Sound: Key Concepts for IB and AQA Science | IB AQA 科学：声考点精讲

📚 Sound: Key Concepts for IB and AQA Science | IB AQA 科学：声考点精讲

Sound is a longitudinal mechanical wave that propagates through a medium by compressions and rarefactions. Understanding its properties, from frequency and amplitude to speed and interference, is essential for both IB Physics and AQA GCSE/AS Level Science. This article covers the core principles and exam-relevant applications of sound, linking theory to real-world contexts such as music, hearing and medical imaging.

声音是一种通过介质以疏密波形式传播的纵向机械波。从频率、振幅到声速和干涉，理解其特性是 IB 物理和 AQA GCSE/AS 科学考试的关键。本文涵盖了声学的核心原理和考点应用，将理论与音乐、听觉、医学成像等实际场景联系起来。

1. The Nature of Sound Waves | 声波的本质

Sound is a mechanical, longitudinal wave consisting of alternating regions of high pressure (compressions) and low pressure (rarefactions). Unlike transverse waves on a string, particles in a sound wave vibrate parallel to the direction of energy transfer.

声音是一种机械纵波，由高压区（密部）和低压区（疏部）交替组成。与弦上的横波不同，声波中粒子的振动方向与能量传播方向平行。

In IB Physics, sound is modelled as a pressure wave. The displacement of particles ahead and behind causes density variations. AQA specifications emphasise that sound cannot travel through a vacuum because there is no medium to transmit the vibrations.

在 IB 物理中，声音被建模为压力波。粒子前后位移导致密度变化。AQA 大纲强调声音无法在真空中传播，因为没有介质传递振动。

Graphically, a sound wave is often represented as a sine curve showing pressure variation against position or time. The crest corresponds to compression and the trough to rarefaction. This visualisation helps link frequency and wavelength.

通常用正弦曲线表示声波，展示压力随位置或时间的变化。波峰对应密部，波谷对应疏部。这种可视化有助于将频率和波长联系起来。

v = f × λ

2. Speed of Sound | 声速

The speed of sound depends on the medium’s density and elasticity. In air at 20 °C, it is approximately 343 m s⁻¹. In water, sound travels about 1480 m s⁻¹, and in steel it can exceed 5000 m s⁻¹. The stiffer and less dense the medium, the faster the sound.

声速取决于介质的密度和弹性。在 20 °C 的空气中，大约为 343 米/秒。在水中约 1480 米/秒，在钢中可超过 5000 米/秒。介质刚度越大、密度越小，声速越快。

For IB and AQA, the relationship v = √(E/ρ) can be used for solids, where E is Young’s modulus and ρ the density. In gases, temperature affects speed: v ∝ √T, where T is in Kelvin. A 1 °C increase in air raises speed by about 0.6 m s⁻¹.

在 IB 和 AQA 内容中，固体中可用 v = √(E/ρ) 表示，其中 E 为杨氏模量，ρ 为密度。在气体中，温度影响速度：v ∝ √T，T 为开氏温度。空气温度每升高 1 °C，声速约增加 0.6 米/秒。

A standard experiment uses a microphone and oscilloscope to measure time delay between two microphones at known separation, giving v = distance / time. AQA required practical: measure speed using an echo from a flat wall.

标准实验利用麦克风和示波器测量相距一定距离的两个麦克风之间的时间延迟，得 v = 距离 / 时间。AQA 必做实验：利用平坦墙面产生的回声测量声速。

3. Frequency, Pitch, and Amplitude | 频率、音调与振幅

Frequency (f) is the number of complete vibrations per second, measured in hertz (Hz). It determines the pitch of a sound: high frequency means high pitch. The audible range for humans is typically 20 Hz to 20 000 Hz.

频率（f）是每秒完整振动的次数，单位赫兹（Hz）。它决定了声音的音调：高频意味着高音调。人类听觉范围通常为 20 Hz 至 20 000 Hz。

Amplitude is the maximum displacement of particles from their rest position. In a pressure wave, it corresponds to the maximum pressure change. A larger amplitude produces a louder sound. Loudness is subjective, while amplitude is physical.

振幅是粒子偏离平衡位置的最大位移。在压力波中，对应最大压力变化。振幅越大，声音越响。响度是主观感受，而振幅是物理量。

On an oscilloscope trace, a sound of higher frequency shows more cycles across the same time base, while greater amplitude shows taller peaks. IB questions often require identifying f and amplitude from such traces.

在示波器波形上，频率较高的声音在相同时间基线内显示更多周期，而振幅较大则波峰更高。IB 考题常要求从这类波形识别频率和振幅。

f = 1 / T

4. Reflection and Echoes | 反射与回声

Sound waves reflect off hard surfaces following the law of reflection: angle of incidence equals angle of reflection. Reflection is used in sonar, ultrasound imaging and auditorium design. An echo is a reflected sound that arrives more than 0.1 s after the direct sound.

声波遇硬表面反射，遵循反射定律：入射角等于反射角。反射用于声呐、超声成像和音乐厅设计。回声是比原声延迟超过 0.1 秒到达的反射声。

Using the speed of sound and the time delay, distance to a reflector can be calculated: d = (v × t) / 2 (the sound travels there and back). AQA exam questions frequently feature this calculation with echoes or sonar.

利用声速和时间延迟，可计算到反射面的距离：d = (v × t) / 2（声音往返）。AQA 考题经常涉及回声或声呐的这种计算。

Multiple reflections can cause reverberation. In concert halls, a short reverberation time is desirable for clarity, while too much causes muddiness. Soundboards and curved ceilings enhance projection.

多次反射会引起混响。在音乐厅中，较短的混响时间有利于清晰度，而过多混音则模糊不清。音板和弧形天花板可增强声音投射。

5. Refraction and Diffraction of Sound | 声的折射与衍射

Sound refracts when it passes from one medium to another with different speeds, or when air temperature changes with height. On a warm day, sound bends upward because the ground is hotter, creating a shadow zone. At night, cooler air near the ground can bend sound downward, helping it travel farther.

声音从一个介质进入另一个不同速度的介质时会发生折射，或当空气温度随高度变化时亦然。白天，地面较热导致声波向上弯曲，形成声影区。夜晚地面附近空气较冷可使声波向下弯曲，传播更远。

Diffraction is the spreading of waves around obstacles or through gaps. Sound waves have relatively long wavelengths (e.g. 340 Hz gives λ ≈ 1 m), so they diffract noticeably around doors and corners. Low-frequency sounds diffract more than high frequencies, which is why bass notes are less directional.

衍射是波绕过障碍物或通过缝隙时扩散的现象。声波波长相对较长（如 340 Hz 时 λ ≈ 1 米），因此能明显绕射门框和拐角。低频声比高频声衍射更显著，这就是低音方向性较弱的原因。

The amount of diffraction depends on the size of the gap relative to wavelength. AQA questions may ask why you can hear someone around a corner but not see them: sound diffracts, light does not (visible λ << gap).

衍射程度取决于缝隙大小与波长的比值。AQA 考题可能问：为何你能听到拐角处的人说话却看不见他们：声音衍射，光不衍射（可见光波长远小于缝隙）。

6. Interference and Beats | 干涉与拍频

When two sound waves of the same frequency meet in phase, they superpose constructively, increasing amplitude (louder). If they meet out of phase (180°), destructive interference reduces amplitude (softer or silence). This is the basis of noise-cancelling headphones.

当两个相同频率的声波同相相遇时，发生相长干涉，振幅增大（更响）。异相（180°）相遇时，相消干涉使振幅减小（更轻或无声）。这是降噪耳机的原理。

Interference patterns can be demonstrated with two loudspeakers connected to the same signal generator. Moving a microphone along a line reveals alternating maxima and minima. Path difference nλ gives constructive; (n + ½)λ gives destructive.

干涉图样可用连接同一信号发生器的两个扬声器演示。沿线移动麦克风可观察到交替出现的极大值和极小值。程差为 nλ 产生相长干涉；(n + ½)λ 产生相消干涉。

Beats occur when two sounds of slightly different frequencies f₁ and f₂ overlap. The beat frequency is |f₁ – f₂|. Musicians use beats to tune instruments: as the pitch approaches unison, the beat frequency decreases until it vanishes.

当两个频率稍有不同的声音 f₁ 和 f₂ 叠加时产生拍频。拍频等于 |f₁ – f₂|。音乐家利用拍频调音：当音高接近同度时，拍频降低直至消失。

f_beat = |f₁ – f₂|

7. The Doppler Effect | 多普勒效应

The Doppler effect is the change in observed frequency when a sound source moves relative to an observer. If the source approaches, waves are compressed, increasing frequency (higher pitch). If it recedes, waves are stretched, decreasing frequency.

多普勒效应是声源与观察者相对运动时观测频率发生变化的现象。声源靠近时，波被压缩，频率升高（音调变高）。远离时，波被拉伸，频率降低。

The observed frequency f’ can be calculated: for a stationary observer and moving source, f’ = f × v / (v ± v_s), where v is sound speed, v_s source speed. Use minus when source moves towards observer, plus when away. This formula appears in IB Physics.

观测频率 f’ 可计算：对静止观察者与运动声源，f’ = f × v / (v ± v_s)，其中 v 为声速，v_s 为声源速度。声源靠近时用减号，远离时用加号。该公式出现于 IB 物理。

Applications include police radar (though uses EM waves), echolocation by bats, and medical ultrasound to measure blood flow. AQA may ask qualitative questions about pitch change for a passing siren or train horn.

应用包括警用雷达（尽管使用电磁波）、蝙蝠回声定位，以及医用超声测量血流。AQA 可能要求定性分析警笛或火车鸣笛经过时的音调变化。

8. Intensity and Decibels | 声强与分贝

Sound intensity (I) is the power per unit area, measured in W m⁻². The threshold of human hearing I₀ is 1 × 10⁻¹² W m⁻². Because the range of audible intensities is enormous, a logarithmic scale is used: sound intensity level in decibels (dB).

声强（I）是单位面积上的功率，单位 W/m²。人耳听阈 I₀ 为 1 × 10⁻¹² W/m²。由于可听强度范围巨大，采用对数标度：声强级以分贝（dB）表示。

L = 10 log₁₀ (I / I₀) dB

Every 10 dB increase represents a tenfold increase in intensity. A 3 dB increase doubles the intensity. IB problems often require calculating intensity ratios from dB differences or finding resulting level from multiple identical sources.

每增加 10 dB 代表强度增加十倍。增加 3 dB 则强度加倍。IB 题目常要求根据分贝差计算强度比，或由多个相同声源求总声强级。

Intensity also drops off with distance from a point source following the inverse square law: I ∝ 1/r². This is because the same power is spread over a spherical surface area 4πr².

点声源的强度随距离平方反比下降：I ∝ 1/r²。因为相同的功率分布在球面面积 4πr² 上。

Example Sound	Intensity Level (dB)
Threshold of hearing	0
Quiet library	30
Normal conversation	60
Busy street	80
Rock concert	110
Jet engine (near)	140

声音例子	声强级 (dB)
听觉阈	0
安静的图书馆	30
正常交谈	60
繁忙街道	80
摇滚音乐会	110
喷气发动机（近处）	140

9. Standing Waves and Resonance | 驻波与共振

Standing waves form when incident and reflected waves of the same frequency interfere in a bounded medium. Nodes are points of zero displacement, antinodes are points of maximum amplitude. In sound columns (pipes), they can be open or closed at ends.

当相同频率的入射波与反射波在有限介质内干涉时，形成驻波。波节是位移为零的点，波腹是振幅最大点。在空气柱（管乐器）中，两端可开可闭。

In a pipe closed at one end, only odd harmonics exist: f = nv/(4L) for n = 1,3,5,… The closed end is a displacement node (pressure antinode), the open end a displacement antinode (pressure node). In a pipe open at both ends, all harmonics: f = nv/(2L) for n = 1,2,3,…

一端封闭的管中只存在奇次谐波：f = nv/(4L)，n = 1,3,5,… 闭端为位移波节（压力波腹），开端为位移波腹（压力波节）。两端开口管中所有谐波均存在：f = nv/(2L)，n = 1,2,3,…

Resonance occurs when a driving frequency matches a natural frequency of an object, causing a dramatic increase in amplitude. A tuning fork over a tube of air can produce loud sound when the column length matches a resonant length.

当驱动频率与物体的固有频率相匹配时，发生共振，振幅显著增大。音叉置于空气柱上方，当空气柱长度等于共振长度时可产生响亮声音。

IB students may be asked to sketch standing wave patterns showing node/antinode positions for different harmonics and to calculate frequencies using wave equations.

IB 学生可能会被要求绘制驻波图样，标示不同谐波的波节/波腹位置，并用波动方程计算频率。

10. Musical Instruments and Harmonics | 乐器和泛音

Musical instruments rely on standing waves in strings or air columns. String instruments (violin, guitar) produce a fundamental frequency f₁ = (1/(2L))√(T/μ), where T is tension and μ mass per unit length. Overtones are integer multiples of f₁.

乐器依赖于弦或空气柱中的驻波。弦乐器（小提琴、吉他）产生基频 f₁ = (1/(2L))√(T/μ)，其中 T 为张力，μ 为单位长度质量。泛音为 f₁ 的整数倍。

Timbre (quality) is determined by the mix of harmonics. A clarinet acts like a closed pipe, favouring odd harmonics, giving a hollow sound. A flute acts like an open pipe, rich in both odd and even harmonics.

音色由泛音组合决定。单簧管类似闭管，强调奇次谐波，声音空洞。长笛类似开管，既有奇次又有偶次谐波，音色丰富。

AQA questions might give oscilloscope traces of different instruments playing the same note, showing same fundamental frequency but different waveforms due to harmonic content. IB may require calculating f₁ from given harmonic frequencies or finding tensions.

AQA 考题可能给出不同乐器演奏同一音符的示波器波形图，显示基频相同但泛音不同导致波形差异。IB 可能要求根据给定谐波频率计算基频或求张力。

11. The Human Ear and Hearing | 人耳与听觉

The outer ear collects sound and channels it through the ear canal to the eardrum, causing it to vibrate. The middle ear bones (ossicles) amplify these vibrations and transmit them to the oval window of the fluid-filled cochlea.

外耳收集声音并经由耳道传导至鼓膜，使其振动。中耳听小骨放大这些振动并传递至充满淋巴液的耳蜗的卵圆窗。

Inside the cochlea, specialised hair cells convert mechanical vibrations into electrical impulses. High-frequency sounds stimulate hair cells near the oval window; low frequencies stimulate those further inside. This is the place theory of hearing.

耳蜗内的毛细胞将机械振动转换为电脉冲。高频声音刺激靠近卵圆窗的毛细胞；低频声音刺激更深处毛细胞。这是听觉的部位学说。

Hearing damage can result from prolonged exposure to sounds above 85 dB. The ear’s response is roughly logarithmic, matching the dB scale. A-weighting is used to approximate the ear’s varying sensitivity to different frequencies.

长时间暴露于 85 dB 以上声音可导致听力损伤。人耳响应大致呈对数特性，与分贝标度匹配。A 计权用于近似人耳对不同频率的敏感度变化。

12. Ultrasound and Applications | 超声波及应用

Ultrasound refers to sound waves with frequencies above 20 kHz, beyond human hearing. Its short wavelength allows it to be focused into narrow beams, making it ideal for imaging and diagnostics.

超声波指频率超过 20 kHz、超出人耳听觉范围的声波。其波长短，可聚焦成窄束，非常适合成像和诊断。

Medical ultrasound uses the pulse-echo technique: a transducer sends short pulses and detects reflections from tissue boundaries. The time delay and intensity give information about depth and tissue type. Frequencies of 2–18 MHz are typical.

医用超声采用脉冲回声技术：换能器发射短脉冲并检测组织界面的反射。时间延迟和强度提供深度和组织类型信息。常用频率为 2–18 MHz。

Industrial applications include sonar for depth sounding, flaw detection in metals, and cleaning. Bats use ultrasound for echolocation, emitting clicks and interpreting returning echoes to navigate and hunt in darkness.

工业应用包括测深声呐、金属探伤和清洗。蝙蝠利用超声波回声定位，发出咔嗒声并解读返回的回声，在黑暗中导航和捕食。

IB and AQA syllabi both cover the principles of ultrasound imaging: v = fλ still applies, and the pulse-echo equation d = (v × t) / 2 is central. Safety aspects, such as the non-ionising nature of ultrasound compared to X-rays, are also examined.

IB 和 AQA 大纲都涵盖超声成像原理：v = fλ 依然适用，脉冲回声公式 d = (v × t) / 2 是核心。与 X 射线相比，超声波无电离辐射的安全性方面也是考点。

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Sound: Key Concepts for IB and AQA Science | IB AQA 科学：声 考点精讲