Companion to the IB-Style Practice SetIB 风格练习题的解析配套
Syllabus C2.1 to C2.6考纲 C2.1 至 C2.6PHYSICS HL
A cork bobs as a water wave of amplitude $4.0\ \mathrm{cm}$ passes. (a) net horizontal displacement and what is transported; (b) factor by which power changes when amplitude is doubled.软木塞随一列振幅 $4.0\ \mathrm{cm}$ 的水波起伏。(a) 净水平位移与所传输之物；(b) 振幅加倍时功率的变化倍数。
In a travelling wave each particle of the medium oscillates about a fixed equilibrium and does not migrate with the wave, so the net horizontal displacement of the cork is essentially zero. (A1)
What crosses the pond is energy (and momentum), not water: this transfer of energy without net transfer of matter is the defining property of a travelling wave. (A1)
The power carried by a wave is proportional to the square of the amplitude, $P \propto A^{2}$. (M1)
Doubling $A$ multiplies the power by $2^{2} = 4$, so the wave carries four times the power. (A1)
行波中介质的每个质点绕固定平衡位置振动，不随波迁移，故软木塞的净水平位移本质上为零。(A1)
横跨池塘的是能量（与动量），不是水：这种传递能量而无物质净传递正是行波的定义性质。(A1)
波所携带的功率与振幅平方成正比，$P \propto A^{2}$。(M1)
$A$ 加倍使功率乘以 $2^{2} = 4$，故波携带的功率为四倍。(A1)
A loudspeaker cone vibrates along the line toward a listener. (a) classify the sound; (b) describe air molecules at compression and rarefaction; (c) a second wave can be polarised, so classify it and explain.扬声器纸盆沿指向听者的方向振动。(a) 对声波分类；(b) 描述压缩与稀疏处的空气分子；(c) 另一列波可被偏振，分类并解释。
The air molecules oscillate along the same line that the energy travels, that is, the particle oscillation is parallel to the direction of energy transfer. (M1)
A wave with parallel oscillation is longitudinal, so the sound is a longitudinal wave. (A1)
At a compression the air molecules are bunched together, giving locally high pressure and density. (A1)
At a rarefaction the molecules are spread apart, giving locally low pressure and density; each molecule merely oscillates about its own mean position. (A1)
Polarisation restricts the oscillation to a single plane, which is only possible when the oscillation is perpendicular to the direction of travel. (R1)
Only a transverse wave has perpendicular oscillation, so a wave that can be polarised must be transverse. (A1)
空气分子沿能量传播的同一方向振动，即质点振动方向与能量传递方向平行。(M1)
平行振动的波是纵波，故该声波为纵波。(A1)
压缩处空气分子密集聚集，局部高压、高密度。(A1)
稀疏处分子彼此散开，局部低压、低密度；每个分子只绕自身平均位置振动。(A1)
偏振将振动限制在单一平面内，只有当振动垂直于传播方向时才可能。(R1)
只有横波具有垂直振动，故可被偏振的波必为横波。(A1)
Sound in air: $f = 512\ \mathrm{Hz}$, $v = 340\ \mathrm{m\,s^{-1}}$. (a) wavelength; (b) period; (c) what happens to $f$ and $\lambda$ on entering water where the sound is faster.空气中声波：$f = 512\ \mathrm{Hz}$、$v = 340\ \mathrm{m\,s^{-1}}$。(a) 波长；(b) 周期；(c) 进入声速更快的水中后 $f$ 与 $\lambda$ 如何变化。
Rearrange the wave equation $v = f\lambda$ for $\lambda$: (M1)
$$ \lambda = \frac{v}{f} = \frac{340}{512} \approx 0.664\ \mathrm{m} \approx 0.66\ \mathrm{m}. $$(A1)
Period is the reciprocal of frequency, $T = 1/f$: (M1)
$$ T = \frac{1}{512} \approx 1.95\times10^{-3}\ \mathrm{s} \approx 2.0\times10^{-3}\ \mathrm{s}. $$(A1)
Frequency is fixed by the source and does not change when the medium changes, so $f$ stays at $512\ \mathrm{Hz}$. (A1)
From $v = f\lambda$ with $f$ constant, an increase in $v$ forces a proportional increase in $\lambda$, so the wavelength increases. (R1)
由波动方程 $v = f\lambda$ 解出 $\lambda$：(M1)
$$ \lambda = \frac{v}{f} = \frac{340}{512} \approx 0.664\ \mathrm{m} \approx 0.66\ \mathrm{m}. $$(A1)
周期是频率的倒数，$T = 1/f$：(M1)
$$ T = \frac{1}{512} \approx 1.95\times10^{-3}\ \mathrm{s} \approx 2.0\times10^{-3}\ \mathrm{s}. $$(A1)
频率由波源决定，换介质时不变，故 $f$ 保持 $512\ \mathrm{Hz}$。(A1)
由 $v = f\lambda$ 且 $f$ 不变，$v$ 增大使 $\lambda$ 成比例增大，故波长增大。(R1)
Ripple tank: circular wavefronts from an $8.0\ \mathrm{Hz}$ point source, adjacent crest wavefronts $2.5\ \mathrm{cm}$ apart. (a) wavelength; (b) speed; (c) ray-wavefront angle and what a ray is; (d) how wavefronts and rays change far from the source.水波槽：$8.0\ \mathrm{Hz}$ 点波源的圆形波前，相邻波峰波前相距 $2.5\ \mathrm{cm}$。(a) 波长；(b) 波速；(c) 射线与波前夹角及射线含义；(d) 远离波源处波前与射线如何变化。
Adjacent crest wavefronts join points one full cycle apart, so they are separated by exactly one wavelength: $\lambda = 2.5\ \mathrm{cm} = 0.025\ \mathrm{m}$. (A1)
Apply the wave equation $v = f\lambda$: (M1)
$$ v = f\lambda = 8.0 \times 0.025 = 0.20\ \mathrm{m\,s^{-1}}. $$(A1)
A ray shows the direction of energy propagation and is by definition perpendicular to the wavefront, so the angle between them is $90^{\circ}$. (A1)
Close to a point source the wavefronts are tightly curved circles and the rays fan out radially. (A1)
Far from the source only a small portion of each circle is seen, so the wavefronts become essentially straight (plane) and the rays become essentially parallel. (A1)
相邻波峰波前连接相差一个完整周期的点，故相距恰好一个波长：$\lambda = 2.5\ \mathrm{cm} = 0.025\ \mathrm{m}$。(A1)
用波动方程 $v = f\lambda$：(M1)
$$ v = f\lambda = 8.0 \times 0.025 = 0.20\ \mathrm{m\,s^{-1}}. $$(A1)
射线表示能量传播方向，按定义垂直于波前，故两者夹角为 $90^{\circ}$。(A1)
靠近点波源处波前是弯曲很紧的圆，射线呈放射状散开。(A1)
远离波源处只看到每个圆的一小段，故波前趋于直线（平面），射线趋于平行。(A1)
EM spectrum from radio to gamma. (a) list seven bands by increasing frequency; (b) frequency of red light $\lambda = 6.5\times10^{-7}\ \mathrm{m}$; (c) compare X-ray and red-light vacuum speed, plus two shared EM properties.电磁波谱从无线电到伽马。(a) 按频率递增列出七个波段；(b) 红光 $\lambda = 6.5\times10^{-7}\ \mathrm{m}$ 的频率；(c) X 射线与红光真空速度比较，及电磁波两条共有性质。
Frequency increases (and wavelength decreases) in the order: radio, microwave, infrared, visible, ultraviolet, X-ray, gamma. (A1 for the correct ends radio and gamma, A1 for the full correct order between them)
In vacuum every EM wave travels at $c$, so use $c = f\lambda$ rearranged for $f$: (M1)
$$ f = \frac{c}{\lambda} = \frac{3.00\times10^{8}}{6.5\times10^{-7}}. $$(M1 for substitution)
$$ f \approx 4.6\times10^{14}\ \mathrm{Hz}. $$(A1)
X-rays and red light travel at exactly the same vacuum speed $c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$; they differ only in frequency and wavelength. (A1)
Two further shared properties: all EM waves are transverse (A1) and all can be polarised. (A1)
频率递增（波长递减）的顺序为：无线电、微波、红外、可见光、紫外、X 射线、伽马。(两端无线电与伽马正确得 A1，中间完整顺序正确得 A1)
真空中每列电磁波都以 $c$ 传播，故用 $c = f\lambda$ 解出 $f$：(M1)
$$ f = \frac{c}{\lambda} = \frac{3.00\times10^{8}}{6.5\times10^{-7}}. $$(代入得 M1)
$$ f \approx 4.6\times10^{14}\ \mathrm{Hz}. $$(A1)
X 射线与红光在真空中速度完全相同，$c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$；二者只在频率与波长上不同。(A1)
另外两条共有性质：所有电磁波都是横波 (A1)，且都可被偏振。(A1)
One wave, two graphs: Graph 1 displacement-distance (repeat $0.80\ \mathrm{m}$), Graph 2 displacement-time (repeat $0.025\ \mathrm{s}$), peak displacement $3.0\ \mathrm{cm}$. (a) which graph gives $\lambda$, which gives $T$; (b) amplitude and wavelength; (c) frequency; (d) speed; (e) up or down for a particle on the leading edge of a crest; (f) error in reading $0.80$ as the period.同一列波，两张图：图 1 位移-距离（重复 $0.80\ \mathrm{m}$），图 2 位移-时间（重复 $0.025\ \mathrm{s}$），峰值位移 $3.0\ \mathrm{cm}$。(a) 哪张图给 $\lambda$、哪张给 $T$；(b) 振幅与波长；(c) 频率；(d) 波速；(e) 波峰前缘质点向上还是向下；(f) 把 $0.80$ 读为周期的错误。
The wavelength is a distance, so it is read from Graph 1, whose horizontal axis is distance $x$. (A1)
The period is a time, so it is read from Graph 2, whose horizontal axis is time $t$. (A1)
The amplitude is the peak displacement, the same on both graphs: $A = 3.0\ \mathrm{cm} = 0.030\ \mathrm{m}$. (A1)
The wavelength is the repeat length on the distance graph: $\lambda = 0.80\ \mathrm{m}$. (A1)
The period from the time graph is $T = 0.025\ \mathrm{s}$, so $f = 1/T$: (M1)
$$ f = \frac{1}{0.025} = 40\ \mathrm{Hz}. $$(A1)
Combine the two readings through $v = f\lambda$: (M1)
$$ v = f\lambda = 40 \times 0.80 = 32\ \mathrm{m\,s^{-1}}. $$(A1)
Shift the whole snapshot a small amount in the direction of travel (increasing $x$). The crest moves toward the particle on its leading edge, so the displacement that arrives at the particle's position is larger than now. (M1)
The particle therefore moves up next. (A1)
The value $0.80$ is read off the distance axis of Graph 1, so it is the wavelength $\lambda = 0.80\ \mathrm{m}$, not the period. (A1)
The period must be read off a time axis (Graph 2), where the repeat is $0.025\ \mathrm{s}$; recording a distance as a time mixes up the two graphs. (R1)
波长是距离量，故从横轴为距离 $x$ 的图 1 读取。(A1)
周期是时间量，故从横轴为时间 $t$ 的图 2 读取。(A1)
振幅是峰值位移，两图相同：$A = 3.0\ \mathrm{cm} = 0.030\ \mathrm{m}$。(A1)
波长是距离图上的重复长度：$\lambda = 0.80\ \mathrm{m}$。(A1)
由时间图得周期 $T = 0.025\ \mathrm{s}$，故 $f = 1/T$：(M1)
$$ f = \frac{1}{0.025} = 40\ \mathrm{Hz}. $$(A1)
用 $v = f\lambda$ 合并两读数：(M1)
$$ v = f\lambda = 40 \times 0.80 = 32\ \mathrm{m\,s^{-1}}. $$(A1)
将整张快照沿传播方向（$x$ 增大）微移一点。波峰向其前缘的质点靠近，故到达该质点位置的位移比现在更大。(M1)
因此该质点下一刻向上运动。(A1)
$0.80$ 读自图 1 的距离轴，故它是波长 $\lambda = 0.80\ \mathrm{m}$，不是周期。(A1)
周期必须从时间轴（图 2）读取，那里的重复为 $0.025\ \mathrm{s}$；把距离记作时间就是混淆了两张图。(R1)
Ripple tank: $5$ wavelengths span $(12.0 \pm 0.2)\ \mathrm{cm}$; dipper does $20$ oscillations in $8.0\ \mathrm{s}$. (a) wavelength and why dividing by $5$ helps; (b) period and frequency; (c) speed; (d) percentage uncertainty in the span.水波槽：$5$ 个波长跨 $(12.0 \pm 0.2)\ \mathrm{cm}$；振子 $8.0\ \mathrm{s}$ 内振动 $20$ 次。(a) 波长及为何除以 $5$ 有帮助；(b) 周期与频率；(c) 波速；(d) 跨距的百分比不确定度。
The span covers $5$ complete wavelengths, so divide the total by $5$: (M1)
$$ \lambda = \frac{12.0\ \mathrm{cm}}{5} = 2.40\ \mathrm{cm} = 0.024\ \mathrm{m}. $$(A1)
Measuring across many wavelengths and dividing spreads the fixed ruler-reading uncertainty over a larger length, reducing the percentage uncertainty in $\lambda$. (R1)
The period is the time for one oscillation: $T = 8.0/20 = 0.40\ \mathrm{s}$, hence $f = 1/T = 2.5\ \mathrm{Hz}$. (M1·A1)
Apply $v = f\lambda$: (M1)
$$ v = f\lambda = 2.5 \times 0.024 = 0.060\ \mathrm{m\,s^{-1}}. $$(A1)
The span is $12.0\ \mathrm{cm}$ with absolute uncertainty $\pm 0.2\ \mathrm{cm}$: (M1)
$$ \frac{0.2}{12.0}\times 100\% \approx 1.67\% \approx 2\%. $$(A1)
Dividing the span by the exact count $5$ does not change the percentage uncertainty, so the wavelength carries the same $\approx 2\%$. (B1)
跨距包含 $5$ 个完整波长，故将总长除以 $5$：(M1)
$$ \lambda = \frac{12.0\ \mathrm{cm}}{5} = 2.40\ \mathrm{cm} = 0.024\ \mathrm{m}. $$(A1)
跨越多个波长再相除，把固定的尺读不确定度摊到更长的距离上，从而降低 $\lambda$ 的百分比不确定度。(R1)
周期是一次振动的时间：$T = 8.0/20 = 0.40\ \mathrm{s}$，故 $f = 1/T = 2.5\ \mathrm{Hz}$。(M1·A1)
用 $v = f\lambda$：(M1)
$$ v = f\lambda = 2.5 \times 0.024 = 0.060\ \mathrm{m\,s^{-1}}. $$(A1)
跨距为 $12.0\ \mathrm{cm}$，绝对不确定度 $\pm 0.2\ \mathrm{cm}$：(M1)
$$ \frac{0.2}{12.0}\times 100\% \approx 1.67\% \approx 2\%. $$(A1)
将跨距除以精确计数 $5$ 不改变百分比不确定度，故波长同样带 $\approx 2\%$。(B1)
Sound at $\approx 340\ \mathrm{m\,s^{-1}}$, light at $c$. (a) two properties from being mechanical longitudinal; (b) wavelength of a $256\ \mathrm{Hz}$ note; (c) distance to a lightning strike heard $6.0\ \mathrm{s}$ after seen; (d) distance to a cliff giving an echo after $1.5\ \mathrm{s}$; (e) why lightning is seen before thunder is heard.声速 $\approx 340\ \mathrm{m\,s^{-1}}$，光速 $c$。(a) 因机械纵波而有的两条性质；(b) $256\ \mathrm{Hz}$ 音的波长；(c) 看到后 $6.0\ \mathrm{s}$ 听到雷声的雷击距离；(d) $1.5\ \mathrm{s}$ 后听到回声的峭壁距离；(e) 为何先见闪电后闻雷声。
It needs a material medium to travel, so it cannot pass through a vacuum. (A1)
Its particles oscillate parallel to the direction of energy transfer, forming compressions and rarefactions, and it cannot be polarised. (A1)
Use the wave equation $v = f\lambda$ rearranged for $\lambda$: (M1)
$$ \lambda = \frac{v}{f} = \frac{340}{256}. $$(M1 for substitution)
$$ \lambda \approx 1.33\ \mathrm{m} \approx 1.3\ \mathrm{m}. $$(A1)
Assume the light reaches the observer effectively instantly (the light travel time is negligible because $c$ is enormous), so the $6.0\ \mathrm{s}$ delay is the time for the sound to travel. (R1)
Distance $= v_{\text{sound}} \times t = 340 \times 6.0$: (M1)
$$ d = 2040\ \mathrm{m} \approx 2.0\ \mathrm{km}. $$(A1)
The echo travels to the cliff and back, a total path of $2d$, in $1.5\ \mathrm{s}$: $2d = v_{\text{sound}} \times t$. (M1)
$$ d = \frac{v_{\text{sound}}\, t}{2} = \frac{340 \times 1.5}{2}. $$(M1 for halving the round trip)
$$ d = 255\ \mathrm{m}. $$(A1)
Light is an electromagnetic wave; it needs no medium and travels at $c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$, so it arrives almost instantly. (M1)
Sound is a mechanical wave that needs the air and travels at only about $340\ \mathrm{m\,s^{-1}}$. (A1)
Because the speeds differ by a factor of nearly a million, the sound lags far behind the light, and the delay (about $3\ \mathrm{s}$ per kilometre) measures the distance. (R1)
它需要物质介质才能传播，故不能穿过真空。(A1)
其质点平行于能量传递方向振动，形成压缩与稀疏，且不能被偏振。(A1)
由波动方程 $v = f\lambda$ 解出 $\lambda$：(M1)
$$ \lambda = \frac{v}{f} = \frac{340}{256}. $$(代入得 M1)
$$ \lambda \approx 1.33\ \mathrm{m} \approx 1.3\ \mathrm{m}. $$(A1)
假设光实际上瞬间到达观察者（因 $c$ 极大，光传播时间可忽略），故 $6.0\ \mathrm{s}$ 延迟即声音传播所用时间。(R1)
距离 $= v_{\text{sound}} \times t = 340 \times 6.0$：(M1)
$$ d = 2040\ \mathrm{m} \approx 2.0\ \mathrm{km}. $$(A1)
回声到峭壁再返回，总路程为 $2d$，用时 $1.5\ \mathrm{s}$：$2d = v_{\text{sound}} \times t$。(M1)
$$ d = \frac{v_{\text{sound}}\, t}{2} = \frac{340 \times 1.5}{2}. $$(往返折半得 M1)
$$ d = 255\ \mathrm{m}. $$(A1)
光是电磁波；无需介质，以 $c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$ 传播，故几乎瞬间到达。(M1)
声是机械波，需要空气，速度仅约 $340\ \mathrm{m\,s^{-1}}$。(A1)
由于两者速度相差近百万倍，声音远落后于光，且延迟（每千米约 $3\ \mathrm{s}$）可量出距离。(R1)
All EM waves obey $c = f\lambda$ in vacuum. (a) frequency of a $0.12\ \mathrm{m}$ radar microwave; (b) its wavelength in a block where $v = 2.00\times10^{8}\ \mathrm{m\,s^{-1}}$; (c) wavelength of a $98\ \mathrm{MHz}$ FM wave and its band; (d) same vacuum speed plus two shared properties; (e) why sound is not in the EM spectrum.所有电磁波在真空中遵循 $c = f\lambda$。(a) $0.12\ \mathrm{m}$ 雷达微波的频率；(b) 在 $v = 2.00\times10^{8}\ \mathrm{m\,s^{-1}}$ 介质块中的波长；(c) $98\ \mathrm{MHz}$ FM 波的波长与波段；(d) 真空速度相同及两条共有性质；(e) 为何声音不属电磁波谱。
In vacuum the microwave travels at $c$, so rearrange $c = f\lambda$ for $f$: (M1)
$$ f = \frac{c}{\lambda} = \frac{3.00\times10^{8}}{0.12}. $$(M1 for substitution)
$$ f = 2.5\times10^{9}\ \mathrm{Hz}. $$(A1)
The source sets the frequency, so $f = 2.5\times10^{9}\ \mathrm{Hz}$ is unchanged inside the block; only $v$ and $\lambda$ change. (R1)
Apply $\lambda = v/f$ with the reduced speed: (M1)
$$ \lambda = \frac{v}{f} = \frac{2.00\times10^{8}}{2.5\times10^{9}} = 0.080\ \mathrm{m}. $$(A1)
With $f = 98\ \mathrm{MHz} = 98\times10^{6}\ \mathrm{Hz}$, use $\lambda = c/f$: (M1)
$$ \lambda = \frac{3.00\times10^{8}}{98\times10^{6}} \approx 3.06\ \mathrm{m} \approx 3.1\ \mathrm{m}. $$(A1)
A wavelength of a few metres lies in the radio band of the spectrum. (A1)
Yes: both the radar microwave and the FM radio wave are electromagnetic, so both travel at exactly $c$ in vacuum. (A1)
Two further shared properties: both are transverse (A1) and both can be polarised. (A1)
Sound is a mechanical longitudinal wave that requires a material medium and travels far slower than $c$. (A1)
The electromagnetic spectrum consists only of transverse oscillations of electric and magnetic fields that propagate through a vacuum at $c$, so sound cannot belong to it. (R1)
真空中微波以 $c$ 传播，故由 $c = f\lambda$ 解出 $f$：(M1)
$$ f = \frac{c}{\lambda} = \frac{3.00\times10^{8}}{0.12}. $$(代入得 M1)
$$ f = 2.5\times10^{9}\ \mathrm{Hz}. $$(A1)
波源决定频率，故进入介质块后 $f = 2.5\times10^{9}\ \mathrm{Hz}$ 不变；只有 $v$ 与 $\lambda$ 改变。(R1)
用减小的速度代入 $\lambda = v/f$：(M1)
$$ \lambda = \frac{v}{f} = \frac{2.00\times10^{8}}{2.5\times10^{9}} = 0.080\ \mathrm{m}. $$(A1)
取 $f = 98\ \mathrm{MHz} = 98\times10^{6}\ \mathrm{Hz}$，用 $\lambda = c/f$：(M1)
$$ \lambda = \frac{3.00\times10^{8}}{98\times10^{6}} \approx 3.06\ \mathrm{m} \approx 3.1\ \mathrm{m}. $$(A1)
几米的波长落在波谱的无线电波段。(A1)
是：雷达微波与 FM 无线电波都是电磁波，故在真空中都恰以 $c$ 传播。(A1)
另外两条共有性质：两者都是横波 (A1)，且都可被偏振。(A1)
声音是机械纵波，需要物质介质，且速度远小于 $c$。(A1)
电磁波谱只由在真空中以 $c$ 传播的电场与磁场的横向振荡组成，故声音不可能属于它。(R1)