📚 Year 11 WJEC Physics: Sound and Hearing Revision | 年级11 WJEC 物理：声音与听力备考专项
Welcome to your focused revision guide for the Sound and Hearing topic in WJEC GCSE Physics. Sound is a fascinating wave phenomenon that influences everything from music and speech to medical imaging and underwater navigation. Mastering this topic will require you to link wave properties, human biology, and practical applications. This article provides you with clear explanations, key equations, and exam-style tips organised into easy-to-digest sections.
欢迎阅读 WJEC GCSE 物理中声音与听力专题的专项备考指南。声音是一种迷人的波动现象，影响着从音乐、语言交流到医学成像和水下导航的方方面面。掌握本专题需要你将波动性质、人体生物学和实际应用联系起来。本文用清晰的解释、关键公式和考试技巧，将内容分解为易于消化的章节。
1. What Is Sound? | 什么是声音？
Sound is a form of energy produced by vibrating objects. It travels as a longitudinal wave, meaning that the particles of the medium vibrate parallel to the direction of energy transfer. These vibrations create a series of compressions (regions of high pressure) and rarefactions (regions of low pressure) that propagate outwards.
声音是由振动物体产生的一种能量。它以纵波的形式传播，即介质粒子的振动方向与能量传播方向平行。这些振动产生一系列密部（高压区域）和疏部（低压区域）向外传播。
To visualise a longitudinal wave, imagine a slinky spring being pushed and pulled at one end. The coils bunch together at compressions and spread apart at rarefactions. In air, sound waves are rapid fluctuations in pressure that our ears detect as sound.
要想象纵波，可以想象一个弹簧玩具在一端被推拉。线圈在密部聚集，在疏部分开。在空气中，声波是压力的快速波动，我们的耳朵将其检测为声音。
2. How Sound Travels | 声音如何传播
Sound requires a medium to travel – it cannot propagate through a vacuum. This is because there are no particles to vibrate and transfer the energy. The three typical media are solids, liquids, and gases, and sound generally travels fastest in solids, slower in liquids, and slowest in gases.
声音需要介质才能传播——它不能在真空中传播。这是因为没有粒子来振动并传递能量。三种典型的介质是固体、液体和气体，声音通常在固体中传播最快，在液体中较慢，在气体中最慢。
The speed depends on how close the particles are and how stiff the bonding is. In solids, particles are tightly packed and vibrations pass rapidly from one particle to the next. In gases, particles are far apart, so the transfer of vibration takes longer.
速度取决于粒子的紧密程度和键连接的刚性。在固体中，粒子紧密排列，振动能迅速从一个粒子传递到下一个。在气体中，粒子相距较远，因此振动的传递需要更长时间。
| Medium | 介质 | Approximate Speed (m/s) | 近似速度 (m/s) |
|---|---|
| Air (20 °C) | 空气 (20 °C) | 343 |
| Water | 水 | 1500 |
| Steel | 钢 | 5000–6000 |
3. Speed of Sound and the Wave Equation | 声速与波动方程
s simple formula is often tested in practical contexts, such as measuring echoes or using two microphones placed a known distance apart.
声速 (v) 可以根据传播距离 (d) 和所用时间 (t) 计算得出：v = d / t。这个简单的公式经常在实验情境中被考查，例如测量回声或使用两个相隔已知距离的麦克风。
e speed in metres per second (m/s), f is frequency in hertz (Hz), and λ is wavelength in metres (m). For a given medium, sound speed is nearly constant, so if frequency increases, wavelength must decrease.
你还需要掌握将速度、频率和波长联系起来的一般波动方程：v = f × λ。其中，v 是波速，单位为米每秒 (m/s)；f 是频率，单位为赫兹 (Hz)；λ 是波长，单位为米 (m)。在给定介质中，声速几乎是恒定的，因此频率增加时，波长必然减小。
v = f × λ
4. Frequency and Pitch | 频率与音高
Frequency is the number of complete vibrations per second, measured in hertz (Hz). In terms of sound, frequency directly determines the perceived pitch: a high-frequency sound has a high pitch (like a whistle), while a low-frequency sound has a low pitch (like a bass drum).
频率是每秒完整振动的次数，单位为赫兹 (Hz)。对于声音而言，频率直接决定了我们感知的音高：高频声音音调高（如哨声），低频声音音调低（如低音鼓）。
The healthy human ear can detect frequencies ranging from about 20 Hz to 20 000 Hz (20 kHz). This range is known as the audible range. Sounds below 20 Hz are called infrasound, and those above 20 kHz are ultrasound. Different animals have different audible ranges; for example, dogs can hear up to about 45 kHz, and bats use ultrasound for echolocation.
健康的人耳可以检测到大约 20 Hz 到 20 000 Hz（20 kHz）的频率范围，这一区间称为可听范围。低于 20 Hz 的声音称为次声波，高于 20 kHz 的称为超声波。不同动物的可听范围不同；例如，狗可以听到高达约 45 kHz 的声音，蝙蝠利用超声波进行回声定位。
On an oscilloscope display, a higher-frequency waveform will show more cycles across the screen for the same time base setting. Recognising this is a classic exam skill.
在示波器显示中，在相同时间基准设置下，频率越高的波形在屏幕上显示的周期数越多。识别这一点是一项经典的考试技能。
5. Amplitude and Loudness | 振幅与响度
Amplitude is the maximum displacement of a particle from its rest position. In a sound wave, a larger amplitude means that the compressions are more compressed and the rarefactions are more spread out, carrying more energy. Loudness is our perception of amplitude – bigger amplitude produces a louder sound.
振幅是粒子离开其平衡位置的最大位移。在声波中，振幅越大意味着密部更密、疏部更疏，携带的能量也更多。响度是我们对振幅的感知——振幅越大，声音越响亮。
Loudness is measured on the decibel (dB) scale. A whisper is around 30 dB, normal conversation is about 60 dB, and a rock concert can exceed 110 dB. The decibel scale is logarithmic, meaning an increase of 10 dB corresponds roughly to a doubling of perceived loudness.
响度用分贝 (dB) 标度来衡量。耳语大约为 30 dB，正常交谈约为 60 dB，摇滚音乐会可能超过 110 dB。分贝标度是对数的，这意味着每增加 10 dB，主观响度大约加倍。
On an oscilloscope trace, a sound with a larger amplitude will show taller peaks and deeper troughs. Be prepared to compare waveforms and explain how they relate to the sound heard.
在示波器迹线上，振幅较大的声音会显示更高的波峰和更深的波谷。准备好在考试中比较波形，并解释它们与所听到声音的关系。
6. The Human Ear and Hearing Mechanism | 人耳与听觉机制
The ear is a complex organ that converts sound waves into electrical signals that the brain interprets. Sound waves enter the outer ear and travel down the ear canal, causing the eardrum (tympanic membrane) to vibrate. These vibrations are passed through three tiny bones in the middle ear – the hammer, anvil, and stirrup – which amplify the vibrations and transmit them to the oval window of the cochlea.
耳朵是一种复杂的器官，能将声波转换成大脑可以解读的电信号。声波进入外耳，沿着耳道传播，引起鼓膜振动。这些振动通过中耳的三块小骨——锤骨、砧骨和镫骨——进行传递，它们放大振动并将其传递到耳蜗的卵圆窗。
Inside the cochlea (in the inner ear), the vibrations create waves in the fluid that bend hair cells. Different frequencies stimulate hair cells at different positions along the cochlea. The hair cells generate nerve impulses that travel via the auditory nerve to the brain, where they are interpreted as sound.
在耳蜗（内耳）内部，振动在液体中产生波动，使毛细胞弯曲。不同频率的声音会刺激耳蜗内不同位置的毛细胞。毛细胞产生神经冲动，经听神经传递到大脑，被解读为声音。
7. Ultrasound and Its Applications | 超声波及其应用
Ultrasound refers to sound waves with frequencies above 20 kHz, beyond the human audible range. It has several important applications in medicine and industry. Because ultrasound can be produced in narrow, directed beams and can reflect off boundaries between different tissues, it is ideal for non-invasive imaging.
超声波指频率高于 20 kHz、超出人类可听范围的声波。它在医学和工业中有若干重要应用。由于超声波能以狭窄指向性波束发射，且能在不同组织的界面反射，它非常适合无创成像。
In medicine, ultrasound scans are used to examine a developing fetus, view internal organs, and diagnose conditions without the risks associated with X-rays. A transducer sends pulses of ultrasound into the body and detects the echoes. The time taken for the echoes to return is used to calculate the depth of the reflecting surface, and a computer builds an image.
在医学中，超声波扫描用于检查发育中的胎儿、观察内部器官和诊断疾病，且没有 X 射线相关的风险。换能器向体内发射超声波脉冲并检测回声。根据回声返回所需的时间可以计算出反射面的深度，然后由计算机构建图像。
Industrial uses include cleaning delicate items (ultrasonic cleaning), detecting flaws in metals (non-destructive testing), and SONAR for underwater navigation and fish finding.
工业用途包括清洗精密物品（超声波清洗）、检测金属内部缺陷（无损检测）以及用于水下导航和鱼群探测的声呐。
8. Echoes and Sonar Calculations | 回声与声呐计算
la: distance = (speed of sound × time) / 2. The division by 2 is crucial because the sound travels to the reflecting surface and back, so the total time must be halved to find the one-way distance.
回声是声音的反射，在直达声之后到达听者。要利用回声测量距离，你可以使用公式：距离 = (声速 × 时间) / 2。除以 2 很关键，因为声音传播到反射面再返回，所以总时间必须减半才能得出单程距离。
SONAR (Sound Navigation and Ranging) uses this principle. A ship or a submarine sends out an ultrasound pulse and listens for the echo. Knowing the speed of sound in water and the time delay, the depth of the sea or the distance to a shoal of fish can be calculated.
声呐（声音导航与测距）就是运用这个原理。船或潜艇发出超声波脉冲并收听回声。知道了声音在水中的速度和延时，就可以计算出海深或鱼群的距离。
d = (v × t) / 2
In WJEC exam questions, be careful with units: convert time in milliseconds to seconds before using it in calculations, and remember to show the division by 2 clearly.
在 WJEC 考试题目中，要注意单位：将毫秒为单位的时间转换为秒后再代入计算，并记得明确写出除以 2 的步骤。
9. Hearing Loss and Noise Protection | 听力丧失与噪声防护
Prolonged exposure to loud sounds, especially above 85 dB, can cause permanent hearing damage. Loud sounds can rupture the eardrum or, more commonly, damage the delicate hair cells in the cochlea. Unlike some other cells, once these hair cells are destroyed, they do not regenerate in humans, leading to irreversible hearing loss.
长时间暴露于大声级声音，尤其是超过 85 dB 的声音，可导致永久性听力损伤。巨大的声音可能导致鼓膜破裂，更常见的是损伤耳蜗内脆弱的毛细胞。与其他细胞不同，这些毛细胞一旦被破坏，在人体内不会再生，从而导致不可逆的听力丧失。
Sources of harmful noise include personal audio players at high volume, loud concerts, and industrial machinery. Protective measures include wearing earplugs or earmuffs that reduce the amplitude of sound waves entering the ear, turning down volume levels, and limiting exposure time.
有害噪声来源包括高音量个人音频播放器、喧闹的音乐会和工业机械。防护措施包括佩戴可降低进入耳朵声波振幅的耳塞或耳罩、降低音量以及限制暴露时间。
10. Comparing Light and Sound Waves | 光波与声波的比较
tively slowly (about 340 m/s in air), and can be reflected as an echo. Light is a transverse electromagnetic wave that can travel through a vacuum, moves at 3 × 10⁸ m/s, and we observe its reflection in mirrors.
WJEC 经常要求你比较声波与光波。主要区别：声音是纵波、属于机械波，需要介质，传播速度相对较慢（空气中约 340 m/s），可产生回声形式的反射。光是一种横波，属于电磁波，可以在真空中传播，速度为 3 × 10⁸ m/s，我们可通过镜子观察到它的反射。
Both can be reflected and refracted, though the conditions and equations differ. In a WJEC physics paper, you might be asked to explain why you see lightning before hearing thunder – this is simply because light travels much faster than sound.
两者都能被反射和折射，但条件和方程不同。在 WJEC 物理试卷中，你可能会被要求解释为什么先看到闪电后听到雷声——这仅仅是因为光传播速度远快于声音。
11. Oscilloscope Traces and Data Interpretation | 示波器迹线与数据解读
s the time for one complete wave cycle, and frequency is its reciprocal: f = 1 / T. If a wave completes 5 cycles in 10 milliseconds, the period is 2 ms (0.002 s), and the frequency is 500 Hz.
考试题目经常给出示波器截图，要求你确定频率、周期或振幅。记住周期 (T) 是一个完整波循环的时间，频率是其倒数：f = 1 / T。如果一个波在 10 毫秒内完成 5 个周期，则周期为 2 毫秒（0.002 秒），频率为 500 Hz。
To read amplitude, measure the height from the centre line to a peak in vertical divisions and multiply by the volts-per-division setting. Superimposed waves can be tricky: identify which trace has a larger amplitude (louder sound) and which has a higher frequency (higher pitch).
要读取振幅，测量从中心线到波峰的高度（按垂直格数），再乘以每格电压设置。叠加的波形可能比较棘手：判断哪条迹线振幅更大（声音更响），哪条频率更高（音调更高）。
12. Exam Tips and Common Pitfalls | 考试技巧与常见误区
1) Always state the medium when describing sound speed. 2) Don't confuse amplitude and frequency – ms = 0.001 s. 5) Label compressions and rarefactions clearly on diagrams. 6) Remember that vacuum experiments (like the bell jar) demonstrate that sound needs a medium.
1) 描述声速时一定要说明介质。2) 不要混淆振幅和频率——振幅与响度有关，频率与音高有关。3) 使用回声公式时，记住要除以 2。4) 展示完整的解题步骤和单位换算：1 毫秒 = 0.001 秒。5) 在图中清楚标出密部和疏部。6) 记住真空实验（如钟罩实验）证明了声音需要介质。
le-check that your final answer has the correct unit and makes physical sense. Practise with WJEC past papers to get used to the phrasing of questions on echoes, ultrasound, and ear structure.
灵活运用波动方程 v = f × λ。若已知速度和频率，变形公式求出波长：λ = v / f。仔细检查最终答案是否有正确单位，并且在物理上是否合理。通过练习 WJEC 历年真题，熟悉回声、超声和人耳结构等问题的出题方式。
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