Companion to the IB-Style Practice SetIB 风格练习题的解析配套
Syllabus C5.1 to C5.6考纲 C5.1 至 C5.6PHYSICS HL
A car sounds a steady horn at constant speed. (a) explain why the observed frequency is higher on approach using wavefront spacing; (b) describe the pitch over the whole passage and at the level instant.汽车匀速鸣笛。(a) 用波前间距解释为何接近时观察频率更高；(b) 描述整个经过及平齐瞬间的音调。
In each period the source moves a little closer to the observer, so each successive wavefront is emitted nearer than the last and the wavefronts ahead of the car bunch together. (M1)
Closer-spaced wavefronts mean a shorter observed wavelength $\lambda'$. The wave speed in the air is unchanged, so from $v = f\lambda$ a shorter $\lambda'$ gives a higher observed frequency $f' > f$. (A1)
While approaching, the observed frequency is steady and above $f$ (a high pitch). After passing, the wavefronts stretch out and the observed frequency falls below $f$ (a low pitch). (A1)
At the instant the car is level, its motion is momentarily at right angles to the line of sight, so the radial speed is zero and $f' \approx f$: the pedestrian hears a sudden drop from high to low at that point. (A1)
每个周期内波源稍稍更靠近观察者，故每个相继波前都比前一个发出得更近，车前方的波前彼此堆积。(M1)
波前间距变小意味着观测波长 $\lambda'$ 变短。空气中波速不变，由 $v = f\lambda$，$\lambda'$ 变短给出更高的观测频率 $f' > f$。(A1)
接近时观测频率稳定且高于 $f$（音调偏高）。经过后波前拉伸，观测频率降到 $f$ 以下（音调偏低）。(A1)
汽车平齐瞬间，其运动暂时垂直于视线，径向速度为零，$f' \approx f$：行人在该点听到音调由高到低骤降。(A1)
Hydrogen line rest $434.0\ \mathrm{nm}$, observed at $434.3\ \mathrm{nm}$ in a star. (a) red or blue, approaching or receding; (b) radial speed; (c) why the low-speed formula is justified.氢线静止 $434.0\ \mathrm{nm}$，在恒星中观测为 $434.3\ \mathrm{nm}$。(a) 红移或蓝移、接近或远离；(b) 径向速率；(c) 为何低速公式合理。
The observed wavelength is longer than the rest wavelength, $\Delta\lambda = +0.3\ \mathrm{nm} > 0$, so the line is redshifted. (A1)
A redshift corresponds to increasing separation, so the star is receding. (A1)
Use $\dfrac{\Delta\lambda}{\lambda} \approx \dfrac{v}{c}$ with $\Delta\lambda = 0.3\ \mathrm{nm}$, $\lambda = 434.0\ \mathrm{nm}$. (M1)
$$ v \approx c\,\frac{\Delta\lambda}{\lambda} = (3.00\times10^{8})\times\frac{0.3}{434.0}. $$(M1 for substitution)
$$ v \approx (3.00\times10^{8})(6.9\times10^{-4}) \approx 2.1\times10^{5}\ \mathrm{m\,s^{-1}}. $$(A1)
The speed found is about $7\times10^{-4}$ of $c$, that is $v \ll c$, so the non-relativistic approximation $\Delta\lambda/\lambda \approx v/c$ is well justified. (B1)
观测波长长于静止波长，$\Delta\lambda = +0.3\ \mathrm{nm} > 0$，故该线为红移。(A1)
红移对应间距增大，故恒星正在远离。(A1)
用 $\dfrac{\Delta\lambda}{\lambda} \approx \dfrac{v}{c}$，$\Delta\lambda = 0.3\ \mathrm{nm}$，$\lambda = 434.0\ \mathrm{nm}$。(M1)
$$ v \approx c\,\frac{\Delta\lambda}{\lambda} = (3.00\times10^{8})\times\frac{0.3}{434.0}. $$(代入得 M1)
$$ v \approx (3.00\times10^{8})(6.9\times10^{-4}) \approx 2.1\times10^{5}\ \mathrm{m\,s^{-1}}. $$(A1)
所求速率约为 $c$ 的 $7\times10^{-4}$，即 $v \ll c$，故非相对论近似 $\Delta\lambda/\lambda \approx v/c$ 完全成立。(B1)
Train horn $480\ \mathrm{Hz}$, $v_s = 25\ \mathrm{m\,s^{-1}}$, $v = 340\ \mathrm{m\,s^{-1}}$. (a) frequency on approach; (b) on recession; (c) explain why the rise exceeds the fall.火车汽笛 $480\ \mathrm{Hz}$，$v_s = 25\ \mathrm{m\,s^{-1}}$，$v = 340\ \mathrm{m\,s^{-1}}$。(a) 接近频率；(b) 远离频率；(c) 解释为何升高大于降低。
Moving source approaching, take the minus sign: $f' = f\dfrac{v}{v - v_s}$. (M1)
$$ f' = 480\times\frac{340}{340 - 25} = 480\times\frac{340}{315} \approx 518\ \mathrm{Hz}. $$(A1)
Moving source receding, take the plus sign: $f' = f\dfrac{v}{v + v_s}$. (M1)
$$ f' = 480\times\frac{340}{340 + 25} = 480\times\frac{340}{365} \approx 447\ \mathrm{Hz}. $$(A1)
The rise is $518 - 480 = +38\ \mathrm{Hz}$, while the fall is $480 - 447 = -33\ \mathrm{Hz}$: the rise is larger. (M1)
This is because $v_s$ sits in the denominator of $f' = fv/(v \pm v_s)$, so the function is nonlinear: $\dfrac{v}{v - v_s}$ grows faster than $\dfrac{v}{v + v_s}$ shrinks for the same $v_s$. The relation is not symmetric in $v_s$. (R1)
波源接近，取减号：$f' = f\dfrac{v}{v - v_s}$。(M1)
$$ f' = 480\times\frac{340}{340 - 25} = 480\times\frac{340}{315} \approx 518\ \mathrm{Hz}. $$(A1)
波源远离，取加号：$f' = f\dfrac{v}{v + v_s}$。(M1)
$$ f' = 480\times\frac{340}{340 + 25} = 480\times\frac{340}{365} \approx 447\ \mathrm{Hz}. $$(A1)
升高为 $518 - 480 = +38\ \mathrm{Hz}$，降低为 $480 - 447 = -33\ \mathrm{Hz}$：升高更大。(M1)
这是因为 $v_s$ 位于 $f' = fv/(v \pm v_s)$ 的分母，函数非线性：对相同 $v_s$，$\dfrac{v}{v - v_s}$ 增大得比 $\dfrac{v}{v + v_s}$ 减小得快。该关系对 $v_s$ 不对称。(R1)
Speaker $600\ \mathrm{Hz}$, $v = 340\ \mathrm{m\,s^{-1}}$. (a) runner toward stationary speaker at $18\ \mathrm{m\,s^{-1}}$; (b) both toward each other at $18\ \mathrm{m\,s^{-1}}$; (c) why sound treats the two cases differently.扬声器 $600\ \mathrm{Hz}$，$v = 340\ \mathrm{m\,s^{-1}}$。(a) 跑者以 $18\ \mathrm{m\,s^{-1}}$ 朝静止扬声器；(b) 两者以 $18\ \mathrm{m\,s^{-1}}$ 相向；(c) 声波为何区别对待两种情形。
Observer moving toward a stationary source, take the plus sign: $f' = f\dfrac{v + v_o}{v}$. (M1)
$$ f' = 600\times\frac{340 + 18}{340} = 600\times\frac{358}{340} \approx 632\ \mathrm{Hz}. $$(A1)
Combined formula: observer approaching (plus on top), source approaching (minus on bottom): $f' = f\dfrac{v + v_o}{v - v_s}$ with $v_o = v_s = 18$. (M1)
$$ f' = 600\times\frac{340 + 18}{340 - 18} = 600\times\frac{358}{322} \approx 667\ \mathrm{Hz}. $$(A1)
A moving observer sweeps through wavefronts faster but does not change their spacing, so $v_o$ appears in the numerator; a moving source compresses the wavelength itself, so $v_s$ appears in the denominator. (R1)
The two differ because the air is a medium that defines a rest frame: "source moving" and "observer moving" are physically distinct relative to that air, even at the same speed. (R1)
观察者朝静止波源运动，取加号：$f' = f\dfrac{v + v_o}{v}$。(M1)
$$ f' = 600\times\frac{340 + 18}{340} = 600\times\frac{358}{340} \approx 632\ \mathrm{Hz}. $$(A1)
组合公式：观察者接近（分子取加），波源接近（分母取减）：$f' = f\dfrac{v + v_o}{v - v_s}$，$v_o = v_s = 18$。(M1)
$$ f' = 600\times\frac{340 + 18}{340 - 18} = 600\times\frac{358}{322} \approx 667\ \mathrm{Hz}. $$(A1)
运动的观察者更快地掠过波前，但不改变波前间距，故 $v_o$ 出现在分子；运动的波源压缩波长本身，故 $v_s$ 出现在分母。(R1)
两者不同是因为空气作为介质定义了静止参考系："波源运动"与"观察者运动"相对该空气在物理上不同，即使速率相同。(R1)
Galaxy H-line rest $656.3\ \mathrm{nm}$, observed $660.0\ \mathrm{nm}$. (a) define redshift and blueshift; (b) recession speed; (c) Andromeda is blueshifted, interpret and reconcile with the general redshift.星系氢线静止 $656.3\ \mathrm{nm}$，观测 $660.0\ \mathrm{nm}$。(a) 定义红移与蓝移；(b) 退行速率；(c) 仙女座蓝移，解释并与普遍红移调和。
Redshift: the observed wavelength is longer than the rest wavelength, $\Delta\lambda = \lambda_{\text{obs}} - \lambda_{\text{rest}} > 0$. (A1)
Blueshift: the observed wavelength is shorter than the rest wavelength, $\Delta\lambda < 0$. (A1)
$\Delta\lambda = 660.0 - 656.3 = 3.7\ \mathrm{nm} > 0$ (a redshift). Use $v \approx c\,\Delta\lambda/\lambda$: (M1)
$$ v \approx (3.00\times10^{8})\times\frac{3.7}{656.3} \approx 1.7\times10^{6}\ \mathrm{m\,s^{-1}}. $$(A1)
A blueshift means $\lambda_{\text{obs}} < \lambda_{\text{rest}}$, so Andromeda is approaching the Milky Way. (A1)
This does not contradict the general redshift: Andromeda is a near neighbour whose local gravitational attraction toward us exceeds the cosmological recession at such close range. The expansion-driven redshift only dominates for galaxies far enough that the recession of space outruns these local motions. (R1)
红移：观测波长长于静止波长，$\Delta\lambda = \lambda_{\text{obs}} - \lambda_{\text{rest}} > 0$。(A1)
蓝移：观测波长短于静止波长，$\Delta\lambda < 0$。(A1)
$\Delta\lambda = 660.0 - 656.3 = 3.7\ \mathrm{nm} > 0$（红移）。用 $v \approx c\,\Delta\lambda/\lambda$：(M1)
$$ v \approx (3.00\times10^{8})\times\frac{3.7}{656.3} \approx 1.7\times10^{6}\ \mathrm{m\,s^{-1}}. $$(A1)
蓝移意味着 $\lambda_{\text{obs}} < \lambda_{\text{rest}}$，故仙女座正在接近银河系。(A1)
这并不与普遍红移矛盾：仙女座是近邻，在如此近的距离上它朝我们的局部引力吸引超过宇宙学退行。膨胀驱动的红移只在足够远、空间退行超过这些局部运动的星系上占主导。(R1)
Four galaxies, rest line $\lambda_0 = 486.1\ \mathrm{nm}$, shifts $\Delta\lambda = 2.4, 4.9, 7.3, 9.8\ \mathrm{nm}$. (a) show $v$ vs $\Delta\lambda$ is linear through the origin, state the gradient; (b) speed of S, confirm against $c/\lambda_0$; (c) percentage uncertainty in $\Delta\lambda$ for S; (d) trend that evidences expansion.四个星系，静止线 $\lambda_0 = 486.1\ \mathrm{nm}$，移动 $\Delta\lambda = 2.4, 4.9, 7.3, 9.8\ \mathrm{nm}$。(a) 证明 $v$ 对 $\Delta\lambda$ 为过原点直线并说明斜率；(b) S 的速率，用 $c/\lambda_0$ 验证；(c) S 中 $\Delta\lambda$ 的百分比不确定度；(d) 支持膨胀的趋势。
From $\dfrac{\Delta\lambda}{\lambda_0} \approx \dfrac{v}{c}$, rearrange for $v$: (M1)
$$ v = \frac{c}{\lambda_0}\,\Delta\lambda. $$This has the form $v = (\text{gradient})\times\Delta\lambda$ with no intercept, so $v$ against $\Delta\lambda$ is a straight line through the origin. (A1)
Comparing with $y = mx$, the gradient is $\dfrac{c}{\lambda_0} = \dfrac{3.00\times10^{8}}{486.1\times10^{-9}} \approx 6.2\times10^{14}\ \mathrm{m^{-1}\,s^{-1}}$ when $\Delta\lambda$ is in metres, equivalently $6.2\times10^{5}\ \mathrm{s^{-1}}$ per nanometre of shift. (A1)
For S, $\Delta\lambda = 9.8\ \mathrm{nm}$. Using $v = c\,\Delta\lambda/\lambda_0$: (M1)
$$ v_S = (3.00\times10^{8})\times\frac{9.8}{486.1} \approx 6.0\times10^{6}\ \mathrm{m\,s^{-1}}. $$(A1)
Check against the gradient: $(6.2\times10^{5}\ \mathrm{s^{-1}})\times(9.8\ \mathrm{nm}) = (6.17\times10^{5})(9.8) \approx 6.0\times10^{6}\ \mathrm{m\,s^{-1}}$, consistent. (A1)
$\Delta\lambda = \lambda_{\text{obs}} - \lambda_0$ is a difference, so absolute uncertainties add: $0.1 + 0.1 = 0.2\ \mathrm{nm}$. (M1)
$$ \frac{0.2}{9.8}\times100\% \approx 2.0\%. $$(A1)
If the galaxies with larger recession speed $v$ are also found, by independent distance measurements, to be farther away, then speed increases with distance. (B1)
A systematic increase of recession speed with distance, with almost all distant galaxies redshifted, is the evidence that the universe is expanding. (B1)
由 $\dfrac{\Delta\lambda}{\lambda_0} \approx \dfrac{v}{c}$，解出 $v$：(M1)
$$ v = \frac{c}{\lambda_0}\,\Delta\lambda. $$此式形如 $v = (\text{斜率})\times\Delta\lambda$，无截距，故 $v$ 对 $\Delta\lambda$ 为过原点直线。(A1)
与 $y = mx$ 比较，斜率为 $\dfrac{c}{\lambda_0} = \dfrac{3.00\times10^{8}}{486.1\times10^{-9}} \approx 6.2\times10^{14}\ \mathrm{m^{-1}\,s^{-1}}$（$\Delta\lambda$ 取米时），即每纳米移动对应 $6.2\times10^{5}\ \mathrm{s^{-1}}$。(A1)
对 S，$\Delta\lambda = 9.8\ \mathrm{nm}$。用 $v = c\,\Delta\lambda/\lambda_0$：(M1)
$$ v_S = (3.00\times10^{8})\times\frac{9.8}{486.1} \approx 6.0\times10^{6}\ \mathrm{m\,s^{-1}}. $$(A1)
用斜率验证：$(6.2\times10^{5}\ \mathrm{s^{-1}})\times(9.8\ \mathrm{nm}) = (6.17\times10^{5})(9.8) \approx 6.0\times10^{6}\ \mathrm{m\,s^{-1}}$，一致。(A1)
$\Delta\lambda = \lambda_{\text{obs}} - \lambda_0$ 是差值，故绝对不确定度相加：$0.1 + 0.1 = 0.2\ \mathrm{nm}$。(M1)
$$ \frac{0.2}{9.8}\times100\% \approx 2.0\%. $$(A1)
若退行速率 $v$ 较大的星系经独立距离测量也更远，则速度随距离增大。(B1)
退行速度随距离系统性增大，且几乎所有遥远星系都红移，这就是宇宙膨胀的证据。(B1)
Siren $f = 500\ \mathrm{Hz}$ approaching at $v_s = 10,20,30,40\ \mathrm{m\,s^{-1}}$; $f' = 515, 531, 548, 567\ \mathrm{Hz}$; $v = 340\ \mathrm{m\,s^{-1}}$. (a) equation and why $f'$ vs $v_s$ is not linear; (b) show $1/f' = 1/f - v_s/(fv)$, state gradient and intercept; (c) compute $1/f'$ at $v_s = 10, 40$ and the gradient; (d) obtain $v$ and compare with $340$.警笛 $f = 500\ \mathrm{Hz}$ 以 $v_s = 10,20,30,40\ \mathrm{m\,s^{-1}}$ 接近；$f' = 515, 531, 548, 567\ \mathrm{Hz}$；$v = 340\ \mathrm{m\,s^{-1}}$。(a) 方程及为何 $f'$ 对 $v_s$ 非线性；(b) 证明 $1/f' = 1/f - v_s/(fv)$，写出斜率与截距；(c) 求 $v_s = 10, 40$ 处的 $1/f'$ 及斜率；(d) 求 $v$ 并与 $340$ 比较。
For an approaching source $f' = f\dfrac{v}{v - v_s}$. (A1)
Because $v_s$ sits in the denominator, $f'$ is a reciprocal function of $v_s$, not a linear one, so a direct plot of $f'$ against $v_s$ curves upward. (R1)
Take the reciprocal of $f' = fv/(v - v_s)$: (M1)
$$ \frac{1}{f'} = \frac{v - v_s}{fv} = \frac{v}{fv} - \frac{v_s}{fv} = \frac{1}{f} - \frac{v_s}{fv}. $$This is linear in $v_s$ with gradient $-\dfrac{1}{fv}$ (A1) and intercept $\dfrac{1}{f}$ (A1).
At $v_s = 10$: $\dfrac{1}{f'} = \dfrac{1}{515} = 1.942\times10^{-3}\ \mathrm{s}$. At $v_s = 40$: $\dfrac{1}{f'} = \dfrac{1}{567} = 1.764\times10^{-3}\ \mathrm{s}$. (M1·A1)
Gradient between these two points: (M1)
$$ \text{gradient} = \frac{1.764\times10^{-3} - 1.942\times10^{-3}}{40 - 10} = \frac{-1.78\times10^{-4}}{30} \approx -5.9\times10^{-6}\ \mathrm{s^{2}\,m^{-1}}. $$(A1)
The gradient equals $-\dfrac{1}{fv}$, so $v = -\dfrac{1}{f\times\text{gradient}}$: (M1)
$$ v = \frac{-1}{(500)(-5.9\times10^{-6})} = \frac{1}{2.95\times10^{-3}} \approx 3.4\times10^{2}\ \mathrm{m\,s^{-1}}. $$(A1)
This is about $337\ \mathrm{m\,s^{-1}}$, within roughly $1\%$ of the stated $340\ \mathrm{m\,s^{-1}}$, so the data are consistent with the accepted value. (R1)
波源接近时 $f' = f\dfrac{v}{v - v_s}$。(A1)
由于 $v_s$ 在分母，$f'$ 是 $v_s$ 的倒数函数而非线性函数，故 $f'$ 对 $v_s$ 的直接作图向上弯曲。(R1)
对 $f' = fv/(v - v_s)$ 取倒数：(M1)
$$ \frac{1}{f'} = \frac{v - v_s}{fv} = \frac{v}{fv} - \frac{v_s}{fv} = \frac{1}{f} - \frac{v_s}{fv}. $$这对 $v_s$ 为线性，斜率为 $-\dfrac{1}{fv}$（A1），截距为 $\dfrac{1}{f}$（A1）。
当 $v_s = 10$：$\dfrac{1}{f'} = \dfrac{1}{515} = 1.942\times10^{-3}\ \mathrm{s}$。当 $v_s = 40$：$\dfrac{1}{f'} = \dfrac{1}{567} = 1.764\times10^{-3}\ \mathrm{s}$。(M1·A1)
两点间斜率：(M1)
$$ \text{斜率} = \frac{1.764\times10^{-3} - 1.942\times10^{-3}}{40 - 10} = \frac{-1.78\times10^{-4}}{30} \approx -5.9\times10^{-6}\ \mathrm{s^{2}\,m^{-1}}. $$(A1)
斜率等于 $-\dfrac{1}{fv}$，故 $v = -\dfrac{1}{f\times\text{斜率}}$：(M1)
$$ v = \frac{-1}{(500)(-5.9\times10^{-6})} = \frac{1}{2.95\times10^{-3}} \approx 3.4\times10^{2}\ \mathrm{m\,s^{-1}}. $$(A1)
约为 $337\ \mathrm{m\,s^{-1}}$，与给定的 $340\ \mathrm{m\,s^{-1}}$ 相差约 $1\%$，故数据与公认值一致。(R1)
Whistle $660\ \mathrm{Hz}$, $v = 340\ \mathrm{m\,s^{-1}}$. (a) source toward cyclist at $30$; (b) cyclist toward stationary whistle at $15$; (c) both toward, $30$ and $15$; (d) derive $f' = fv/(v - v_s)$; (e) would a light source need the source/observer distinction.汽笛 $660\ \mathrm{Hz}$，$v = 340\ \mathrm{m\,s^{-1}}$。(a) 波源以 $30$ 朝骑车者；(b) 骑车者以 $15$ 朝静止汽笛；(c) 两者相向，$30$ 与 $15$；(d) 推导 $f' = fv/(v - v_s)$；(e) 光源是否需区分波源/观察者。
Moving source approaching, minus sign: $f' = f\dfrac{v}{v - v_s}$. (M1)
$$ f' = 660\times\frac{340}{340 - 30} = 660\times\frac{340}{310} \approx 724\ \mathrm{Hz}. $$(A1)
Moving observer approaching, plus sign: $f' = f\dfrac{v + v_o}{v}$. (M1)
$$ f' = 660\times\frac{340 + 15}{340} = 660\times\frac{355}{340} \approx 689\ \mathrm{Hz}. $$(A1)
Combined formula, plus on top (observer approaches), minus on bottom (source approaches): (M1)
$$ f' = f\frac{v + v_o}{v - v_s} = 660\times\frac{340 + 15}{340 - 30}. $$(M1 for both signs correct)
$$ f' = 660\times\frac{355}{310} \approx 756\ \mathrm{Hz}. $$(A1)
In one period $T = 1/f$ the source emits one wavefront and itself advances $v_s T$ toward the observer, so the wavefronts ahead are separated by (M1)
$$ \lambda' = vT - v_s T = (v - v_s)T = \frac{v - v_s}{f}. $$The observer still measures the wave travelling at speed $v$, so $f' = v/\lambda'$: (M1)
$$ f' = \frac{v}{(v - v_s)/f} = f\,\frac{v}{v - v_s}. $$(A1)
No. Light needs no medium, so there is no preferred rest frame and only the relative velocity is defined. (A1)
A single symmetric formula $\Delta f/f \approx v/c$ then applies whether the source or the observer is taken to move, so the source-versus-observer distinction disappears. (R1)
波源接近，取减号：$f' = f\dfrac{v}{v - v_s}$。(M1)
$$ f' = 660\times\frac{340}{340 - 30} = 660\times\frac{340}{310} \approx 724\ \mathrm{Hz}. $$(A1)
观察者接近，取加号：$f' = f\dfrac{v + v_o}{v}$。(M1)
$$ f' = 660\times\frac{340 + 15}{340} = 660\times\frac{355}{340} \approx 689\ \mathrm{Hz}. $$(A1)
组合公式，分子取加（观察者接近），分母取减（波源接近）：(M1)
$$ f' = f\frac{v + v_o}{v - v_s} = 660\times\frac{340 + 15}{340 - 30}. $$(两符号都正确得 M1)
$$ f' = 660\times\frac{355}{310} \approx 756\ \mathrm{Hz}. $$(A1)
在一个周期 $T = 1/f$ 内，波源发出一个波前并自身朝观察者前进 $v_s T$，故前方波前的间距为 (M1)
$$ \lambda' = vT - v_s T = (v - v_s)T = \frac{v - v_s}{f}. $$观察者测得波速仍为 $v$，故 $f' = v/\lambda'$：(M1)
$$ f' = \frac{v}{(v - v_s)/f} = f\,\frac{v}{v - v_s}. $$(A1)
否。光不需要介质，故无优先静止参考系，只定义相对速度。(A1)
此时单一对称公式 $\Delta f/f \approx v/c$ 适用，无论取波源还是观察者运动，故波源与观察者之分消失。(R1)
Radar $24.0\ \mathrm{GHz}$, return shifted up by $3.6\ \mathrm{kHz}$, $c = 3.00\times10^{8}$. (a) why two shifts, hence $\Delta f/f \approx 2v/c$; (b) car speed in m/s and km/h; (c) ultrasound $2.0\ \mathrm{MHz}$ in tissue $1540\ \mathrm{m\,s^{-1}}$, cells at $0.30\ \mathrm{m\,s^{-1}}$, find echo shift; (d) consequence of forgetting the factor of 2.雷达 $24.0\ \mathrm{GHz}$，返回升高 $3.6\ \mathrm{kHz}$，$c = 3.00\times10^{8}$。(a) 为何两次频移，故 $\Delta f/f \approx 2v/c$；(b) 车速 m/s 与 km/h；(c) 超声 $2.0\ \mathrm{MHz}$，组织 $1540\ \mathrm{m\,s^{-1}}$，细胞 $0.30\ \mathrm{m\,s^{-1}}$，求回波频移；(d) 漏掉因子 2 的后果。
The car first receives the microwaves while moving toward the gun, so it acts as a moving observer and detects a blueshifted frequency. (M1)
It then re-emits (reflects) that already-shifted wave while still moving toward the gun, now acting as a moving source, adding a second shift in the same sense. (M1)
The two shifts add, so to first order the round-trip fractional shift is twice the one-way value: $\dfrac{\Delta f}{f} \approx \dfrac{2v}{c}$. (A1)
Rearrange $\dfrac{\Delta f}{f} \approx \dfrac{2v}{c}$ for $v$: $v \approx \dfrac{c\,\Delta f}{2f}$. (M1)
$$ v = \frac{(3.00\times10^{8})(3.6\times10^{3})}{2(24.0\times10^{9})} = \frac{1.08\times10^{12}}{4.8\times10^{10}} = 22.5\ \mathrm{m\,s^{-1}}. $$(A1)
Converting: $22.5\times3.6 \approx 81\ \mathrm{km\,h^{-1}}$. (A1)
Reflection off moving cells gives the double shift $\Delta f \approx \dfrac{2 f v}{v_{\text{sound}}}$: (M1)
$$ \Delta f = \frac{2(2.0\times10^{6})(0.30)}{1540} \approx 7.8\times10^{2}\ \mathrm{Hz}. $$(A1)
Treating the echo as a single one-way shift uses $\Delta f/f \approx v/c$, which solves to $v = c\,\Delta f/f$, exactly twice the true value. (A1)
The reported speed would be doubled, here $45\ \mathrm{m\,s^{-1}}$ instead of $22.5\ \mathrm{m\,s^{-1}}$, a serious error for a speed-enforcement device. (R1)
汽车先在朝雷达运动时接收微波，故作为运动的观察者，检测到蓝移频率。(M1)
随后它在仍朝雷达运动时重新发射（反射）这已频移的波，此时作为运动的波源，沿同一方向叠加第二次频移。(M1)
两次频移相加，故一阶近似下往返的相对频移是单程值的两倍：$\dfrac{\Delta f}{f} \approx \dfrac{2v}{c}$。(A1)
把 $\dfrac{\Delta f}{f} \approx \dfrac{2v}{c}$ 解出 $v$：$v \approx \dfrac{c\,\Delta f}{2f}$。(M1)
$$ v = \frac{(3.00\times10^{8})(3.6\times10^{3})}{2(24.0\times10^{9})} = \frac{1.08\times10^{12}}{4.8\times10^{10}} = 22.5\ \mathrm{m\,s^{-1}}. $$(A1)
换算：$22.5\times3.6 \approx 81\ \mathrm{km\,h^{-1}}$。(A1)
从运动细胞反射给出双重频移 $\Delta f \approx \dfrac{2 f v}{v_{\text{声}}}$：(M1)
$$ \Delta f = \frac{2(2.0\times10^{6})(0.30)}{1540} \approx 7.8\times10^{2}\ \mathrm{Hz}. $$(A1)
把回波当作单程频移会用 $\Delta f/f \approx v/c$，解得 $v = c\,\Delta f/f$，恰为真实值的两倍。(A1)
报出的速率会翻倍，此处为 $45\ \mathrm{m\,s^{-1}}$ 而非 $22.5\ \mathrm{m\,s^{-1}}$，对执法测速装置是严重错误。(R1)
Calcium line rest $397.0\ \mathrm{nm}$, observed $405.0\ \mathrm{nm}$, $c = 3.00\times10^{8}$. (a) shift and direction; (b) radial speed and fraction of $c$; (c) how many galaxies evidence expansion; (d) why $v$ from $c\Delta\lambda/\lambda$ needs caution for very distant galaxies.钙线静止 $397.0\ \mathrm{nm}$，观测 $405.0\ \mathrm{nm}$，$c = 3.00\times10^{8}$。(a) 移动与方向；(b) 径向速率与 $c$ 的分数；(c) 多星系如何证明膨胀；(d) 为何对极远星系 $c\Delta\lambda/\lambda$ 求 $v$ 需谨慎。
$\Delta\lambda = 405.0 - 397.0 = +8.0\ \mathrm{nm}$. (A1)
The wavelength has increased ($\Delta\lambda > 0$), a redshift, so the galaxy is receding. (A1)
Use $v \approx c\,\dfrac{\Delta\lambda}{\lambda}$ with $\lambda = 397.0\ \mathrm{nm}$: (M1)
$$ v = (3.00\times10^{8})\times\frac{8.0}{397.0} \approx 6.0\times10^{6}\ \mathrm{m\,s^{-1}}. $$(A1)
As a fraction of $c$: $\dfrac{v}{c} = \dfrac{8.0}{397.0} \approx 0.020$, about $2\%$ of $c$. (A1)
Measuring many galaxies shows almost all are redshifted, so almost all are receding from us. (B1)
The more distant the galaxy, the greater its redshift and hence recession speed; this systematic increase of speed with distance is the evidence that the universe is expanding. (B1)
The formula $v \approx c\,\Delta\lambda/\lambda$ is the non-relativistic $v \ll c$ limit; for very distant galaxies the recession speed approaches $c$, so the simple formula breaks down and the redshift is better understood as the stretching of space itself. (R1)
$\Delta\lambda = 405.0 - 397.0 = +8.0\ \mathrm{nm}$。(A1)
波长增大（$\Delta\lambda > 0$），为红移，故星系正在远离。(A1)
用 $v \approx c\,\dfrac{\Delta\lambda}{\lambda}$，$\lambda = 397.0\ \mathrm{nm}$：(M1)
$$ v = (3.00\times10^{8})\times\frac{8.0}{397.0} \approx 6.0\times10^{6}\ \mathrm{m\,s^{-1}}. $$(A1)
以 $c$ 的分数表示：$\dfrac{v}{c} = \dfrac{8.0}{397.0} \approx 0.020$，约为 $c$ 的 $2\%$。(A1)
测量许多星系显示几乎全部红移，故几乎全部都在远离我们。(B1)
星系越远，红移越大、退行速度越快；速度随距离系统性增大正是宇宙膨胀的证据。(B1)
公式 $v \approx c\,\Delta\lambda/\lambda$ 是非相对论的 $v \ll c$ 极限；对极遥远星系退行速度接近 $c$，简单公式失效，红移更应理解为空间本身的拉伸。(R1)