📚 IB Physics: Relationship Between Wavelength, Frequency and Wave Speed | IB物理：波长、频率与波速关系及应用
Wave motion is a fundamental concept in IB Physics. Understanding how wavelength, frequency and wave speed are related allows students to describe sound, light, water waves and electromagnetic radiation with one simple equation.
波动是 IB 物理的核心概念。理解波长、频率与波速之间的关系，可以帮助我们用一条简单方程描述声波、光波、水波和电磁辐射。
1. What Are Wavelength, Frequency and Wave Speed? | 什么是波长、频率与波速？
Wavelength (λ) is the distance between two consecutive identical points on a wave, such as crest-to-crest or compression-to-compression. In SI units it is measured in metres (m).
波长（λ）是波上两个相邻同相点之间的距离，例如波峰到波峰或压缩区到压缩区。国际单位制中单位为米（m）。
ations passing a fixed point per second. It is measured in hertz (Hz), where 1 Hz = 1 s⁻¹.
频率（f）是每秒通过固定点的完整振动次数，单位为赫兹（Hz），1 Hz = 1 s⁻¹。
riod (T) is the time for one complete oscillation, and is related to frequency by T = 1/f.
周期（T）是一次完整振动所需时间，与频率互为倒数，即 T = 1/f。
Wave speed (v) is the distance travelled by a wave disturbance per unit time, measured in metres per second (m s⁻¹).
波速（v）是波扰动在单位时间内传播的距离，单位为米每秒（m s⁻¹）。
2. The Wave Equation v = f × λ | 波的方程 v = f × λ
a distance λ in time T, so the speed is v = λ / T. Since T = 1/f, this becomes v = f × λ.
在一个波长内，波在时间 T 内传播距离 λ，因此速度为 v = λ / T。由 T = 1/f，可得 v = f × λ。
v = f × λ
This equation applies to all types of periodic waves, including transverse waves on strings and longitudinal sound waves.
该方程适用于所有周期性波，包括弦上的横波和声波等纵波。
3. Understanding From Wavefronts | 从波前理解
Imagine a wave source emitting crests at regular intervals. The distance between adjacent crests is λ, and each crest takes time T to reach the next position. The pattern moves forward at the wave speed.
想象一个波源以固定时间间隔发出波峰。相邻波峰之间的距离为 λ，每个波峰经过时间 T 前进一个波长。整个图样以波速向前移动。
If the source vibrates faster, frequency increases and adjacent crests become closer, so wavelength decreases when speed is fixed.
若波源振动更快，频率增大，相邻波峰间隔变小；在波速不变时，波长会减小。
4. Wave Speed Depends on the Medium | 波速取决于介质
For mechanical waves, wave speed is set by the properties of the medium, not by the frequency or wavelength. On a string, speed increases with greater tension and decreases with greater linear density.
对于机械波，波速由介质性质决定，而非频率或波长。在弦上，张力越大波速越大；线密度越大波速越小。
For sound, speed depends on temperature and medium. In air at 20 °C it is about 343 m s⁻¹; in water it is about 1480 m s⁻¹.
声速与温度和介质有关。在 20 °C 空气中约为 343 m s⁻¹；在水中约为 1480 m s⁻¹。
dium fixes v, changing the source frequency changes the wavelength according to λ = v / f.
由于介质决定 v，改变波源频率会改变波长，即 λ = v / f。
5. Electromagnetic Waves in Vacuum | 真空中的电磁波
ravel through a vacuum. They all travel at the same speed in vacuum, c = 3.00 × 10⁸ m s⁻¹.
电磁波是可以在真空中传播的横波。所有电磁波在真空中的速度相同，c = 3.00 × 10⁸ m s⁻¹。
es c = f × λ. Since c is constant, a higher frequency corresponds to a shorter wavelength.
波速方程化为 c = f × λ。由于 c 恒定，频率越高，波长越短。
This explains why radio waves have long wavelengths and gamma rays have very short wavelengths.
这解释了为什么无线电波波长较长，而 γ 射线波长非常短。
6. Applications in Sound and Music | 声学与音乐中的应用
Musical instruments produce specific frequencies. A guitar string with fixed tension and length produces a fundamental frequency determined by the wave speed and the string length.
乐器产生特定频率。在张力和长度一定时，吉他弦发出的基频由波速和弦长决定。
In a pipe organ, the resonant wavelength fits into the pipe length, so changing the effective length changes the frequency heard.
在管风琴中，共振波长与管长匹配，因此改变有效管长会改变听到的频率。
7. Applications in Communication and Medicine | 通信与医学中的应用
Mobile phones, Wi-Fi and radio use electromagnetic waves of different frequencies. Antenna designs depend on the corresponding wavelength.
手机、Wi-Fi 和无线电使用不同频率的电磁波。天线设计取决于对应的波长。
Ultrasound imaging uses high-frequency mechanical waves. Higher frequency gives better resolution but reduces penetration depth.
超声成像使用高频机械波。频率越高分辨率越好，但穿透深度会减小。
Laser light with a smaller wavelength can focus to a smaller spot, which is essential in eye surgery and optical data storage.
波长更小的激光可以聚焦到更小的光斑，这对眼科手术和光存储非常重要。
8. Worked Example 1 – Sound Wave | 例题 1：声波
A sound wave has a frequency of 680 Hz in air where the wave speed is 340 m s⁻¹. Find its wavelength.
空气中一列声波的频率为 680 Hz，波速为 340 m s⁻¹。求波长。
λ = v / f = 340 / 680 = 0.50 m
The wavelength is 0.50 m.
波长为 0.50 m。
9. Worked Example 2 – Light Wave | 例题 2：光波
A laser emits visible light with wavelength 500 nm in vacuum. Calculate its frequency.
一束激光在真空中的波长为 500 nm。计算其频率。
f = c / λ = (3.00 × 10⁸) / (500 × 10⁻⁹) = 6.00 × 10¹⁴ Hz
The frequency is approximately 6.00 × 10¹⁴ Hz.
该光波的频率约为 6.00 × 10¹⁴ Hz。
10. Common Misconceptions and Exam Tips | 常见误区与考试技巧
- Misconception: Changing frequency changes wave speed. In fact, wave speed is determined by the medium for a fixed state.
- 误区：改变频率会改变波速。事实上，在介质状态不变时，波速由介质决定。
- Use standard units: nm should be converted to metres before substitution.
- 使用标准单位：代入公式前应把纳米换算成米。
- When comparing waves, state whether the medium is the same or different.
- 比较波时，需说明介质是否相同。
11. Quick Practice Questions | 快速练习
| Question / 问题 | Answer / 答案 |
| A wave has λ = 2.0 m and f = 5.0 Hz. What is v? / 一列波的 λ = 2.0 m，f = 5.0 Hz，v 是多少？ | v = 10 m s⁻¹ |
| A wave has v = 1500 m s⁻¹ and λ = 0.75 m. Find f. / 一列波的 v = 1500 m s⁻¹，λ = 0.75 m，求 f。 | f = 2000 Hz |
| What is the wavelength of a 100 MHz radio wave in air? / 空气中频率为 100 MHz 的无线电波波长是多少？ | λ = 3.0 m |
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