Acoustic Doppler echoes and standing-wave boundaries
GRE Subject Test · GRE Physics Topic 16 5:20 English narration · English + 中文 subtitles burned in
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A driver hears the echo from a stationary wall at a shifted pitch, because the outbound and the returning sound each pick up a Doppler factor.
司机听到静止墙壁反射回来的回声音调发生了变化，因为去程和回程的声音各自带上一个多普勒因子。
This topic applies source and observer Doppler factors in the frame of the air, derives the allowed pipe frequencies from the boundary conditions, and keeps frequency, wavelength and boundary changes apart.
本主题在空气参考系中应用声源和观察者的多普勒因子，由边界条件推导管中允许的频率，并区分频率、波长和边界的变化。
For sound in a stationary medium, separate source motion from observer motion.
对于静止介质中的声波，需将波源运动与观测者运动分开处理。
A source approaching at speed u sub s compresses wavefront spacing and produces received frequency f equals c divided by c minus u sub s at a stationary observer.
当波源以速度 u_s 靠近时，会压缩波前间距，使静止观测者接收到的频率为 f 等于 c 除以 c 减 u_s。
An observer approaching a stationary source at speed u sub o meets more wavefronts per second and measures f equals c plus u sub o divided by c.
当观测者以速度 u_o 靠近静止波源时，每秒遇到的波前数增多，测得的频率为 f 等于 c 加 u_o 除以 c。
Speeds are measured relative to the medium; the acoustic source and observer formulas are not symmetric under exchanging roles.
所有速度均相对于介质测量；声源公式与观测者公式在交换角色后并不对称。
Receding motion reverses the appropriate sign.
远离运动则取相反的符号。
These expressions assume subsonic motion along the propagation line.
上述表达式假设沿传播方向作亚音速运动。
For a siren moving toward a stationary reflecting wall with speed u, the wall first receives f sub wall equals f c divided by c minus u.
对于以速度 u 向静止反射墙移动的警笛，墙壁首先接收到的频率为 f_wall = fc / (c − u)。
Reflection from a stationary wall preserves frequency in the medium frame.
来自静止墙壁的反射在介质参考系中保持频率不变。
The moving driver then approaches the returning wavefronts and receives f echo equals f sub wall times c plus u divided by c, which equals f times c plus u divided by c minus u.
随后，移动的信号源靠近返回的波前，接收到的回波频率为 f_echo = f_wall(c + u) / c = f(c + u) / (c − u)。
At small u over c, the fractional shift is approximately 2u over c, but the exact expression has different numerator and denominator.
当 u/c 很小时，相对频移近似为 2u/c，但精确表达式的分子和分母不同。
A moving reflecting surface needs its own Doppler step; do not reuse the stationary-wall result blindly.
移动反射面需要其自身的多普勒步骤；不要盲目套用静止墙壁的结果。
At an ideal open pipe end, air displacement is an antinode and pressure variation is a node.
在理想开管端口处，空气位移为波腹，压强变化为波节。
At a rigid closed end, displacement is a node and pressure is an antinode.
在刚性闭口端处，位移为波节，压强为波腹。
For both ends open, length L contains n half-wavelengths, giving f sub n equals n c over 2 L, where n equals 1, 2, and so on.
两端开口时，长度L包含n个半波长，得出f下标n等于nc除以2L，其中n等于1、2等。
For one end closed, it contains an odd number of quarter-wavelengths, giving f sub n equals 2 n minus 1, times c, all over 4 L.
一端封闭时，包含奇数个四分之一波长，得出f下标n等于2n减1乘以c，再除以4L。
Real pipes may need end corrections; the ideal GRE model uses the stated length without inventing an adjustment.
实际管道可能需要端口修正；理想GRE模型直接使用给定长度，不引入额外调整。
Closing one end of a previously both-open pipe halves its fundamental and leaves only odd multiples of that new fundamental.
将原本两端开口的管道一端封闭，其基频减半，仅保留该新基频的奇数倍。
Its old frequencies were integer multiples of c/(2L), which are even multiples of c/(4L), so none is an allowed frequency of the new ideal one-closed spectrum.
旧频率是c除以2L的整数倍，即c除以4L的偶数倍，因此均不属于新的理想一端闭口管道的允许频率。
This differs from simply deleting even harmonics while retaining the old fundamental.
这与单纯删除偶次谐波而保留原基频的情况不同。
When the medium is unchanged, wave speed is unchanged; frequency and wavelength change together to satisfy the new boundary geometry.
当介质不变时，波速保持不变；频率与波长共同变化以满足新的边界几何条件。
A 600 hertz siren approaches the wall at 20 metres per second, with sound at 340.
一个 600 赫兹的警笛以每秒 20 米的速度驶向墙壁，声速为每秒 340 米。
The driver hears 600 times 360 over 320: 675 hertz.
司机听到的频率是 600 乘 360 除以 320：675 赫兹。
These speeds are measured relative to the air; this is not the formula for light.
这些速度都相对于空气测量；这不是光的多普勒公式。
A stationary source and receiver face a reflector approaching at speed u in still air.
静止的波源和接收器正对一个在静空气中以速度u靠近的反射器。
Derive the reflected frequency and invert the relation to find u from measured ratio r.
推导反射频率，并反解该关系式，根据测得的比值r求出u。
Check the low-speed limit.
检验低速极限情况。
Four marks.
本题4分。
The reflector receives f times c plus u, all over c, as a moving observer and re-emits toward the receiver as a moving source.
反射器作为运动观测者接收 f(c+u)/c，并作为运动波源向接收器重新发射。
Therefore, f sub r equals f times c plus u, all over c minus u.
因此，f_r = f(c+u)/(c-u)。
R times c minus u equals c plus u gives u equals c times r minus 1, all over r plus 1.
由 r(c-u) = c+u 可得 u = c(r-1)/(r+1)。
For u much less than c, r is approximately equal to 1 plus 2u divided by c.
当 u ≪ c 时，r ≈ 1 + 2u/c。
Award two factors, product, inversion and limit.
评分点包括两个因子、乘积、倒数和极限。
The equality with the moving-siren echo ratio does not make source and observer roles interchangeable.
该比值与移动警报器回波比值的相等性，并不意味着波源与观测者的角色可以互换。
An echo needs both Doppler factors.
回声需要两个多普勒因素。
Closing a pipe changes the fundamental; odd harmonics are counted from the new fundamental, not the old one.
封闭管道会改变基频；奇次谐波从新的基频开始计数，而非旧的基频。
displacement antinode: A standing-wave position where the air displacement amplitude is maximal.
位移波腹。 空气位移振幅达到最大的驻波位置。
Doppler factor: Frequency multiplier produced by a specified source or observer motion.
多普勒因子。 由特定声源或观测者运动产生的频率乘数。
Retrieve the meanings without reading the definitions, then change a parameter in the original transfer and test the limiting case.
在不阅读定义的情况下回忆其含义，然后更改原始传递中的一个参数并检验极限情况。