The Doppler effect · 多普勒效应
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| pitch/pɪtʃ/ | 音调 | yīn diào |
| frequency/ˈfriːkwənsi/ | 频率 | pín lǜ |
| Doppler effect/ˈdɒplə ɪˈfekt/ | 多普勒效应 | duō pǔ lè xiào yìng |
| source/sɔːs/ | 波源 | bō yuán |
| wavefronts/ˈweɪvfrʌnts/ | 波前 | bō qián |
| observer/ɒbˈzɜːvə/ | 观察者 | guān chá zhě |
The passing siren
- An ambulance races past and its siren suddenly drops in pitch 音调.
- The siren itself never changed — your ear heard a different frequency 频率.
- This is the Doppler effect 多普勒效应.
驶过的警笛
- 一辆救护车飞驰而过,它的警笛音调突然降低。
- 警笛本身从未改变——是你的耳朵听到了不同的频率。
- 这就是 多普勒效应(Doppler effect)。
Why the pitch changes
- A source moving toward you bunches the wavefronts 波前 ahead → shorter $\lambda$ → higher pitch.
- Moving away, the wavefronts spread out → longer $\lambda$ → lower pitch.
音调为什么改变
- 声源 朝 你运动时,把前方的波前挤在一起 → 更短的 $\lambda$ → 更高 的音调。
- 远离 时,波前散开 → 更长的 $\lambda$ → 更低 的音调。
Doppler effect · 多普勒效应
Send the source moving and watch the wavefronts bunch up ahead (higher pitch) and stretch out behind — the siren effect, controlled by the source's speed. · 让波源移动，观察前方的波前聚集（音调更高）和后方的波前拉伸——这是多普勒效应，由波源的速度控制。
As a sound source moves toward you, the pitch you hear is: · 当声源 朝 你运动时,你听到的音调:
Moving toward you bunches the wavefronts, shortening the wavelength and raising the frequency you hear. · 朝你运动把波前挤在一起,缩短波长并提高你听到的频率。
A source moving toward an observer bunches the wavefronts ahead of it. · 朝观察者运动的声源把它前方的波前挤在一起。
Yes — each new wavefront is sent from a point a little closer, so they crowd together in front. · 是的——每个新的波前都从更近一点的位置发出,所以它们在前方挤在一起。
What does not change
- The source 波源 still emits $f_{\text{s}}$ waves every second — its frequency is fixed.
- The waves still travel through the air at the same speed $v$; the moving source cannot push them faster.
- Only the spacing of the wavefronts changes, so only the observed wavelength and frequency change.
什么不变
- 波源(source) 每秒仍然发出 $f_{\text{s}}$ 个波——它的频率是固定的。
- 波在空气中仍以 相同的速率 $v$ 传播;运动的波源不能把它们推得更快。
- 只有波前的 间距 改变了,所以只有观察到的波长和频率改变。
A source of sound moves towards a stationary listener. Which are unchanged? Select all · 所有 that apply. · 一个声源朝静止的听者运动。哪些量不变?选出所有适用的。
The medium fixes the speed and the source fixes what it emits. What the motion changes is the WAVELENGTH ahead of the source, and hence the frequency heard. · 介质决定速率,源决定它发出什么。运动改变的是源前方的波长,因而改变听到的频率。
Where the formula comes from
- In one period $T = \dfrac{1}{f_{\text{s}}}$ the source moves $v_{\text{s}}T$, so the wavelength ahead of it is squeezed to $\lambda' = \lambda - v_{\text{s}}T = \dfrac{v - v_{\text{s}}}{f_{\text{s}}}$.
- The observer 观察者 hears $f_{\text{o}} = \dfrac{v}{\lambda'}$.
- Behind the source the wavelength is stretched instead: $\lambda' = \dfrac{v + v_{\text{s}}}{f_{\text{s}}}$.
公式从哪里来
- 在一个周期 $T = \dfrac{1}{f_{\text{s}}}$ 内,波源移动了 $v_{\text{s}}T$,所以它前方的波长被压缩为 $\lambda' = \lambda - v_{\text{s}}T = \dfrac{v - v_{\text{s}}}{f_{\text{s}}}$。
- 观察者听到 $f_{\text{o}} = \dfrac{v}{\lambda'}$。
- 在波源后方,波长反而被拉长:$\lambda' = \dfrac{v + v_{\text{s}}}{f_{\text{s}}}$。
The formula
- For a moving source and a still observer: $f_{\text{o}} = \dfrac{v\,f_{\text{s}}}{v \pm v_{\text{s}}}$.
- Use minus when approaching (higher pitch), plus when receding (lower pitch).
A moving source squashes the wavefronts ahead of it, raising the observed frequency
公式
- 对于运动的声源和静止的观察者:$f_{\text{o}} = \dfrac{v\,f_{\text{s}}}{v \pm v_{\text{s}}}$。
- 接近时用 减号(更高音调),远离时用 加号(更低音调)。
一个运动的声源把它前方的波前挤压,提高了观察到的频率
In $f_{\text{o}} = \dfrac{v\,f_{\text{s}}}{v \pm v_{\text{s}}}$, for a source moving toward the observer you use: · 在 $f_{\text{o}} = \dfrac{v\,f_{\text{s}}}{v \pm v_{\text{s}}}$ 中,声源 朝 观察者运动时,你用:
A smaller denominator gives a larger $f_{\text{o}}$ — the higher pitch you expect when approaching. · 更小的分母给出更大的 $f_{\text{o}}$——这正是接近时你预期的更高音调。
Match each situation to what a stationary listener hears. · 把每种情形与静止的听者听到的结果配对。
Approaching → minus in the formula → higher; receding → plus → lower; the switch between the two is the drop you hear as it passes. · 接近 → 公式里用减号 → 更高;远离 → 加号 → 更低;两者之间的切换就是它经过时你听到的音调下降。
Put the Doppler calculation in order. · 把多普勒计算按顺序排列。
The last step catches every sign error for free: an approaching source must give a HIGHER observed frequency. · 最后一步能白捡地抓住所有符号错误:靠近的源必定给出更高的观测频率。
Worked example: approaching
- A horn at $f_{\text{s}} = 800\ \text{Hz}$ moves at $30\ \dfrac{\text{m}}{\text{s}}$ toward you; $v = 340\ \dfrac{\text{m}}{\text{s}}$.
- Approaching → minus sign: $f_{\text{o}} = \dfrac{340 \times 800}{340 - 30}$.
- $f_{\text{o}} = \dfrac{272000}{310} \approx 877\ \text{Hz}$ — higher, as expected.
例题:接近
- 一个 $f_{\text{s}} = 800\ \text{Hz}$ 的喇叭以 $30\ \dfrac{\text{m}}{\text{s}}$ 朝你运动;$v = 340\ \dfrac{\text{m}}{\text{s}}$。
- 接近 → 减号:$f_{\text{o}} = \dfrac{340 \times 800}{340 - 30}$。
- $f_{\text{o}} = \dfrac{272000}{310} \approx 877\ \text{Hz}$——如预期更高。
A horn ($f_{\text{s}} = 400\ \text{Hz}$) moves at $20\ \dfrac{\text{m}}{\text{s}}$ toward you; $v = 340\ \dfrac{\text{m}}{\text{s}}$. What frequency do you hear? · 一个喇叭($f_{\text{s}} = 400\ \text{Hz}$)以 $20\ \dfrac{\text{m}}{\text{s}}$ 朝你运动;$v = 340\ \dfrac{\text{m}}{\text{s}}$。你听到的频率是多少?
$f_{\text{o}} = \dfrac{340 \times 400}{340 - 20} = \dfrac{136000}{320} = 425\ \text{Hz}$. · $f_{\text{o}} = \dfrac{340 \times 400}{340 - 20} = \dfrac{136000}{320} = 425\ \text{Hz}$。
A source moving away from you gives a ____ frequency than the source. · 远离 你运动的声源给出比声源 ____ 的频率。
Moving away spreads the wavefronts out, lengthening the wavelength and lowering the frequency. · 远离运动把波前散开,拉长波长并降低频率。
A siren emitting 800 Hz approaches at 30 m/s. What frequency is heard, in Hz? (speed of sound 340 m/s) · 一个发出 800 Hz 的警笛以 30 m/s 靠近。听到的频率是多少 Hz?(声速 340 m/s)
f = 800 x 340/(340 - 30) = 877 Hz. Approaching means minus on the bottom and an answer ABOVE 800 Hz; getting a lower number is the sign error announcing itself. · f = 800 x 340/(340 - 30) = 877 Hz。靠近意味着分母用减号、答案高于 800 Hz;算出更低的数,就是符号错误在自报家门。
Worked example: a source going round in a circle
A siren on a car driving round a circular track is heard by someone standing outside the track. The frequency they hear varies between a maximum of $1000\ \text{Hz}$ and a minimum of $818\ \text{Hz}$. Take $v = 340\ \dfrac{\text{m}}{\text{s}}$. Find the speed of the car and the siren's own frequency.
- Maximum is heard when the car moves straight toward the listener: $1000 = \dfrac{340\,f_{\text{s}}}{340 - v_{\text{s}}}$.
- Minimum when it moves straight away: $818 = \dfrac{340\,f_{\text{s}}}{340 + v_{\text{s}}}$.
- Divide the two equations so $f_{\text{s}}$ cancels: $\dfrac{1000}{818} = \dfrac{340 + v_{\text{s}}}{340 - v_{\text{s}}}$, giving $v_{\text{s}} = 34\ \dfrac{\text{m}}{\text{s}}$.
- Source frequency: $f_{\text{s}} = \dfrac{1000 \times (340 - 34)}{340} = 900\ \text{Hz}$.
- Check: the true frequency lies between the two heard values, and the car's speed is about a tenth of the speed of sound — sensible for a racing car.
例题:绕圈行驶的声源
一辆汽车在环形赛道上行驶,车上的警笛被站在赛道外的人听到。听到的频率在最大值 $1000\ \text{Hz}$ 和最小值 $818\ \text{Hz}$ 之间变化。取 $v = 340\ \dfrac{\text{m}}{\text{s}}$。求汽车的速率和警笛本身的频率。
- 最大值 在汽车正 朝向 听者运动时听到:$1000 = \dfrac{340\,f_{\text{s}}}{340 - v_{\text{s}}}$。
- 最小值 在它正 远离 时听到:$818 = \dfrac{340\,f_{\text{s}}}{340 + v_{\text{s}}}$。
- 把两个方程 相除,消去 $f_{\text{s}}$:$\dfrac{1000}{818} = \dfrac{340 + v_{\text{s}}}{340 - v_{\text{s}}}$,得 $v_{\text{s}} = 34\ \dfrac{\text{m}}{\text{s}}$。
- 声源频率: $f_{\text{s}} = \dfrac{1000 \times (340 - 34)}{340} = 900\ \text{Hz}$。
- 检查: 真实频率在两个听到的值之间,汽车速率约为声速的十分之一——对赛车来说合理。
A siren of frequency $900\ \text{Hz}$ moves directly away from you at $34\ \dfrac{\text{m}}{\text{s}}$. The speed of sound is $340\ \dfrac{\text{m}}{\text{s}}$. What frequency do you hear, in Hz? · 一个频率为 $900\ \text{Hz}$ 的警笛以 $34\ \dfrac{\text{m}}{\text{s}}$ 正离你远去。声速为 $340\ \dfrac{\text{m}}{\text{s}}$。你听到的频率是多少(单位 Hz)?
Receding → plus sign: $f_{\text{o}} = \dfrac{340 \times 900}{340 + 34} = \dfrac{306000}{374} = 818\ \text{Hz}$. · 远离 → 加号:$f_{\text{o}} = \dfrac{340 \times 900}{340 + 34} = \dfrac{306000}{374} = 818\ \text{Hz}$。
Three traps. The $v$ on top is the speed of the wave ($340\ \dfrac{\text{m}}{\text{s}}$ for sound), never the speed of the source. The $\pm$ sign belongs to the source speed. And while a source drives straight at you at constant speed, the pitch you hear is constant — it does not climb as the car gets nearer; it changes only as the car passes and its direction relative to you changes. Louder is not higher.
三个陷阱。分子上的 $v$ 是 波的速率(声音是 $340\ \dfrac{\text{m}}{\text{s}}$),绝不是波源的速率。$\pm$ 号属于 波源 的速率。而且当一个波源以恒定速率 正对着 你驶来时,你听到的音调是 恒定的——它不会随着汽车靠近而升高;只有当汽车经过、它相对于你的方向改变时,音调才变化。更响不等于更高。
A police car drives straight toward you at a constant speed with its siren on. The pitch you hear rises steadily as it gets closer. · 一辆警车开着警笛以恒定速率正对着你驶来。随着它靠近,你听到的音调稳步升高。
While it approaches head-on at constant speed the observed frequency is constant — it depends on the speed, not the distance. The sound gets louder, not higher. The pitch only drops as the car passes. · 当它以恒定速率正对着你驶来时,观察到的频率是恒定的——它取决于速率,而不是距离。声音变得更响,而不是更高。只有当汽车经过时音调才下降。
You've got it
- a moving source changes the frequency you hear — the Doppler effect; the source frequency and wave speed do not change
- toward → bunched wavefronts → higher; away → spread out → lower
- $f_{\text{o}} = \dfrac{v\,f_{\text{s}}}{v \pm v_{\text{s}}}$ (minus for approaching); divide the max and min equations to find $v_{\text{s}}$
你掌握了
- 运动的声源改变你听到的频率——多普勒效应;声源频率和波速不变
- 朝向 → 波前挤在一起 → 更高;远离 → 散开 → 更低
- $f_{\text{o}} = \dfrac{v\,f_{\text{s}}}{v \pm v_{\text{s}}}$(接近用减号);把最大值和最小值的方程相除可求 $v_{\text{s}}$