PART I · SHORT RESPONSE第一部分 · 短答题AP-style MCQ + ON/BC short answer · 25 marksAP 风格选择题 + 安/卑省考短答 · 共 25 分
Section A · Short ResponseA 部分 · 短答题
Mix of multiple-choice and short-answer items. For MCQs, circle the letter; show enough work in the margin that a marker could verify. For short-answer items, state units in every answer. Take the speed of sound in air as $v = 340\ \text{m/s}$ unless told otherwise. No calculator on Q1-Q2; calculator permitted on Q3-Q6.本节包含选择题与短答题。选择题请圈出字母答案，并在空白处写下足以让阅卷人核对的过程。短答题每道都要写出单位。除非另有说明，空气中声速取 $v = 340\ \text{m/s}$。Q1-Q2 不可使用计算器；Q3-Q6 可用计算器。
In a longitudinal wave such as sound in air, how do the medium particles move relative to the direction the wave travels?在纵波（如空气中的声波）中，介质质点相对于波的传播方向如何运动？
(A)Perpendicular to the wave direction垂直于波的传播方向
(B)Parallel to the wave direction平行于波的传播方向
(C)In circles around the wave direction围绕波的传播方向做圆周运动
(D)They do not move at all完全不运动
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 a period of $T = 0.025$ s. What is its frequency?某波的周期为 $T = 0.025$ s。其频率是多少？
A transverse wave on a rope has a crest-to-crest distance of $0.40$ m and completes $5.0$ full oscillations per second.一条绳上的横波相邻波峰间距为 $0.40$ m，每秒完成 $5.0$ 次完整振动。
(a)State the wavelength and the frequency, with units.写出波长与频率，并标注单位。[1]
Two sound sources produce waves that overlap at a point.两个声源产生的波在某点重叠。
(a)Two waves of amplitude $3.0$ cm each arrive perfectly in phase. State the resultant amplitude and name the type of interference.两列振幅均为 $3.0$ cm 的波完全同相到达。写出合振幅并说明干涉类型。[2]
(b)The same two waves now arrive exactly out of phase (path difference $\lambda/2$). State the resultant amplitude.同样两列波现在恰好反相到达（路径差为 $\lambda/2$）。写出合振幅。[2]
(c)Two tuning forks sound together at $512$ Hz and $508$ Hz. Find the beat frequency and explain in one sentence what a listener hears.两个音叉同时以 $512$ Hz 和 $508$ Hz 发声。求拍频，并用一句话说明听者听到什么。[2]
A sound has intensity $I = 10^{-6}\ \text{W/m}^2$. Take the threshold of hearing $I_0 = 10^{-12}\ \text{W/m}^2$ and use $\beta = 10\log_{10}(I/I_0)$.某声音声强 $I = 10^{-6}\ \text{W/m}^2$。取听阈 $I_0 = 10^{-12}\ \text{W/m}^2$，使用 $\beta = 10\log_{10}(I/I_0)$。
(a)Calculate the sound level in decibels.计算声级（分贝）。[3]
(b)The intensity is now increased tenfold to $10^{-5}\ \text{W/m}^2$. State the new decibel level.声强现增大十倍至 $10^{-5}\ \text{W/m}^2$。写出新的分贝值。[2]
(c)State in one sentence why the decibel scale is logarithmic rather than linear.用一句话说明分贝标度为何是对数而非线性的。[1]
PART II · EXTENDED RESPONSE第二部分 · 简答题AP-feeder FRQ + honors · 30 marksAP 衔接简答题 + 荣誉级 · 共 30 分
Section B · Extended ResponseB 部分 · 简答题
Show every step of reasoning. Identify the wave relationship used before substituting values. State units in every final answer. Take the speed of sound in air as $v = 340\ \text{m/s}$ unless told otherwise. Calculator permitted on Q7-Q10.每一步推理都要写出。在代入数值前先注明所用波动关系式。每个最终答案都要写单位。除非另有说明，空气中声速取 $v = 340\ \text{m/s}$。Q7-Q10 可用计算器。
A tuning fork emits a sound wave of frequency $440$ Hz (concert A). Sound travels at $340$ m/s in air and at $1480$ m/s in water.一个音叉发出频率为 $440$ Hz（音乐会 A 音）的声波。声音在空气中速率为 $340$ m/s，在水中为 $1480$ m/s。
(a)Find the wavelength of the wave in air.求该波在空气中的波长。[2]
(b)The same wave passes into water. State which of $f$, $v$, $\lambda$ stays the same, with a reason.同一波进入水中。说明 $f$、$v$、$\lambda$ 中哪一个保持不变，并说明理由。[2]
(c)Find the wavelength of the wave in water.求该波在水中的波长。[2]
(d)Explain in one or two sentences why the wavelength changes as the wave crosses into water.用一两句话解释波进入水中时波长为何改变。[2]
A pipe of length $L = 0.40$ m is closed at one end and open at the other. Sound speed in air is $340$ m/s.一根长 $L = 0.40$ m 的管，一端封闭、一端开口。空气中声速为 $340$ m/s。
(a)Find the fundamental frequency using $f_1 = v/(4L)$.用 $f_1 = v/(4L)$ 求基频。[2]
(b)State which harmonics this closed-open pipe supports and give the next two frequencies above the fundamental.说明该一端封闭一端开口的管支持哪些谐波，并给出基频以上的下两个频率。[3]
(c)State, with a reason, whether an open-open pipe of the same length would have a higher or lower fundamental.说明同长度的两端开口管基频是更高还是更低，并给出理由。[1]
A fire truck emits a siren at $f_s = 800$ Hz while travelling at $v_s = 30$ m/s. The speed of sound in air is $v = 340$ m/s. Use $f_{\text{obs}} = f_s \cdot v/(v \mp v_s)$.一辆消防车以 $v_s = 30$ m/s 行驶，发出 $f_s = 800$ Hz 的警报声。空气中声速为 $v = 340$ m/s。使用 $f_{\text{obs}} = f_s \cdot v/(v \mp v_s)$。
(a)Find the frequency heard by a stationary observer as the truck approaches.求消防车靠近时静止观察者听到的频率。[3]
(b)Find the frequency heard as the truck recedes.求消防车远离时听到的频率。[2]
(c)Explain physically, in terms of wavefronts, why the approaching pitch is higher.从波阵面的角度，解释靠近时音调为何更高。[2]
(d)State whether the wave speed in air changes as the truck moves, and why.说明消防车运动时空气中波速是否改变，并说明原因。[1]
PART III · MODELING / APPLIED第三部分 · 建模与应用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
Define symbols (with units) at the start of each question. State the wave relationship used before substituting. Take the speed of sound in air as $v = 340\ \text{m/s}$ unless told otherwise. Conclude each question with a one-sentence answer in context. Calculator permitted throughout Part III.每题开始时定义符号（含单位）。代入数值前先写出所用波动关系式。除非另有说明，空气中声速取 $v = 340\ \text{m/s}$。每题以一句结合情境的完整句子作答。第三部分全程可用计算器。
A loudspeaker behaves as a point source. At a distance of $r_1 = 2.0$ m it produces a sound intensity of $I_1 = 0.080\ \text{W/m}^2$.一个扬声器可视为点声源。在 $r_1 = 2.0$ m 处它产生声强 $I_1 = 0.080\ \text{W/m}^2$。
(a)State the inverse-square law relating intensity and distance from a point source.写出点声源声强与距离的平方反比关系。[1]
(b)Find the intensity at $r_2 = 6.0$ m.求 $r_2 = 6.0$ m 处的声强。[3]
(c)At what distance from the speaker has the intensity dropped to one-quarter of $I_1$?在距扬声器多远处声强降为 $I_1$ 的四分之一？[3]
(d)Explain in one sentence why the intensity spreads as $1/r^2$ rather than $1/r$.用一句话解释为何声强按 $1/r^2$ 而非 $1/r$ 衰减。[2]
An ambulance siren emits $f_s = 660$ Hz. The ambulance travels at $v_s = 25$ m/s past a stationary pedestrian. Take $v = 340$ m/s and use $f_{\text{obs}} = f_s \cdot v/(v \mp v_s)$.救护车警报器发出 $f_s = 660$ Hz。救护车以 $v_s = 25$ m/s 驶过一名静止的行人。取 $v = 340$ m/s，使用 $f_{\text{obs}} = f_s \cdot v/(v \mp v_s)$。
(a)Find the frequency the pedestrian hears as the ambulance approaches.求救护车靠近时行人听到的频率。[3]
(b)Find the frequency heard after the ambulance has passed and is receding.求救护车驶过并远离后行人听到的频率。[2]
(c)Find the total drop in frequency the pedestrian hears as the ambulance passes.求救护车驶过时行人听到的频率总下降量。[2]
(d)State in one sentence what the pedestrian hears at the instant the ambulance is directly alongside.用一句话说明救护车正好与行人并排的瞬间行人听到什么。[1]
A guitarist plucks a string that should sound $440$ Hz (concert A). Held next to a $440$ Hz tuning fork, the player hears $3$ beats per second. The player then tightens the string slightly (which raises its frequency) and the beat rate increases to $5$ per second.一位吉他手拨动一根应发出 $440$ Hz（音乐会 A 音）的弦。与 $440$ Hz 音叉同时发声，弹奏者每秒听到 $3$ 次拍。随后弹奏者略微拉紧弦（这会升高其频率），拍率增加到每秒 $5$ 次。
(a)State the beat-frequency relationship and the two possible string frequencies before tightening.写出拍频关系式以及拉紧前弦的两种可能频率。[3]
(b)Use the fact that tightening increased the beat rate to determine which value was the true original frequency. Justify your reasoning.利用拉紧后拍率增加这一事实，判断哪个数值是真正的原始频率。说明推理过程。[3]
(c)State the string's frequency in the final (tightened) state.写出弦在最终（拉紧后）状态的频率。[2]
🇨🇦 Alberta阿尔伯塔Physics 20 Unit D · 20-D2.3k · 20-D2.6k · 20-D2.8k · 20-D2.9k
Full Syllabus Map lives in ../Study Guides/Unit_6_Waves_and_Sound.html. Note: NGSS HS-PS4-1 anchors the wave model and $v = f\lambda$; the decibel and Doppler quantitative items (Q6, Q10, Q12) sit above the NGSS floor but are core for AB Physics 20 (20-D2.9k) and BC Physics 11.完整大纲对照表见 ../Study Guides/Unit_6_Waves_and_Sound.html。注：NGSS HS-PS4-1 锚定波模型与 $v = f\lambda$；分贝与多普勒定量题（Q6、Q10、Q12）超出 NGSS 范围，但为阿省物理 20（20-D2.9k）与卑诗物理 11 的核心内容。