_{\text{总}} = \frac12 m\omega^2 A^2 = \frac12 kA^2$ | $E \propto A^2$ |
| 波速公式 / Wave Speed | $v = f\lambda = \lambda/T$ | 频率由波源决定，波速由介质决定 |
| 相位差 / Phase Diff. | $\Delta\phi = 2\pi\Delta x/\lambda = 2\pi\Delta t/T$ | $2\pi$ 整数倍同相，$\pi$ 奇数倍反相 |
| 叠加原理 / Superposition | $y_{\text{总}} = y_1 + y_2$ | 适用于所有波 |
| 相长干涉 / Constructive | $\Delta r = n\lambda$ | $n = 0, 1, 2, \ldots$ |
| 相消干涉 / Destructive | $\Delta r = (n + \frac12)\lambda$ | $n = 0, 1, 2, \ldots$ |
| 反射定律 / Reflection | $\theta_i = \theta_r$ | 固定端反相，自由端同相 |
| 折射定律 / Refraction | $\sin\theta_1/\sin\theta_2 = v_1/v_2$ | 频率不变 |
| 衍射 / Diffraction | $\theta \approx \lambda/b$ | $\lambda \approx b$ 时显著 |
| 驻波方程 / Standing Wave | $y = 2A\cos(kx)\sin(\omega t)$ | 相邻波节间距 $\lambda/2$ |
| 弦振动 / String | $f_n = nv/2L$, $v = \sqrt{T/\mu}$ | $n = 1, 2, 3, \ldots$ |
| 两端开口管 / Open Pipe | $f_n = nv/2L$ | $n = 1, 2, 3, \ldots$ |
| 一端闭口管 / Closed Pipe | $f_n = nv/4L$ | $n = 1, 3, 5, \ldots$（奇数） |
| 多普勒效应 / Doppler | $f_o = f_s(v \pm v_o)/(v \mp v_s)$ | 分子：观朝+；分母：源朝- |
📝 分节练习 / Section Practice
按知识点逐节练习，每题即时反馈。完成一节后自动记录进度。
Practice section by section with instant feedback. Progress is automatically saved.
由易到难 / Easy→Hard即时反馈 / Instant记录进度 / Progress Saved
🎯 随机测试 / Random Quiz
从题库随机抽取题目，模拟考试环境。提交后统一判分。
Randomly drawn from the question bank. Simulates exam conditions. Graded after submission.
测试设置 / Quiz Settings
📊 成绩分析 / Score Analysis
📭
尚无测试记录 / No Records Yet