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. Convert frequency to angular frequency before using any SHM equation, and take the positive root for $\omega$. 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。使用任何简谐方程前先把频率换算为角频率，且 $\omega$ 取正根。数值答案保留适当的有效数字。
A small object oscillates so that its acceleration $a$ and displacement $x$ (in SI units) are related by $a = -100\,x$.一个小物体振动，其加速度 $a$ 与位移 $x$（SI 单位）满足 $a = -100\,x$。
(a)State the two features the relation must show for the motion to be simple harmonic, and confirm that this relation has both.写出该关系须具备哪两个特征才表明运动为简谐运动，并确认此关系两者皆有。[2]
(b)Determine the angular frequency, period and frequency of the oscillation.求振动的角频率、周期与频率。[3]
A simple pendulum completes $24$ full swings in $30\ \mathrm{s}$ at small amplitude.一个单摆在小振幅下于 $30\ \mathrm{s}$ 内完成 $24$ 次完整摆动。
(a)Calculate the period, the frequency and the angular frequency of the pendulum.计算该单摆的周期、频率与角频率。[3]
(b)The amplitude of the swing is then doubled, the motion remaining simple harmonic. State and justify the effect on the period.随后摆幅加倍，运动仍为简谐。写出对周期的影响并说明理由。[2]
Q3MEDIUMPaper 1mass-spring period弹簧振子周期[6 marks]
A $0.30\ \mathrm{kg}$ mass on a spring of spring constant $k = 120\ \mathrm{N\,m^{-1}}$ oscillates horizontally on a frictionless surface.一个 $0.30\ \mathrm{kg}$ 的质量挂在弹簧常数 $k = 120\ \mathrm{N\,m^{-1}}$ 的弹簧上，在无摩擦水平面上振动。
(a)Calculate the period of oscillation.计算振动周期。[2]
(b)The mass is replaced by one four times heavier on the same spring. Determine the new period.在同一弹簧上换为四倍重的质量。求新周期。[2]
(c)State how the period would change if a stiffer spring (larger $k$) were used with the original mass, and explain why.说明若用原质量配更硬的弹簧（$k$ 更大）周期将如何变化，并解释原因。[2]
A simple pendulum is to be built with a period of exactly $2.0\ \mathrm{s}$ on Earth, where $g = 9.81\ \mathrm{m\,s^{-2}}$.要在地球（$g = 9.81\ \mathrm{m\,s^{-2}}$）上制作一个周期恰为 $2.0\ \mathrm{s}$ 的单摆。
(a)Calculate the required length of the pendulum, and state one assumption needed for the period formula to apply.计算所需摆长，并写出周期公式适用所需的一个假设。[3]
(b)The same pendulum is taken to the Moon, where $g$ is about one sixth of its Earth value. Determine, with reasoning, the factor by which its period changes.把同一单摆带到月球，那里 $g$ 约为地球值的六分之一。求其周期变化的倍数并说明推理。[3]
Q5HARDPaper 1HL ONLY$v_{\max}$, $a_{\max}$ and $v(x)$$v_{\max}$、$a_{\max}$ 与 $v(x)$[8 marks]
A particle performs simple harmonic motion with amplitude $x_{0} = 0.050\ \mathrm{m}$ and angular frequency $\omega = 20\ \mathrm{rad\,s^{-1}}$.一质点作简谐运动，振幅 $x_{0} = 0.050\ \mathrm{m}$、角频率 $\omega = 20\ \mathrm{rad\,s^{-1}}$。
(a)Calculate the maximum speed of the particle and state where in the motion it occurs.计算质点的最大速率，并说明出现在运动的何处。[2]
(b)Calculate the maximum magnitude of the acceleration and state where it occurs.计算加速度的最大值及其出现位置。[2]
(c)Using $v = \pm\,\omega\sqrt{x_{0}^{2} - x^{2}}$, calculate the speed when the displacement is $x = 0.030\ \mathrm{m}$, and explain why this is not $60\%$ of the maximum speed.用 $v = \pm\,\omega\sqrt{x_{0}^{2} - x^{2}}$ 计算位移为 $x = 0.030\ \mathrm{m}$ 时的速率，并解释为何它不是最大速率的 $60\%$。[4]
PART II · PAPER 1B / DATA ANALYSIS第二部分 · 第一卷 B / 数据分析Graphs · data · uncertainties · 20 marks图像 · 数据 · 不确定度 · 20 分
Graph and Data Questions图像与数据题
These items reward careful linearising, reading of gradients, and correct handling of uncertainties. Quote uncertainties to one significant figure and round the value to match.这些题考查细致的线性化、斜率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 1B$T^{2}$ vs $L$ pendulum + uncertainty$T^{2}$-$L$ 单摆图与不确定度[11 marks]
A student measures the period $T$ of a simple pendulum for several lengths $L$ and tabulates $T^{2}$ against $L$:学生测量单摆在若干摆长 $L$ 下的周期 $T$，并将 $T^{2}$ 对 $L$ 列表：
$L\ /\ \mathrm{m}$
$0.200$
$0.400$
$0.600$
$0.800$
$1.000$
$T^{2}\ /\ \mathrm{s^{2}}$
$0.805$
$1.610$
$2.415$
$3.219$
$4.024$
(a)Starting from the period formula for a simple pendulum, show that a graph of $T^{2}$ against $L$ should be a straight line through the origin, and state what the gradient represents.从单摆周期公式出发，证明 $T^{2}$ 对 $L$ 的图应为过原点的直线，并说明斜率代表什么。[3]
(b)Calculate the gradient of the line, and hence determine a value for the acceleration of free fall $g$.计算该直线的斜率，由此求自由落体加速度 $g$ 的一个值。[3]
(c)At $L = 1.000\ \mathrm{m}$ the period $T = 2.01\ \mathrm{s}$ has an absolute uncertainty of $\pm 0.02\ \mathrm{s}$. Calculate the percentage uncertainty in $T$ and hence the percentage uncertainty in $T^{2}$ at this point.在 $L = 1.000\ \mathrm{m}$ 处周期 $T = 2.01\ \mathrm{s}$，绝对不确定度为 $\pm 0.02\ \mathrm{s}$。计算 $T$ 的百分比不确定度，进而求该点 $T^{2}$ 的百分比不确定度。[2]
(d)State one source of systematic error that would make the line miss the origin, and state in which direction (above or below) the intercept would lie if the measured lengths were all too short.写出一个会使直线不过原点的系统误差来源，并说明若测得的摆长全部偏短，截距会位于（上方还是下方）哪一侧。[3]
The displacement-time graph of an oscillator is a cosine curve that starts at its positive maximum. The peak displacement is $0.060\ \mathrm{m}$ and one complete cycle takes $0.40\ \mathrm{s}$.某振子的位移-时间图为从正最大值开始的余弦曲线。峰位移为 $0.060\ \mathrm{m}$，一个完整周期为 $0.40\ \mathrm{s}$。
(a)State the amplitude and the period, and calculate the angular frequency.写出振幅与周期，并计算角频率。[2]
(b)Using the relation between the slope of the $x$-$t$ graph and velocity, state where in the cycle the speed is greatest and calculate that maximum speed.利用 $x$-$t$ 图斜率与速度的关系，说明速率在循环何处最大，并计算该最大速率。[3]
(c)On the same time axis, the velocity-time and acceleration-time graphs are also sinusoidal. State the phase relationship of each, relative to the displacement, and the shape (sine, cosine or inverted cosine) of each curve.在同一时间轴上，速度-时间图与加速度-时间图也是正弦型。分别写出二者相对位移的相位关系，以及各曲线的形状（正弦、余弦或倒余弦）。[2]
(d)State, with a reason, the displacement at which the acceleration has its greatest magnitude.写出加速度大小最大处的位移，并说明理由。[2]
PART III · PAPER 2 STYLE第三部分 · 第二卷风格Extended structured · calculator · 30 marks长结构题 · 可用计算器 · 30 分
Extended Structured Problems长结构问题
Set up each problem clearly, stating your phase reference where relevant. Method marks dominate the longer items; carry intermediate values to extra figures and round only the final answer.每题清晰列式，必要时写明相位参考。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 2HL ONLYmass-spring energy + max speed + phase弹簧振子能量 + 最大速率 + 相位[12 marks]
A $0.40\ \mathrm{kg}$ mass on a spring of spring constant $k = 160\ \mathrm{N\,m^{-1}}$ oscillates horizontally with amplitude $x_{0} = 0.060\ \mathrm{m}$ on a frictionless surface.一个 $0.40\ \mathrm{kg}$ 的质量挂在弹簧常数 $k = 160\ \mathrm{N\,m^{-1}}$ 的弹簧上，在无摩擦水平面上以振幅 $x_{0} = 0.060\ \mathrm{m}$ 振动。
(a)Calculate the angular frequency of the oscillation.计算振动的角频率。[2]
(b)Calculate the total energy of the oscillator and hence its maximum speed.计算振子的总能量，进而求其最大速率。[3]
(c)Calculate the kinetic energy and the potential energy when the displacement is $x = 0.030\ \mathrm{m}$, and confirm that they sum to the total energy.计算位移为 $x = 0.030\ \mathrm{m}$ 时的动能与势能，并确认二者之和等于总能量。[3]
(d)State how the total energy would change if the amplitude were doubled, with the same mass and spring, and justify your answer.说明在同一质量与弹簧下，若振幅加倍，总能量将如何变化，并说明理由。[2]
(e)On a single time axis, sketch how the kinetic energy and the potential energy vary during one oscillation, and state how many times per oscillation each reaches its maximum.在同一时间轴上画出一次振动中动能与势能的变化，并说明二者每次振动各达到最大几次。[2]
Q9HARDPaper 2restoring force to SHM + damping恢复力推简谐 + 阻尼[10 marks]
A trolley of mass $m = 0.25\ \mathrm{kg}$ is held between two springs on a horizontal track. When displaced a small distance $x$ from equilibrium, the net restoring force on it is $F = -k x$ with $k = 100\ \mathrm{N\,m^{-1}}$.一辆质量 $m = 0.25\ \mathrm{kg}$ 的小车被两根弹簧夹在水平轨道之间。当从平衡位置偏移小位移 $x$ 时，所受合恢复力为 $F = -k x$，$k = 100\ \mathrm{N\,m^{-1}}$。
(a)By applying Newton's second law to the trolley, show that its motion is simple harmonic and that $\omega^{2} = k/m$.对小车应用牛顿第二定律，证明其运动为简谐运动且 $\omega^{2} = k/m$。[3]
(b)Calculate the period of the oscillation.计算振动周期。[2]
(c)The trolley is released from rest at a displacement of $0.040\ \mathrm{m}$. Calculate the magnitude of its acceleration at the instant of release.小车在位移 $0.040\ \mathrm{m}$ 处由静止释放。计算释放瞬间其加速度的大小。[2]
(d)A small frictional force is now introduced, so the oscillation is lightly damped. Describe how the amplitude and the total energy change with time, and state what happens to the period compared with the undamped case.现在引入一个小摩擦力，使振动为轻阻尼。描述振幅与总能量随时间如何变化，并说明周期与无阻尼情形相比有何变化。[3]
A mass on a spring oscillates with simple harmonic motion of amplitude $x_{0} = 0.080\ \mathrm{m}$ and period $T = 0.50\ \mathrm{s}$. It is released from rest at maximum displacement at time $t = 0$.一个弹簧上的质量作简谐运动，振幅 $x_{0} = 0.080\ \mathrm{m}$、周期 $T = 0.50\ \mathrm{s}$。在 $t = 0$ 时由最大位移处静止释放。
(a)Write down the displacement equation $x(t)$ for this motion, justifying the choice of cosine rather than sine, and calculate the angular frequency.写出此运动的位移方程 $x(t)$，说明为何选余弦而非正弦，并计算角频率。[2]
(b)Calculate the displacement and the velocity of the mass at $t = 0.10\ \mathrm{s}$.计算 $t = 0.10\ \mathrm{s}$ 时质量的位移与速度。[3]
(c)Calculate the acceleration at $t = 0.10\ \mathrm{s}$, and verify that it is consistent with the defining condition $a = -\omega^{2}x$.计算 $t = 0.10\ \mathrm{s}$ 时的加速度，并验证其与定义条件 $a = -\omega^{2}x$ 一致。[3]