PART I · PAPER 1 STYLE第一部分 · 第一卷风格Short structured · calculator · 30 marks短结构题 · 可用计算器 · 30 分
Short Structured Items短结构题
Show all working in the space below each question. Marks are awarded for correct method as well as final answers. Draw the standing-wave pattern before reaching for a formula, and read the wavelength off the picture. Give numerical answers to an appropriate number of significant figures.在每题下方空白处写出全部解题过程。方法分（method marks）与最终答案同等重要。套公式前先画出驻波图样，并从图中读出波长。数值答案保留适当的有效数字。
Q1MEDIUMPaper 1standing vs travelling wave驻波与行波对比[4 marks]
A standing wave is set up on a stretched string when an incident travelling wave reflects from a fixed end and superposes with itself.在拉紧的弦上，一列入射行波从固定端反射并与自身叠加，形成驻波。
(a)State the two conditions the two superposing waves must satisfy for a standing wave to form.写出两列叠加波要形成驻波必须满足的两个条件。[2]
(b)State two ways in which the resulting standing wave differs from a travelling wave.写出所形成的驻波与行波之间的两点不同。[2]
Q2MEDIUMPaper 1node and antinode geometry波节与波腹的几何[6 marks]
A standing wave on a string has adjacent nodes $0.18\ \mathrm{m}$ apart. The travelling waves that produced it move along the string at $144\ \mathrm{m\,s^{-1}}$.弦上一列驻波相邻波节相距 $0.18\ \mathrm{m}$。产生它的行波沿弦以 $144\ \mathrm{m\,s^{-1}}$ 传播。
(a)State the relationship between the node spacing and the wavelength, and hence find the wavelength.写出波节间距与波长的关系，并由此求波长。[2]
(b)Calculate the frequency of the wave.计算该波的频率。[2]
(c)State the distance from a node to the nearest antinode, and state how the phase of two points either side of a node compare.写出波节到最近波腹的距离，并说明波节两侧两点的相位关系。[2]
Q3HARDPaper 1string fixed both ends: harmonics两端固定弦的谐波[6 marks]
A wire of length $0.80\ \mathrm{m}$ is fixed at both ends. Transverse waves travel along it at $320\ \mathrm{m\,s^{-1}}$.一根长 $0.80\ \mathrm{m}$ 的弦两端固定。横波沿弦以 $320\ \mathrm{m\,s^{-1}}$ 传播。
(a)State the boundary condition at each end, and hence write the allowed wavelengths in terms of $L$ and the harmonic number $n$.写出两端的边界条件，并由此用 $L$ 与谐波次数 $n$ 表示允许的波长。[2]
(b)Calculate the fundamental frequency and the frequency of the third harmonic.计算基频与第三谐波的频率。[2]
(c)State the number of nodes and the number of antinodes present in the third-harmonic pattern.写出第三谐波图样中波节与波腹的数目。[2]
Q4HARDPaper 1open and closed pipes开管与闭管[6 marks]
An organ pipe closed at one end has length $0.50\ \mathrm{m}$. The speed of sound in air is $340\ \mathrm{m\,s^{-1}}$.一端闭口的风琴管长 $0.50\ \mathrm{m}$。空气中声速为 $340\ \mathrm{m\,s^{-1}}$。
(a)State the displacement boundary condition at the open end and at the closed end, and hence calculate the fundamental frequency.写出开口端与闭口端的位移边界条件，并由此计算基频。[2]
(b)State and calculate the frequency of the next harmonic the closed pipe can produce above the fundamental.写出并计算闭管在基频之上能产生的下一个谐波频率。[2]
(c)Calculate the fundamental frequency if instead the pipe were open at both ends, and state in words how it compares with the closed-pipe value in (a).计算若该管改为两端开口时的基频，并用文字说明它与 (a) 中闭管值的关系。[2]
Q5HARDPaper 1resonance and damping共振与阻尼[8 marks]
A mass on a spring has a natural frequency $f_{0}$. It is driven by a periodic force whose frequency $f_{d}$ can be varied. The steady-state amplitude is recorded against $f_{d}$ to give a resonance curve.弹簧上的物块有固有频率 $f_{0}$。它被一个频率 $f_{d}$ 可变的周期力驱动。记录稳态振幅对 $f_{d}$ 的关系，得到共振曲线。
(a)Define resonance and state the condition on $f_{d}$ for it to occur.给出共振的定义，并写出发生共振时对 $f_{d}$ 的条件。[2]
(b)State the frequency at which the system oscillates in the steady state when $f_{d} \neq f_{0}$.写出当 $f_{d} \neq f_{0}$ 时系统在稳态下振动的频率。[2]
(c)Describe how the resonance curve changes as the damping of the system is increased.描述当系统阻尼增大时共振曲线如何变化。[2]
(d)Give one example where resonance is useful and one where it is a hazard that engineers design against.举出共振有用的一个例子，以及工程师需加以防范的有害的一个例子。[2]
PART II · PAPER 1B / DATA ANALYSIS第二部分 · 第一卷 B / 数据分析Graphs · data · uncertainties · 24 marks图像 · 数据 · 不确定度 · 24 分
Graph and Data Questions图像与数据题
These items reward correct linearisation, careful reading of gradients, and proper handling of uncertainties. Quote uncertainties to one significant figure and round the value to match.这些题考查正确的线性化、对斜率的细致读取以及对不确定度的正确处理。不确定度保留 1 位有效数字，并使数值的末位与之对齐。
Q6HARDPaper 1Blinearised f vs 1/L for a pipe管的 f 对 1/L 线性化[12 marks]
A student investigates the fundamental frequency $f$ of an air column open at both ends as its length $L$ is varied. For each length the fundamental frequency is measured:学生研究两端开口空气柱的基频 $f$ 随其长度 $L$ 改变的情况。对每个长度测量其基频：
$L\ /\ \mathrm{m}$
$1.00$
$0.50$
$0.25$
$0.20$
$1/L\ /\ \mathrm{m^{-1}}$
$1.00$
$2.00$
$4.00$
$5.00$
$f\ /\ \mathrm{Hz}$
$170$
$340$
$680$
$850$
(a)Starting from the harmonic relation for an open–open pipe, show that a graph of $f$ against $1/L$ should be a straight line through the origin, and state what its gradient represents.从开–开管的谐波关系出发，证明 $f$ 对 $1/L$ 的图应为过原点的直线，并说明其斜率代表什么。[3]
(b)Calculate the gradient of the line, and hence determine the speed of sound implied by the data.计算该直线的斜率，并由此求数据所给出的声速。[3]
(c)The frequency at $L = 0.20\ \mathrm{m}$ has an absolute uncertainty of $\pm 10\ \mathrm{Hz}$. Calculate the percentage uncertainty in this frequency.$L = 0.20\ \mathrm{m}$ 处的频率绝对不确定度为 $\pm 10\ \mathrm{Hz}$。计算该频率的百分比不确定度。[2]
(d)The same air column is now stopped at one end (closed pipe). State, with reasoning, by what factor the fundamental frequency at each length would change.现将同一空气柱一端封闭（闭管）。说明并解释每个长度下基频将改变的倍数。[2]
(e)State one source of systematic error in this experiment that would make the measured frequencies consistently differ from the ideal values.写出该实验中会使所测频率系统性地偏离理想值的一个系统误差来源。[2]
Q7HARDPaper 1Bharmonic series: f vs n graph谐波列：f 对 n 图[12 marks]
A string of length $0.60\ \mathrm{m}$ is fixed at both ends. The string is driven at a series of resonant frequencies and the harmonic number $n$ of each is identified:一根长 $0.60\ \mathrm{m}$ 的弦两端固定。该弦被驱动在一系列共振频率上，并辨认出每个的谐波次数 $n$：
$n$
$1$
$2$
$3$
$4$
$5$
$f\ /\ \mathrm{Hz}$
$200$
$400$
$600$
$800$
$1000$
(a)Explain why a graph of resonant frequency $f$ against harmonic number $n$ is a straight line through the origin.解释为何共振频率 $f$ 对谐波次数 $n$ 的图是一条过原点的直线。[2]
(b)Calculate the gradient of the line and state what it represents physically.计算该直线的斜率，并说明它的物理意义。[3]
(c)Use the gradient to determine the speed of the transverse waves on the string.用斜率求弦上横波的波速。[3]
(d)A second string, identical except that it is held under four times the tension, replaces the first. Given that the wave speed on a string is $v = \sqrt{T/\mu}$, state the new fundamental frequency and explain how the $f$–$n$ graph changes.用第二根弦替换第一根，它除张力为四倍外完全相同。已知弦上波速 $v = \sqrt{T/\mu}$，写出新的基频并解释 $f$–$n$ 图如何变化。[4]
PART III · PAPER 2 STYLE第三部分 · 第二卷风格Extended structured · calculator · 26 marks长结构题 · 可用计算器 · 26 分
Extended Structured Problems长结构问题
Set up each problem with a clear sketch of the standing-wave pattern and labelled nodes and antinodes. Method marks dominate the longer items; carry intermediate values to extra figures and round only the final answer.每题先画清晰的驻波图样并标注波节与波腹。长题中方法分占比最大；中间值多保留几位，仅在最终答案处取舍有效数字。
Q8HARDPaper 2guitar string: superposition to harmonics吉他弦：从叠加到谐波[14 marks]
A guitar string of length $0.65\ \mathrm{m}$ is fixed at both ends. Transverse waves travel along it at $260\ \mathrm{m\,s^{-1}}$.一根长 $0.65\ \mathrm{m}$ 的吉他弦两端固定。横波沿弦以 $260\ \mathrm{m\,s^{-1}}$ 传播。
(a)Two identical waves $y_{1} = A\sin(kx - \omega t)$ and $y_{2} = A\sin(kx + \omega t)$ travel in opposite directions along the string. Show that their superposition can be written $y = 2A\sin(kx)\cos(\omega t)$, and state which factor fixes the amplitude at each point.两列相同的波 $y_{1} = A\sin(kx - \omega t)$ 与 $y_{2} = A\sin(kx + \omega t)$ 沿弦反向传播。证明它们的叠加可写为 $y = 2A\sin(kx)\cos(\omega t)$，并指出哪个因子决定各点的振幅。[3]
(b)Using the result of (a), show that the permanently stationary points (nodes) are spaced half a wavelength apart.利用 (a) 的结果，证明始终静止的点（波节）间隔为半个波长。[3]
(c)Calculate the fundamental frequency of the string and the wavelength of the third harmonic.计算弦的基频与第三谐波的波长。[3]
(d)Explain, in terms of energy, why a standing wave on the string transfers no net energy along its length.从能量角度解释为何弦上的驻波沿其长度不传递净能量。[3]
(e)The player presses a fret so that the vibrating length is reduced to $0.49\ \mathrm{m}$ at the same wave speed. Calculate the new fundamental frequency.演奏者按品使振动长度在波速不变下减为 $0.49\ \mathrm{m}$。计算新的基频。[2]
A vertical glass tube is partly filled with water, leaving an air column closed at the bottom (water surface) and open at the top. A tuning fork of frequency $480\ \mathrm{Hz}$ is held over the open end and the water level is slowly lowered. The speed of sound in air is $340\ \mathrm{m\,s^{-1}}$.一根竖直玻璃管部分注水，留下底部（水面）闭口、顶部开口的空气柱。一个频率 $480\ \mathrm{Hz}$ 的音叉置于开口端上方，水位缓慢下降。空气中声速为 $340\ \mathrm{m\,s^{-1}}$。
(a)Explain why a loud sound is heard only at certain lengths of the air column.解释为何只在空气柱的某些长度处才听到响亮的声音。[2]
(b)Calculate the shortest air-column length at which resonance occurs.计算发生共振的最短空气柱长度。[3]
(c)Calculate the next two longer air-column lengths at which resonance is heard, and state the distance between successive resonant lengths.计算接下来两个能听到共振的较长空气柱长度，并写出相邻共振长度之间的距离。[4]
(d)As the water level is lowered further, the resonances become quieter and harder to locate precisely. State the role of damping here, and explain whether light or heavy damping makes a resonance easier to pinpoint.随着水位继续下降，共振变得更弱、更难精确定位。说明阻尼在此的作用，并解释轻阻尼还是重阻尼更易精确定位共振。[3]