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Waves and Sound · Solutions波与声 · 详解

Companion to the Practice Set · Mark-by-mark walkthroughs · AP-Feeder / ON / BC / AB styles练习题配套答案 · 逐分讲解 · AP 衔接 / 安 / 卑 / 阿省考风格

EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 🇨🇦 AB阿 AP-style MCQAP 风格选择题 AP-feeder FRQAP 衔接简答题 ON Provincial-style安大略省考风格 BC Provincial-style卑诗省考风格 AB Diploma-style阿尔伯塔毕业考风格 Honors荣誉级
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 详解AP MCQ + ON/BC short answer · 25 marksAP 选择题 + 安/卑省考短答 · 共 25 分

Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §1 Wave types波的类型 · HS-PS4-1 [3 marks][3 分]

In a longitudinal wave (sound in air), how do the medium particles move relative to the wave direction?在纵波（空气中的声波）中，介质质点相对于波的传播方向如何运动？

Answer:答案：  (B)  Parallel to the wave direction平行于波的传播方向

(a) Classify the wave by particle motion按质点运动分类波 M1·A1·A1

In a longitudinal wave the medium particles oscillate back and forth parallel to the direction the wave travels, creating alternating compressions (particles bunched) and rarefactions (particles spread out). Sound in air is the standard example. Option (B).在纵波中，介质质点沿波的传播方向平行地来回振动，形成交替的压缩区（质点密集）与稀疏区（质点稀疏）。空气中的声波是标准例子。选 (B)。
Why the distractors fail.干扰项分析。
(A): perpendicular oscillation describes a transverse wave (a wave on a string, or light), not a longitudinal one.垂直振动描述的是横波（绳波或光），而非纵波。
(C): circular particle motion describes surface water waves, a special mixed case, not a pure longitudinal wave.圆周质点运动描述的是水面波这种特殊混合情形，并非纯纵波。
(D): particles must oscillate, otherwise no energy would be transmitted.质点必须振动，否则无法传递能量。
Classify a wave by the angle between particle oscillation and energy travel.通过质点振动方向与能量传播方向的夹角来分类波。 A wave transports energy without transporting matter; the medium particles only oscillate about a fixed equilibrium. The single distinguishing question is the angle between that oscillation and the direction of energy transfer: parallel gives a longitudinal wave (sound, P-waves in earthquakes), perpendicular gives a transverse wave (string waves, light, S-waves). Memorise one example of each and reason from the picture; do not memorise the words alone, because exam questions often describe an unfamiliar wave and ask you to classify it from its motion.波传递能量而不传递物质；介质质点只在固定平衡位置附近振动。唯一的区分问题是该振动方向与能量传播方向之间的夹角：平行为纵波（声波、地震 P 波），垂直为横波（绳波、光、地震 S 波）。每类记住一个例子并据图推理；不要只记词，因为考题常描述一种陌生的波并要你据其运动来分类。
Q2EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §1 Period and frequency周期与频率 · HS-PS4-1 [3 marks][3 分]

A wave has period $T = 0.025$ s. Frequency?某波周期 $T = 0.025$ s。频率为多少？

Answer:答案：  (C)  $40\ \text{Hz}$

(a) Frequency is the reciprocal of period频率是周期的倒数 M1·A1·A1

$$ f \;=\; \frac{1}{T} \;=\; \frac{1}{0.025} \;=\; 40 \;\text{Hz.} $$ Option (C).选 (C)。
Why the distractors fail.干扰项分析。
(A) $0.025\ \text{Hz}$: simply restates $T$ as if it were $f$, ignoring the reciprocal.直接把 $T$ 当作 $f$，忽略了取倒数。
(B) $250\ \text{Hz}$: comes from $1/0.004$ or a misplaced decimal point.来自 $1/0.004$ 或小数点错位。
(D) $4.0\ \text{Hz}$: divides by $0.25$ instead of $0.025$, a one-decimal slip.误除以 $0.25$ 而非 $0.025$，差一位小数。
Period and frequency are reciprocals: $f = 1/T$, $T = 1/f$.周期与频率互为倒数：$f = 1/T$，$T = 1/f$。 Period $T$ is the time for one full oscillation (seconds); frequency $f$ is the number of oscillations per second (hertz). They carry the same information in inverted form. A quick sanity check on the arithmetic: $0.025$ s per cycle means $40$ cycles fit into one second ($40 \times 0.025 = 1.00$), so $f = 40$ Hz. Always confirm the decimal placement by checking that $f \times T = 1$. This reciprocal relationship is the foundation for the wave equation $v = f\lambda$ that follows.周期 $T$ 是完成一次完整振动所需的时间（秒）；频率 $f$ 是每秒振动次数（赫兹）。两者以倒数形式携带相同信息。快速核算：每个周期 $0.025$ s，意味着一秒内能容纳 $40$ 个周期（$40 \times 0.025 = 1.00$），故 $f = 40$ Hz。始终用 $f \times T = 1$ 来确认小数位。这一倒数关系是后续波动方程 $v = f\lambda$ 的基础。
Q3EASY易 🇨🇦 ON安 ON Provincial-style安大略省考风格 §1, §2 Wave anatomy波的解剖 · SPH3U E2 [4 marks][4 分]

Rope wave: crest-to-crest $0.40$ m, $5.0$ oscillations per second. (a) Wavelength and frequency. (b) Wave speed. (c) Period.绳波：波峰间距 $0.40$ m，每秒 $5.0$ 次振动。(a) 波长与频率。(b) 波速。(c) 周期。

Answer:答案：  (a) $\lambda = 0.40\ \text{m},\ f = 5.0\ \text{Hz}$  ·  (b) $v = 2.0\ \text{m/s}$  ·  (c) $T = 0.20\ \text{s}$

(a) Read off wavelength and frequency读出波长与频率 A1

The crest-to-crest distance is one full wavelength, so $\lambda = 0.40$ m. "$5.0$ oscillations per second" is the definition of frequency, so $f = 5.0$ Hz.波峰间距即一个完整波长，故 $\lambda = 0.40$ m。"每秒 $5.0$ 次振动"正是频率的定义，故 $f = 5.0$ Hz。

(b) Wave speed using $v = f\lambda$用 $v = f\lambda$ 求波速 M1·A1

$$ v \;=\; f\lambda \;=\; 5.0 \times 0.40 \;=\; 2.0 \;\text{m/s.} $$

(c) Period as the reciprocal of frequency周期为频率的倒数 A1

$$ T \;=\; \frac{1}{f} \;=\; \frac{1}{5.0} \;=\; 0.20 \;\text{s.} $$
Wavelength, frequency, period and speed are four views of one wave, linked by $v = f\lambda$ and $T = 1/f$.波长、频率、周期与波速是同一列波的四种视角，由 $v = f\lambda$ 和 $T = 1/f$ 联系。 The most common error here is to confuse "crest-to-crest" (one wavelength) with "crest-to-trough" (half a wavelength). Read the geometry carefully: crest to the next crest is a full cycle in space. Once $\lambda$ and $f$ are known, every other wave quantity follows: speed from $v = f\lambda$, period from $T = 1/f$. Note that speed here ($2.0$ m/s) is a property of the rope, not the source, while frequency is set by whoever shakes the rope. Provincial markers expect units stated on every line.此处最常见的错误是把"波峰到波峰"（一个波长）与"波峰到波谷"（半个波长）混淆。仔细读图：从一个波峰到下一个波峰是空间上的一个完整周期。一旦知道 $\lambda$ 与 $f$，其余波动量随之而来：由 $v = f\lambda$ 得速率，由 $T = 1/f$ 得周期。注意此处波速（$2.0$ m/s）是绳的性质，而非声源的性质，而频率由抖绳的人决定。省考阅卷人要求每一行都标注单位。
Q4MEDIUM中 🇺🇸 US美 AP-style MCQAP 风格选择题 §2 Wave equation波动方程 · HS-PS4-1 [3 marks][3 分]

Wave: speed $300$ m/s, wavelength $0.60$ m. Frequency?波：速率 $300$ m/s，波长 $0.60$ m。频率为多少？

Answer:答案：  (D)  $500\ \text{Hz}$

(a) Rearrange $v = f\lambda$ to solve for $f$整理 $v = f\lambda$ 求 $f$ M1·A1·A1

$$ f \;=\; \frac{v}{\lambda} \;=\; \frac{300}{0.60} \;=\; 500 \;\text{Hz.} $$ Option (D).选 (D)。
Why the distractors fail.干扰项分析。
(A) $180\ \text{Hz}$: multiplies $300 \times 0.60$ instead of dividing.把 $300 \times 0.60$ 相乘而非相除。
(B) $0.002\ \text{Hz}$: inverts the ratio, computing $\lambda / v$.将比值颠倒，算成 $\lambda / v$。
(C) $300\ \text{Hz}$: copies the speed value, ignoring the wavelength entirely.直接照抄速率值，完全忽略波长。
The wave equation $v = f\lambda$ rearranges three ways; pick the form with your unknown isolated.波动方程 $v = f\lambda$ 可三向变形；选未知量被单独隔出的形式。 The single relation $v = f\lambda$ gives $f = v/\lambda$, $\lambda = v/f$, and $v = f\lambda$. A units check resolves any doubt about which to use: $f$ must come out in hertz ($1/\text{s}$), and $(\text{m/s}) / (\text{m}) = 1/\text{s}$ confirms $f = v/\lambda$ is dimensionally correct, while $(\text{m/s}) \times (\text{m}) = \text{m}^2/\text{s}$ (option A) is not a frequency. Whenever you are unsure whether to multiply or divide, let the units decide.单一关系 $v = f\lambda$ 给出 $f = v/\lambda$、$\lambda = v/f$ 和 $v = f\lambda$。单位检验可消除选哪一个的疑虑：$f$ 必须以赫兹（$1/\text{s}$）为单位，而 $(\text{m/s}) / (\text{m}) = 1/\text{s}$ 证实 $f = v/\lambda$ 量纲正确，而 $(\text{m/s}) \times (\text{m}) = \text{m}^2/\text{s}$（选项 A）不是频率。每当不确定该乘还是该除时，让单位来决定。
Q5MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §3 + §7 Superposition & beats叠加与拍 · Physics 11 [6 marks][6 分]

(a) Two $3.0$ cm waves in phase: resultant amplitude and interference type. (b) Same waves out of phase ($\lambda/2$): resultant amplitude. (c) $512$ Hz and $508$ Hz forks: beat frequency and what is heard.(a) 两列 $3.0$ cm 波同相：合振幅与干涉类型。(b) 同样两列波反相（$\lambda/2$）：合振幅。(c) $512$ Hz 与 $508$ Hz 音叉：拍频与听到什么。

Answer:答案：  (a) $6.0\ \text{cm}$ (constructive)（相长）  ·  (b) $0\ \text{cm}$  ·  (c) $f_{\text{beat}} = 4\ \text{Hz}$

(a) In-phase superposition (constructive)同相叠加（相长） M1·A1

When two waves arrive in phase, their displacements add: $A_{\text{result}} = A_1 + A_2 = 3.0 + 3.0 = 6.0$ cm. This is constructive interference (crest meets crest).两列波同相到达时，位移相加：$A_{\text{result}} = A_1 + A_2 = 3.0 + 3.0 = 6.0$ cm。这是相长干涉（波峰遇波峰）。

(b) Out-of-phase superposition (destructive)反相叠加（相消） M1·A1

A path difference of $\lambda/2$ makes the waves exactly $180^\circ$ out of phase; displacements subtract: $A_{\text{result}} = |A_1 - A_2| = |3.0 - 3.0| = 0$ cm. The crests of one fill the troughs of the other, cancelling completely.路径差 $\lambda/2$ 使两波恰好反相 $180^\circ$；位移相减：$A_{\text{result}} = |A_1 - A_2| = |3.0 - 3.0| = 0$ cm。一列波的波峰填入另一列波的波谷，完全抵消。

(c) Beat frequency拍频 M1·A1

$$ f_{\text{beat}} \;=\; |f_1 - f_2| \;=\; |512 - 508| \;=\; 4 \;\text{Hz.} $$ The listener hears the average pitch ($510$ Hz) pulsing in loudness four times per second.听者听到平均音调（$510$ Hz）的响度每秒脉动四次。
Superposition is just algebraic addition of displacements; beats are superposition spread out over time.叠加只是位移的代数相加；拍是叠加在时间上的展开。 Parts (a) and (b) are the two extremes of the superposition principle at one instant: in phase gives the maximum $A_1 + A_2$, exactly out of phase gives the minimum $|A_1 - A_2|$. Part (c) is the same idea unfolding in time: two close frequencies slowly drift between in-phase (loud) and out-of-phase (soft), and the loud-soft cycle repeats $|f_1 - f_2|$ times per second. This is why piano tuners listen for beats and tighten a string until the beats vanish, the signature that two frequencies have become equal.(a) 与 (b) 是某一瞬间叠加原理的两个极端：同相给出最大值 $A_1 + A_2$，恰好反相给出最小值 $|A_1 - A_2|$。(c) 是同一思想在时间上的展开：两个相近频率在同相（响）与反相（弱）之间缓慢漂移，响-弱循环每秒重复 $|f_1 - f_2|$ 次。这就是为何钢琴调音师聆听拍声，并拉紧琴弦直到拍声消失，这正是两频率已相等的标志。
Q6MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §5 Sound intensity (dB)声强（分贝） · SPH3U E3 [6 marks][6 分]

$I = 10^{-6}\ \text{W/m}^2$, $I_0 = 10^{-12}\ \text{W/m}^2$. (a) Decibel level. (b) Level after tenfold increase. (c) Why the scale is logarithmic.$I = 10^{-6}\ \text{W/m}^2$，$I_0 = 10^{-12}\ \text{W/m}^2$。(a) 分贝级。(b) 增大十倍后的分贝级。(c) 为何标度是对数的。

Answer:答案：  (a) $\beta = 60\ \text{dB}$  ·  (b) $70\ \text{dB}$  ·  (c) to compress a huge range of intensities into manageable numbers将极大范围的声强压缩为便于处理的数值

(a) Apply the decibel formula套用分贝公式 M1·A1·A1

$$ \beta \;=\; 10\log_{10}\!\left(\frac{I}{I_0}\right) \;=\; 10\log_{10}\!\left(\frac{10^{-6}}{10^{-12}}\right) \;=\; 10\log_{10}(10^{6}) \;=\; 10 \times 6 \;=\; 60 \;\text{dB.} $$ This is a normal conversation level.这是正常交谈的级别。

(b) Tenfold intensity increase声强增大十倍 M1·A1

Each factor of $10$ in intensity adds $10\log_{10}(10) = 10$ dB. So the level rises from $60$ to $\mathbf{70}$ dB. (Check: $10\log_{10}(10^{-5}/10^{-12}) = 10 \times 7 = 70$ dB.)声强每增大 $10$ 倍，分贝增加 $10\log_{10}(10) = 10$ dB。故级别从 $60$ 升至 $\mathbf{70}$ dB。（验证：$10\log_{10}(10^{-5}/10^{-12}) = 10 \times 7 = 70$ dB。）

(c) Why logarithmic为何取对数 A1

Audible intensities span roughly $10^{12}$ from the threshold of hearing to the threshold of pain; a logarithmic scale turns this enormous span into a convenient $0$ to $120$ dB range and matches how the ear perceives loudness.可听声强从听阈到痛阈跨越约 $10^{12}$ 倍；对数标度把这一巨大跨度变为方便的 $0$ 到 $120$ dB 区间，并与耳朵对响度的感知方式相符。
Every $+10$ dB is a tenfold intensity jump; every $+3$ dB roughly doubles intensity.每 $+10$ dB 对应声强增大十倍；每 $+3$ dB 约使声强翻倍。 The decibel is a ratio measure, not an absolute one: it always compares $I$ to the reference $I_0 = 10^{-12}\ \text{W/m}^2$. The two facts worth memorising are the $+10$ dB $\Rightarrow \times 10$ rule (used directly in part b) and the $+3$ dB $\Rightarrow \times 2$ rule (since $10\log_{10} 2 \approx 3.0$). These let you reason about loudness changes without a calculator. A common error is to think doubling the intensity doubles the decibel reading; in fact it adds only about $3$ dB. The logarithm is the reason a $120$ dB jet is not "twice as loud" as a $60$ dB conversation, it is $10^{6}$ times more intense.分贝是比值量度，而非绝对量度：它始终把 $I$ 与参考值 $I_0 = 10^{-12}\ \text{W/m}^2$ 相比。值得记住的两个事实是 $+10$ dB $\Rightarrow \times 10$ 规则（(b) 中直接使用）与 $+3$ dB $\Rightarrow \times 2$ 规则（因为 $10\log_{10} 2 \approx 3.0$）。它们让你无需计算器即可推断响度变化。常见错误是以为声强加倍则分贝读数加倍；实际上只增加约 $3$ dB。对数正是为何 $120$ dB 的喷气机不是 $60$ dB 交谈的"两倍响"，而是声强强 $10^{6}$ 倍的原因。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  简答题 · 详解AP-feeder FRQ + honors · 30 marksAP 衔接简答题 + 荣誉级 · 共 30 分

Section B · Worked SolutionsB 部分 · 详细解答

Q7MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §2 Wave equation (medium change)波动方程（介质改变） · HS-PS4-1 [8 marks][8 分]

$440$ Hz tuning fork. Air $v = 340$ m/s, water $v = 1480$ m/s. (a) Wavelength in air. (b) Which of $f, v, \lambda$ is unchanged, with reason. (c) Wavelength in water. (d) Why $\lambda$ changes.$440$ Hz 音叉。空气 $v = 340$ m/s，水 $v = 1480$ m/s。(a) 空气中波长。(b) $f, v, \lambda$ 中哪个不变，并说明理由。(c) 水中波长。(d) 波长为何改变。

Answer:答案：  (a) $\lambda_{\text{air}} \approx 0.77\ \text{m}$  ·  (b) frequency is unchanged频率不变  ·  (c) $\lambda_{\text{water}} \approx 3.36\ \text{m}$

(a) Wavelength in air using $\lambda = v/f$用 $\lambda = v/f$ 求空气中波长 M1·A1

$$ \lambda_{\text{air}} \;=\; \frac{v}{f} \;=\; \frac{340}{440} \;\approx\; 0.77 \;\text{m.} $$

(b) Which quantity stays the same哪个量保持不变 M1·A1

The frequency stays the same. Frequency is set by the source (the tuning fork) and does not change when the wave crosses into a new medium. The speed is fixed by the medium, so the wavelength must adjust to keep $v = f\lambda$ satisfied.频率保持不变。频率由声源（音叉）决定，波进入新介质时不改变。波速由介质固定，因此波长必须调整以维持 $v = f\lambda$。

(c) Wavelength in water水中波长 M1·A1

$$ \lambda_{\text{water}} \;=\; \frac{v}{f} \;=\; \frac{1480}{440} \;\approx\; 3.36 \;\text{m.} $$

(d) Why the wavelength changes波长为何改变 A1·A1

Water transmits sound much faster than air ($1480$ vs $340$ m/s). With the frequency held fixed, a faster speed forces a proportionally longer wavelength so that $v = f\lambda$ still holds; the wavelength stretches by the same factor the speed increases.水传播声音远快于空气（$1480$ 对 $340$ m/s）。频率固定时，更快的速度迫使波长按比例变长以维持 $v = f\lambda$；波长按速度增大的相同倍数被拉长。
At a medium boundary the source frequency is conserved; speed is set by the medium, and wavelength is the dependent variable.在介质边界处声源频率守恒；速度由介质决定，波长是因变量。 This is the single most tested idea in the wave-equation section. The source dictates how many wavefronts leave per second (frequency), and that count cannot change as the wave enters water, otherwise wavefronts would pile up or vanish at the boundary. The medium dictates how fast each wavefront travels (speed). The wavelength is therefore forced: $\lambda = v/f$. A quick consistency check: the speed increased by a factor $1480/340 \approx 4.35$, and the wavelength grew by the same factor ($3.36/0.77 \approx 4.36$), confirming $f$ is constant. Students who instead assume "the wavelength stays the same" get the physics backwards.这是波动方程一节中最常考的单一概念。声源决定每秒发出多少个波阵面（频率），波进入水中时这个数目不能改变，否则波阵面会在边界处堆积或消失。介质决定每个波阵面传播多快（速度）。因此波长被迫定下：$\lambda = v/f$。快速一致性检验：速度增大了 $1480/340 \approx 4.35$ 倍，波长也增大了相同倍数（$3.36/0.77 \approx 4.36$），证实 $f$ 不变。若有学生反过来假设"波长不变"，则把物理搞反了。
Q8MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §4 Standing waves (string)驻波（弦） · SPH3U E3 [8 marks][8 分]

Guitar string $L = 0.65$ m, fixed both ends, $v = 320$ m/s. (a) Why fixed ends are nodes. (b) Fundamental frequency. (c) Third harmonic. (d) Half-wavelengths at the third harmonic.吉他弦 $L = 0.65$ m，两端固定，$v = 320$ m/s。(a) 固定端为何是波节。(b) 基频。(c) 第三谐波。(d) 第三谐波的半波长数。

Answer:答案：  (a) the string cannot move there弦在该处无法运动  ·  (b) $f_1 \approx 246\ \text{Hz}$  ·  (c) $f_3 \approx 738\ \text{Hz}$  ·  (d) $3$

(a) Why the fixed ends are nodes固定端为何是波节 A1

A node is a point of permanently zero displacement. Because the string is clamped at each end, it physically cannot move there, so each fixed end is forced to be a node.波节是位移恒为零的点。由于弦的两端被夹紧，它在该处物理上无法运动，因此每个固定端被迫成为波节。

(b) Fundamental frequency using $f_1 = v/(2L)$用 $f_1 = v/(2L)$ 求基频 M1·A1·A1

$$ f_1 \;=\; \frac{v}{2L} \;=\; \frac{320}{2 \times 0.65} \;=\; \frac{320}{1.30} \;\approx\; 246 \;\text{Hz.} $$

(c) Third harmonic第三谐波 M1·A1

$$ f_3 \;=\; 3 f_1 \;=\; 3 \times 246 \;\approx\; 738 \;\text{Hz.} $$

(d) Half-wavelengths on the string弦上的半波长数 A1·A1

For a string fixed at both ends, the $n$th harmonic fits exactly $n$ half-wavelengths in the length. The third harmonic ($n = 3$) fits $\mathbf{3}$ half-wavelengths.对两端固定的弦，第 $n$ 个谐波恰好在长度内容纳 $n$ 个半波长。第三谐波（$n = 3$）容纳 $\mathbf{3}$ 个半波长。
A string fixed at both ends supports all harmonics: $f_n = nv/(2L)$ with $n = 1, 2, 3, \ldots$两端固定的弦支持所有谐波：$f_n = nv/(2L)$，$n = 1, 2, 3, \ldots$ The boundary conditions set the allowed wavelengths: nodes at both ends mean an integer number of half-wavelengths must fit, $L = n(\lambda_n/2)$, so $\lambda_n = 2L/n$ and $f_n = v/\lambda_n = nv/(2L)$. Every harmonic is a whole-number multiple of the fundamental, which is why a plucked string sounds musical (a rich blend of $f_1, 2f_1, 3f_1, \ldots$). Note the wave speed $v = 320$ m/s here is the speed along the string (set by tension and mass per length), not the $340$ m/s speed of the resulting sound in air; do not mix the two.边界条件确定了允许的波长：两端为波节意味着必须容纳整数个半波长，$L = n(\lambda_n/2)$，故 $\lambda_n = 2L/n$，$f_n = v/\lambda_n = nv/(2L)$。每个谐波都是基频的整数倍，这就是为何拨动的弦听起来悦耳（$f_1, 2f_1, 3f_1, \ldots$ 的丰富混合）。注意此处波速 $v = 320$ m/s 是弦上的波速（由张力与线密度决定），并非由此产生的声音在空气中 $340$ m/s 的速度；切勿混淆两者。
Q9HARD难 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §4 Standing waves (pipes)驻波（管） · Physics 11 [6 marks][6 分]

Pipe $L = 0.40$ m, closed at one end, open at the other, $v = 340$ m/s. (a) Fundamental frequency. (b) Which harmonics, and the next two frequencies. (c) Open-open pipe of same length: higher or lower fundamental?管 $L = 0.40$ m，一端封闭、一端开口，$v = 340$ m/s。(a) 基频。(b) 哪些谐波，及下两个频率。(c) 同长度两端开口管：基频更高还是更低？

Answer:答案：  (a) $f_1 \approx 213\ \text{Hz}$  ·  (b) odd only;仅奇次； $f_3 \approx 638,\ f_5 \approx 1063\ \text{Hz}$  ·  (c) higher更高

(a) Fundamental using $f_1 = v/(4L)$用 $f_1 = v/(4L)$ 求基频 M1·A1

$$ f_1 \;=\; \frac{v}{4L} \;=\; \frac{340}{4 \times 0.40} \;=\; \frac{340}{1.60} \;=\; 212.5 \;\approx\; 213 \;\text{Hz.} $$

(b) Allowed harmonics允许的谐波 M1·A1·A1

A closed-open pipe has a node at the closed end and an antinode at the open end, so only odd harmonics fit: $f_n = nv/(4L)$ with $n = 1, 3, 5, \ldots$ The next two above the fundamental are一端封闭一端开口的管在封闭端有波节、开口端有波腹，故只容纳奇次谐波：$f_n = nv/(4L)$，$n = 1, 3, 5, \ldots$ 基频以上的下两个为 $$ f_3 \;=\; 3 \times 212.5 \;\approx\; 638 \;\text{Hz}, \qquad f_5 \;=\; 5 \times 212.5 \;\approx\; 1063 \;\text{Hz.} $$

(c) Open-open pipe of the same length同长度两端开口管 A1

An open-open pipe uses $f_1 = v/(2L)$, which is twice $v/(4L)$, so its fundamental is higher (here $425$ Hz, one octave above $213$ Hz).两端开口管用 $f_1 = v/(2L)$，是 $v/(4L)$ 的两倍，故基频更高（此处 $425$ Hz，比 $213$ Hz 高一个八度）。
Closed-open pipes drop an octave and skip even harmonics; the boundary geometry sets everything.一端封闭的管降低一个八度并跳过偶次谐波；边界几何决定一切。 The single fact to internalise: a closed end is a displacement node, an open end is an antinode. A node-to-antinode span is a quarter wavelength, so the fundamental of a closed-open pipe fits $\lambda/4$ in $L$ (hence $f_1 = v/4L$), while an open-open or string-fixed-both-ends fits $\lambda/2$ (hence $f_1 = v/2L$). Because only odd quarter-wavelength counts give an antinode at the open end, even harmonics are absent in the closed-open case. This is exactly why a clarinet (closed-open) sounds an octave lower and "hollower" than a flute (open-open) of the same length, the missing even harmonics change the timbre.需内化的单一事实：封闭端是位移波节，开口端是波腹。波节到波腹的跨度是四分之一波长，故一端封闭管的基频在 $L$ 内容纳 $\lambda/4$（因此 $f_1 = v/4L$），而两端开口管或两端固定弦容纳 $\lambda/2$（因此 $f_1 = v/2L$）。由于只有奇数个四分之一波长才能在开口端得到波腹，一端封闭一端开口的情形缺失偶次谐波。这正是为何同长度的单簧管（一端封闭）比长笛（两端开口）低一个八度且音色更"空"，缺失的偶次谐波改变了音色。
Q10HARD难Honors荣誉级 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §6 Doppler effect多普勒效应 · HS-PS4-1 (above NGSS floor)（超出 NGSS 基准） [8 marks][8 分]

Fire truck siren $f_s = 800$ Hz, $v_s = 30$ m/s, $v = 340$ m/s, $f_{\text{obs}} = f_s \cdot v/(v \mp v_s)$. (a) Frequency heard approaching. (b) Frequency heard receding. (c) Why the approaching pitch is higher (wavefronts). (d) Does the wave speed in air change?消防车警报 $f_s = 800$ Hz，$v_s = 30$ m/s，$v = 340$ m/s，$f_{\text{obs}} = f_s \cdot v/(v \mp v_s)$。(a) 靠近时听到的频率。(b) 远离时听到的频率。(c) 靠近时音调为何更高（波阵面）。(d) 空气中波速是否改变？

Answer:答案：  (a) $f_{\text{app}} \approx 877\ \text{Hz}$  ·  (b) $f_{\text{rec}} \approx 735\ \text{Hz}$  ·  (c) wavefronts bunch up ahead of the source波阵面在声源前方被压密  ·  (d) no不变

(a) Approaching source (subtract $v_s$ in the denominator)声源靠近（分母减 $v_s$） M1·A1·A1

As the source approaches, the observed frequency rises, so use the minus sign:声源靠近时观测频率升高，故取减号： $$ f_{\text{app}} \;=\; f_s\,\frac{v}{v - v_s} \;=\; 800 \times \frac{340}{340 - 30} \;=\; 800 \times \frac{340}{310} \;\approx\; 877 \;\text{Hz.} $$

(b) Receding source (add $v_s$ in the denominator)声源远离（分母加 $v_s$） M1·A1

$$ f_{\text{rec}} \;=\; f_s\,\frac{v}{v + v_s} \;=\; 800 \times \frac{340}{340 + 30} \;=\; 800 \times \frac{340}{370} \;\approx\; 735 \;\text{Hz.} $$

(c) Why the approaching pitch is higher靠近时音调为何更高 A1·A1

As the truck moves toward the observer, each successive wavefront is emitted from a position a little closer than the last, so the crests pile up in front of the source. Shorter spacing between crests means a shorter wavelength, and since the air still carries them at $340$ m/s, more crests pass the ear per second, that is, a higher frequency.消防车向观察者运动时，每个相继发出的波阵面都比上一个发出位置稍近一些，故波峰在声源前方堆积。波峰间距变小意味着波长变短，而空气仍以 $340$ m/s 携带它们，于是每秒经过耳朵的波峰更多，即频率更高。

(d) Does the wave speed change波速是否改变 A1

No. The speed of sound in air ($340$ m/s) is a property of the air alone, set by its temperature; the motion of the source changes the wavelength and frequency, not the propagation speed.不变。空气中声速（$340$ m/s）只是空气本身的性质，由其温度决定；声源运动改变的是波长与频率，而非传播速度。
In the Doppler formula the moving source changes wavelength; the medium fixes the speed; the observed frequency follows.多普勒公式中，运动的声源改变波长；介质固定波速；观测频率随之确定。 The sign in $v/(v \mp v_s)$ is the whole exam: approaching uses the minus sign (smaller denominator, higher frequency), receding uses the plus sign (larger denominator, lower frequency). A reliable check is that the approaching value must exceed $f_s$ and the receding value must fall below it; here $877 > 800 > 735$, as required. The physical picture in part (c) is what makes the sign memorable: the source chases its own wavefronts forward and runs away from them backward. Crucially, the wave speed never changes, only the spacing of the crests does, so the medium and the source play strictly separate roles.公式 $v/(v \mp v_s)$ 中的符号就是整道题的关键：靠近取减号（分母更小，频率更高），远离取加号（分母更大，频率更低）。可靠的核对方法是：靠近值必须大于 $f_s$，远离值必须小于 $f_s$；此处 $877 > 800 > 735$，符合要求。(c) 中的物理图像让符号易记：声源向前追赶自己的波阵面，向后则远离它们。关键在于波速从不改变，改变的只是波峰间距，因此介质与声源各司其职、互不相干。
PART III  ·  MODELING / APPLIED · SOLUTIONS第三部分  ·  建模与应用 · 详解AB Diploma + Universal · 25 marks阿省毕业考 + 通用题型 · 共 25 分

Section C · Modeling and ApplicationsC 部分 · 建模与应用

Q11MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §5 Intensity (inverse-square)声强（平方反比） · 20-D2.6k [9 marks][9 分]

Loudspeaker as a point source: at $r_1 = 2.0$ m, $I_1 = 0.080\ \text{W/m}^2$. (a) State the inverse-square law. (b) Intensity at $r_2 = 6.0$ m. (c) Distance at which the intensity is one-quarter of $I_1$. (d) Why intensity falls as $1/r^2$ not $1/r$.扬声器视为点声源：$r_1 = 2.0$ m 处 $I_1 = 0.080\ \text{W/m}^2$。(a) 写出平方反比定律。(b) $r_2 = 6.0$ m 处的声强。(c) 声强降为 $I_1$ 四分之一处的距离。(d) 声强为何按 $1/r^2$ 而非 $1/r$ 衰减。

Answer:答案：  (a) $I \propto 1/r^2$  ·  (b) $I_2 \approx 8.9\times10^{-3}\ \text{W/m}^2$  ·  (c) $r = 4.0\ \text{m}$  ·  (d) power spreads over a sphere of area $4\pi r^2$功率铺展在面积为 $4\pi r^2$ 的球面上

(a) State the inverse-square law写出平方反比定律 A1

For a point source the intensity falls off as the inverse square of the distance:对点声源，声强按距离的平方反比衰减： $$ I \;\propto\; \frac{1}{r^2}, \qquad \text{so} \qquad I_1 r_1^2 \;=\; I_2 r_2^2. $$

(b) Intensity at $r_2 = 6.0$ m$r_2 = 6.0$ m 处的声强 M1·A1·A1

$$ I_2 \;=\; I_1\!\left(\frac{r_1}{r_2}\right)^{\!2} \;=\; 0.080 \times \left(\frac{2.0}{6.0}\right)^{\!2} \;=\; 0.080 \times \frac{1}{9} \;\approx\; 8.9\times10^{-3} \;\text{W/m}^2. $$

(c) Distance for one-quarter intensity声强降为四分之一处的距离 M1·A1·A1

We need $I = I_1/4$. Since $I \propto 1/r^2$, quartering the intensity means $(r_1/r)^2 = 1/4$, so $r = 2 r_1$:需要 $I = I_1/4$。因 $I \propto 1/r^2$，声强变为四分之一意味着 $(r_1/r)^2 = 1/4$，故 $r = 2 r_1$： $$ r \;=\; 2 \times 2.0 \;=\; 4.0 \;\text{m.} $$

(d) Why $1/r^2$ rather than $1/r$为何是 $1/r^2$ 而非 $1/r$ A1·A1

A point source radiates its power equally in all directions, so at distance $r$ that fixed power is spread over the surface of a sphere of area $4\pi r^2$. Intensity is power per unit area, so it falls as $1/r^2$, the area, not the radius, sets the dilution.点声源向各方向均匀辐射功率，故在距离 $r$ 处这份固定功率铺展在面积为 $4\pi r^2$ 的球面上。声强是单位面积功率，故按 $1/r^2$ 衰减，决定稀释程度的是面积而非半径。
Intensity from a point source obeys $I = P/(4\pi r^2)$, so $I_1 r_1^2 = I_2 r_2^2$; tripling the distance cuts intensity to one-ninth.点声源声强遵循 $I = P/(4\pi r^2)$，故 $I_1 r_1^2 = I_2 r_2^2$；距离变三倍则声强降为九分之一。 The cleanest way to handle inverse-square problems is the ratio form $I_1 r_1^2 = I_2 r_2^2$, which sidesteps ever needing the source power $P$. The geometry is the whole story: sound energy from a point source is conserved but smeared over an ever-larger sphere, and a sphere's area grows as $r^2$. A useful mental table follows: double the distance, quarter the intensity; triple it, one-ninth. Part (c) inverts this, asking for the distance at a target intensity. Note this is a geometric spreading law and ignores absorption by the air, which would make real-world fall-off slightly faster.处理平方反比问题最简洁的方式是比值形式 $I_1 r_1^2 = I_2 r_2^2$，它无需用到声源功率 $P$。几何就是全部：点声源的声能守恒，却被抹散在越来越大的球面上，而球面面积按 $r^2$ 增长。一张实用的心算表：距离加倍，声强变四分之一；距离三倍，变九分之一。(c) 反过来问达到目标声强所需的距离。注意这是几何扩散定律，忽略了空气的吸收，实际衰减会略快一些。
Q12MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §6 Doppler (applied)多普勒（应用） · 20-D2.9k [8 marks][8 分]

Ambulance siren $f_s = 660$ Hz, $v_s = 25$ m/s past a stationary pedestrian, $v = 340$ m/s, $f_{\text{obs}} = f_s \cdot v/(v \mp v_s)$. (a) Frequency approaching. (b) Frequency receding. (c) Total drop in frequency. (d) What is heard at the instant alongside.救护车警报 $f_s = 660$ Hz，以 $v_s = 25$ m/s 驶过静止行人，$v = 340$ m/s，$f_{\text{obs}} = f_s \cdot v/(v \mp v_s)$。(a) 靠近时频率。(b) 远离时频率。(c) 频率总下降量。(d) 并排瞬间听到什么。

Answer:答案：  (a) $f_{\text{app}} \approx 712\ \text{Hz}$  ·  (b) $f_{\text{rec}} \approx 615\ \text{Hz}$  ·  (c) $\Delta f \approx 98\ \text{Hz}$  ·  (d) the true $660$ Hz真实的 $660$ Hz

(a) Approaching (minus sign in the denominator)靠近（分母取减号） M1·A1·A1

$$ f_{\text{app}} \;=\; f_s\,\frac{v}{v - v_s} \;=\; 660 \times \frac{340}{340 - 25} \;=\; 660 \times \frac{340}{315} \;\approx\; 712 \;\text{Hz.} $$

(b) Receding (plus sign in the denominator)远离（分母取加号） M1·A1

$$ f_{\text{rec}} \;=\; f_s\,\frac{v}{v + v_s} \;=\; 660 \times \frac{340}{340 + 25} \;=\; 660 \times \frac{340}{365} \;\approx\; 615 \;\text{Hz.} $$

(c) Total drop the pedestrian hears行人听到的频率总下降量 M1·A1

$$ \Delta f \;=\; f_{\text{app}} - f_{\text{rec}} \;=\; 712 - 615 \;\approx\; 98 \;\text{Hz.} $$ The pitch drops abruptly by about $98$ Hz as the ambulance sweeps past.救护车驶过的瞬间，音调骤降约 $98$ Hz。

(d) Instant the ambulance is alongside救护车并排的瞬间 A1

At that instant the ambulance is moving neither toward nor away from the pedestrian (zero radial velocity), so there is no Doppler shift and the pedestrian hears the true emitted frequency, $660$ Hz.在该瞬间救护车既不靠近也不远离行人（径向速度为零），故无多普勒频移，行人听到真实发出的频率 $660$ Hz。
The Doppler shift depends only on the radial velocity, the component of motion along the line to the listener.多普勒频移只取决于径向速度，即沿听者连线方向的运动分量。 A passing siren is the everyday Doppler demonstration: high while approaching, low while receding, with a sharp fall as it goes by. The "alongside" instant in part (d) is the conceptual crux, students expect the maximum effect when the source is closest, but in fact the shift is zero there because the radial velocity momentarily vanishes; the source is moving purely sideways. The total drop in part (c) is what the ear actually registers as the characteristic "neeee-yowww." Real sirens never reach the idealised $712$ or $615$ Hz exactly, because the radial component only equals the full $v_s$ when the source is far away along the road.驶过的警报声是日常生活中的多普勒演示：靠近时高、远离时低，经过时骤降。(d) 中"并排"瞬间是概念关键：学生以为声源最近时效应最大，但实际上此处频移为零，因为径向速度瞬间消失，声源纯粹做侧向运动。(c) 中的总下降量正是耳朵实际听到的那种典型"呜--哟--"声。真实警报声永远不会精确达到理想的 $712$ 或 $615$ Hz，因为只有当声源沿道路远在两端时，径向分量才等于完整的 $v_s$。
Q13HARD难 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §7 Beats (tuning logic)拍（调音推理） · HS-PS4-1 [8 marks][8 分]

String near a $440$ Hz fork gives $3$ beats/s. Tightening the string (raising its frequency) increases the beat rate to $5$ beats/s. (a) Beat relationship and the two possible string frequencies. (b) Which value was the true original frequency, with justification. (c) String frequency in the tightened state.弦与 $440$ Hz 音叉同时发声产生每秒 $3$ 次拍。拉紧弦（升高其频率）使拍率增加到每秒 $5$ 次。(a) 拍频关系式与弦的两种可能频率。(b) 哪个数值是真正的原始频率，并说明理由。(c) 拉紧后弦的频率。

Answer:答案：  (a) $437\ \text{Hz}$ or或 $443\ \text{Hz}$  ·  (b) $443\ \text{Hz}$  ·  (c) $445\ \text{Hz}$

(a) Beat-frequency relationship and the two candidates拍频关系式与两个候选值 M1·A1·A1

The beat frequency is the absolute difference of the two frequencies:拍频是两频率之差的绝对值： $$ f_{\text{beat}} \;=\; |f_{\text{string}} - f_{\text{fork}}| \;=\; 3 \;\text{Hz} \;\Longrightarrow\; f_{\text{string}} \;=\; 440 \pm 3. $$ So the string is either $437$ Hz or $443$ Hz; the beat rate alone cannot decide which.故弦为 $437$ Hz 或 $443$ Hz；仅凭拍率无法判定是哪一个。

(b) Use the change in beat rate to break the tie用拍率变化打破二选一 M1·A1·A1

Tightening raises the string's frequency. If the string had been $437$ Hz, raising it would move it toward $440$ Hz, shrinking the difference and lowering the beat rate. Instead the beat rate rose from $3$ to $5$ Hz, so the string must have started above $440$ Hz and moved further away. The true original frequency is therefore $\mathbf{443}$ Hz.拉紧会升高弦的频率。若弦原为 $437$ Hz，升高它会使其趋近 $440$ Hz，差值减小、拍率降低。但拍率反而从 $3$ 升到 $5$ Hz，故弦原本必定高于 $440$ Hz，并进一步远离。因此真正的原始频率为 $\mathbf{443}$ Hz。

(c) Frequency after tightening拉紧后的频率 M1·A1

A beat rate of $5$ Hz against the $440$ Hz fork, with the string above the fork, gives相对 $440$ Hz 音叉的拍率为 $5$ Hz，且弦高于音叉，故 $$ f_{\text{tight}} \;=\; 440 + 5 \;=\; 445 \;\text{Hz.} $$
A single beat reading is ambiguous ($f_0 \pm f_{\text{beat}}$); a deliberate small change in one frequency resolves which side you are on.单次拍读数是二义的（$f_0 \pm f_{\text{beat}}$）；故意微调一个频率即可判定你在哪一侧。 This is exactly how musicians tune by ear. The beat rate gives the size of the mismatch but not its sign, so the unknown frequency could lie on either side of the reference. The tuner then nudges the string one way and listens: if the beats slow, you moved toward the reference and were on that side; if they speed up, you moved away and were on the other side. Here tightening (which raises pitch) sped the beats up, proving the string was already sharp at $443$ Hz, and continuing to tighten only makes it worse. The correct fix would be to loosen the string back down toward $440$ Hz until the beats vanish.这正是音乐家凭耳朵调音的方式。拍率给出失谐的大小却不给方向，故未知频率可能落在参考值任一侧。调音者随后把弦朝一个方向轻微调动并聆听：若拍变慢，说明你向参考值靠近、原本在那一侧；若拍变快，说明你在远离、原本在另一侧。此处拉紧（升高音调）使拍变快，证明弦原本已偏高，处于 $443$ Hz，继续拉紧只会更糟。正确做法应是把弦放松回降向 $440$ Hz，直到拍声消失。