Unit C.1: Simple Harmonic Motion单元 C.1：简谐运动
The opening unit of Theme C "Wave behaviour". Simple harmonic motion (SHM) is the periodic motion whose acceleration is proportional to displacement and always points back toward equilibrium: $a = -\omega^{2} x$. We build from that defining condition to period and frequency, the mass–spring and simple-pendulum systems, then (HL) the full displacement, velocity and acceleration equations, the energy interchange between kinetic and potential, and the phase relationships visible on $x$-$t$, $v$-$t$ and $a$-$t$ graphs. SHM is the template for every oscillation and underpins the wave model that follows.主题 C"波动行为"的开篇。简谐运动（SHM）是加速度与位移成正比、始终指向平衡位置的周期运动：$a = -\omega^{2} x$。我们从这一定义条件出发，推进到周期与频率、弹簧振子与单摆系统，再到（HL）完整的位移、速度与加速度方程、动能与势能的能量互换，以及 $x$-$t$、$v$-$t$、$a$-$t$ 图上可见的相位关系。简谐运动是一切振动的范本，也是其后波模型的基础。
How to use this guide本指南使用说明
C.1 is the unit where one idea does all the work: the defining condition $a = -\omega^{2} x$. Everything else, the period formulas, the energy split, the sinusoidal graphs, follows from it. Marks are lost not on hard algebra but on confusing amplitude with displacement, frequency with angular frequency, and on sloppy sign or phase. Train the defining condition until you can read $\omega^{2}$ off any restoring-force expression.C.1 是"一个想法撑起全局"的单元：定义条件 $a = -\omega^{2} x$。其余的一切，周期公式、能量分配、正弦图像，都由它推出。失分往往不在难代数，而在混淆振幅与位移、频率与角频率，以及符号或相位的疏忽。把定义条件练熟，直到能从任何恢复力表达式中读出 $\omega^{2}$。
Memorise $a = -\omega^{2} x$, $\omega = 2\pi f$, $T = 1/f$, and the two period formulas $T = 2\pi\sqrt{m/k}$ (spring) and $T = 2\pi\sqrt{L/g}$ (pendulum). Know that velocity is maximum at the centre and zero at the extremes, while acceleration is the opposite. Energy is constant: $E = \tfrac{1}{2} m \omega^{2} x_{0}^{2}$.
背熟 $a = -\omega^{2} x$、$\omega = 2\pi f$、$T = 1/f$，以及两个周期公式 $T = 2\pi\sqrt{m/k}$（弹簧）与 $T = 2\pi\sqrt{L/g}$（单摆）。记住速度在中心最大、两端为零，加速度恰好相反。总能量恒定：$E = \tfrac{1}{2} m \omega^{2} x_{0}^{2}$。
Be able to derive $\omega^{2} = k/m$ by substituting $F = -kx$ into Newton's second law and matching it to $a = -\omega^{2} x$. Master the HL equations $x = x_{0}\cos\omega t$, $v = \pm\omega\sqrt{x_{0}^{2} - x^{2}}$, and the energy split $E_{K} = \tfrac{1}{2} m \omega^{2}(x_{0}^{2} - x^{2})$. State your phase reference (cosine for release from rest, sine for launch through equilibrium) every time.
能把 $F = -kx$ 代入牛顿第二定律并与 $a = -\omega^{2} x$ 对照，推出 $\omega^{2} = k/m$。掌握 HL 方程 $x = x_{0}\cos\omega t$、$v = \pm\omega\sqrt{x_{0}^{2} - x^{2}}$ 与能量分配 $E_{K} = \tfrac{1}{2} m \omega^{2}(x_{0}^{2} - x^{2})$。每次都写明相位（phase）参考（从静止释放用余弦，过平衡位置出发用正弦）。
The Defining Condition: $a = -\omega^{2} x$定义条件：$a = -\omega^{2} x$ C.1 SL+HL
Restoring force. Multiplying by mass, $F = m a = -m \omega^{2} x$. So the net force is proportional and opposite to displacement: $F = -k x$ with $k = m \omega^{2}$. Any system with a linear restoring force oscillates with SHM.
Key vocabulary. Displacement $x$ (from equilibrium), amplitude $x_{0}$ (maximum $|x|$), equilibrium (where net force is zero).
恢复力（restoring force）。两边乘质量，$F = m a = -m \omega^{2} x$。故合力与位移成正比且方向相反：$F = -k x$，其中 $k = m \omega^{2}$。任何具有线性恢复力的系统都作简谐运动。
关键术语。位移 $x$（相对平衡位置）、振幅（amplitude）$x_{0}$（$|x|$ 的最大值）、平衡位置（合力为零处）。
An object oscillates so that its acceleration and displacement are related by $a = -64\,x$ (SI units). Show this is SHM, and find the angular frequency, period, and frequency.某物体振动时加速度与位移满足 $a = -64\,x$（SI 单位）。证明这是简谐运动，并求角频率、周期与频率。
Identify. The form $a = -(\text{positive constant}) \times x$ matches the defining condition $a = -\omega^{2} x$. Acceleration is proportional to displacement and opposite in sign, so the motion is SHM.
识别。形式 $a = -(\text{正常数}) \times x$ 与定义条件 $a = -\omega^{2} x$ 吻合。加速度与位移成正比且反号，故为简谐运动。
Set up. Match coefficients: $\omega^{2} = 64\ \mathrm{s^{-2}}$.
列式。比较系数：$\omega^{2} = 64\ \mathrm{s^{-2}}$。
Execute. $\omega = 8.0\ \mathrm{rad\,s^{-1}}$. Using $T = 2\pi/\omega$ and $f = 1/T$:
计算。$\omega = 8.0\ \mathrm{rad\,s^{-1}}$。用 $T = 2\pi/\omega$、$f = 1/T$：
$$ T = \frac{2\pi}{8.0} \approx 0.785\ \mathrm{s}, \qquad f = \frac{1}{T} \approx 1.27\ \mathrm{Hz}. $$Evaluate. Always take the positive square root for $\omega$; a negative angular frequency has no physical meaning. The units check: $[\omega^{2}] = \mathrm{s^{-2}}$ because $[a/x] = (\mathrm{m\,s^{-2}})/\mathrm{m}$.
评估。$\omega$ 总取正平方根；负角频率没有物理意义。单位核对：$[\omega^{2}] = \mathrm{s^{-2}}$，因为 $[a/x] = (\mathrm{m\,s^{-2}})/\mathrm{m}$。
Going deeper: why a linear restoring force gives SHM深入：线性恢复力为何给出简谐运动
Take any system with restoring force $F = -k x$. Newton's second law gives $m a = -k x$, hence
取任何具有恢复力 $F = -k x$ 的系统。牛顿第二定律给出 $m a = -k x$，故
$$ a = -\frac{k}{m}\, x. $$Comparing with $a = -\omega^{2} x$ identifies $\omega^{2} = k/m$. This is the single most useful move in C.1: whenever you can write the net force as $-(\text{constant}) \times x$, that constant divided by the mass is $\omega^{2}$, and the period follows immediately.
与 $a = -\omega^{2} x$ 对照得 $\omega^{2} = k/m$。这是 C.1 中最有用的一步：只要能把合力写成 $-(\text{常数}) \times x$，该常数除以质量即 $\omega^{2}$，周期随即得出。
Most real restoring forces are only approximately linear. For small displacements, expanding the force about equilibrium gives a leading linear term, which is why so many systems behave as SHM for small oscillations (the pendulum's small-angle approximation is one example).
多数真实恢复力只是近似线性。对小位移，将力在平衡位置展开会得到主导的线性项，这正是众多系统在小振幅下表现为简谐运动的原因（单摆的小角度近似就是一例）。
Period, Frequency and Angular Frequency周期、频率与角频率 C.1 SL+HL
- Period $T$: time for one complete oscillation, in $\mathrm{s}$.
- Frequency $f$: oscillations per second, in $\mathrm{Hz} = \mathrm{s^{-1}}$.
- Angular frequency $\omega$: in $\mathrm{rad\,s^{-1}}$.
- 周期（period）$T$：完成一次完整振动的时间，单位 $\mathrm{s}$。
- 频率（frequency）$f$：每秒振动次数，单位 $\mathrm{Hz} = \mathrm{s^{-1}}$。
- 角频率（angular frequency）$\omega$：单位 $\mathrm{rad\,s^{-1}}$。
A pendulum completes $30$ full swings in $48\ \mathrm{s}$. Find its period, frequency and angular frequency.某单摆在 $48\ \mathrm{s}$ 内完成 $30$ 次完整摆动。求其周期、频率与角频率。
Identify. "$30$ swings in $48\ \mathrm{s}$" gives the period directly as time per oscillation.
识别。"$48\ \mathrm{s}$ 内 $30$ 次"直接给出每次振动所用时间，即周期。
Set up & Execute.
列式与计算。
$$ T = \frac{48}{30} = 1.6\ \mathrm{s}, \qquad f = \frac{1}{T} = 0.625\ \mathrm{Hz}, \qquad \omega = 2\pi f \approx 3.93\ \mathrm{rad\,s^{-1}}. $$Evaluate. Cross-check: $\omega = 2\pi/T = 2\pi/1.6 \approx 3.93\ \mathrm{rad\,s^{-1}}$. Consistent. Note $\omega$ is about $2\pi$ times $f$, so it is always the larger number.
评估。互校：$\omega = 2\pi/T = 2\pi/1.6 \approx 3.93\ \mathrm{rad\,s^{-1}}$，一致。注意 $\omega$ 约为 $f$ 的 $2\pi$ 倍，故总是更大的数。
Going deeper: why amplitude does not affect the period深入：为何振幅不影响周期
A larger amplitude means the body travels farther in each cycle, but the restoring force, and hence the acceleration, is also proportionally larger ($a = -\omega^{2} x$). The extra distance and the extra acceleration cancel exactly, leaving the round-trip time unchanged. Formally, $\omega = \sqrt{k/m}$ contains no amplitude term.
振幅更大意味着每个周期内行程更远，但恢复力（因而加速度）也按比例更大（$a = -\omega^{2} x$）。多出的距离与多出的加速度恰好抵消，往返时间不变。形式上，$\omega = \sqrt{k/m}$ 不含振幅项。
This isochronism is exactly the property that makes pendulum clocks possible: the timekeeping does not drift as the swing amplitude slowly decreases through friction (provided the amplitude stays small enough for SHM to hold).
这种等时性正是钟摆能计时的原因：随着摆幅因摩擦缓慢减小，计时不会漂移（前提是振幅足够小，使简谐运动近似成立）。
Two Standard SHM Systems两个标准简谐系统 C.1 SL+HL
Simple pendulum (small angles). A bob of mass $m$ on a light string of length $L$ has, for small angular displacement $\theta$, restoring force component $-mg\sin\theta \approx -mg\theta$, giving $\omega^{2} = g/L$ and, from the data booklet, $$ T = 2\pi \sqrt{\frac{L}{g}}. $$ Notice the pendulum period is independent of the bob mass.
Small-angle approximation. $T = 2\pi\sqrt{L/g}$ holds only while $\sin\theta \approx \theta$ (roughly $\theta \lesssim 10^{\circ}$). At larger amplitudes the motion is periodic but no longer simple harmonic.
单摆（pendulum，小角度）。长为 $L$ 的轻绳上质量 $m$ 的摆球，对小角位移 $\theta$，恢复力分量 $-mg\sin\theta \approx -mg\theta$，得 $\omega^{2} = g/L$，由数据手册 $$ T = 2\pi \sqrt{\frac{L}{g}}. $$ 注意单摆周期与摆球质量无关。
小角度近似。$T = 2\pi\sqrt{L/g}$ 仅在 $\sin\theta \approx \theta$（约 $\theta \lesssim 10^{\circ}$）时成立。振幅更大时运动仍周期，但不再是简谐。
A $0.50\ \mathrm{kg}$ mass hangs from a spring of spring constant $k = 200\ \mathrm{N\,m^{-1}}$ and is set oscillating vertically. Find the period and frequency.$0.50\ \mathrm{kg}$ 的质量挂在弹簧常数 $k = 200\ \mathrm{N\,m^{-1}}$ 的弹簧上，竖直振动。求周期与频率。
Identify. Known $m = 0.50\ \mathrm{kg}$, $k = 200\ \mathrm{N\,m^{-1}}$. Use the data-booklet result T = 2π√(m/k).
识别。已知 $m = 0.50\ \mathrm{kg}$、$k = 200\ \mathrm{N\,m^{-1}}$。用数据手册公式 T = 2π√(m/k)。
Execute.
计算。
$$ T = 2\pi \sqrt{\frac{0.50}{200}} = 2\pi \sqrt{2.5\times 10^{-3}} \approx 0.314\ \mathrm{s}, \qquad f = \frac{1}{T} \approx 3.18\ \mathrm{Hz}. $$Evaluate. For a vertical spring, gravity only shifts the equilibrium position; it does not change $k$, so the same period formula applies. The result is amplitude-independent.
评估。竖直弹簧中重力只移动平衡位置，不改变 $k$，故同一周期公式适用。结果与振幅无关。
Find the length of a simple pendulum whose period is exactly $2.0\ \mathrm{s}$ (a "seconds pendulum"). Take $g = 9.81\ \mathrm{m\,s^{-2}}$.求周期恰为 $2.0\ \mathrm{s}$ 的单摆长度（"秒摆"）。取 $g = 9.81\ \mathrm{m\,s^{-2}}$。
Set up. Rearrange the data-booklet result T = 2π√(L/g) for $L$:
列式。将数据手册公式 T = 2π√(L/g) 对 $L$ 变形：
Execute.
计算。
$$ L = \frac{(9.81)(2.0)^{2}}{4\pi^{2}} = \frac{39.24}{39.48} \approx 0.994\ \mathrm{m}. $$Evaluate. A seconds pendulum is close to $1\ \mathrm{m}$ long, a useful memory check. The mass of the bob does not enter, so it is irrelevant.
评估。秒摆长度接近 $1\ \mathrm{m}$，可作记忆核对。摆球质量不出现，故无关。
Going deeper: deriving $\omega^{2} = g/L$ for the pendulum深入：推导单摆的 $\omega^{2} = g/L$
Displace the bob by arc length $x = L\theta$. The tangential restoring force is the component of gravity along the arc: $F = -mg\sin\theta$. For small $\theta$, $\sin\theta \approx \theta = x/L$, so
将摆球沿弧长 $x = L\theta$ 偏移。切向恢复力为重力沿弧的分量：$F = -mg\sin\theta$。小 $\theta$ 时 $\sin\theta \approx \theta = x/L$，故
$$ F \approx -mg\,\frac{x}{L} = -\left(\frac{mg}{L}\right) x. $$This is a linear restoring force with effective constant $k_{\text{eff}} = mg/L$. Then $\omega^{2} = k_{\text{eff}}/m = g/L$, and $T = 2\pi/\omega = 2\pi\sqrt{L/g}$. The mass cancels, which is why the bob mass never appears.
这是线性恢复力，有效常数 $k_{\text{eff}} = mg/L$。则 $\omega^{2} = k_{\text{eff}}/m = g/L$，$T = 2\pi/\omega = 2\pi\sqrt{L/g}$。质量约去，这正是摆球质量从不出现的原因。
The Full SHM Equations完整的简谐运动方程 HL only C.1 AHL
Velocity. Differentiating (cosine case) gives $v = -x_{0}\omega\sin\omega t$. Eliminating $t$ yields the data-booklet form valid at any displacement: $$ v = \pm\,\omega \sqrt{x_{0}^{2} - x^{2}}, \qquad v_{\max} = \omega x_{0}\ \text{(at } x = 0). $$ Acceleration. Differentiating again: $a = -x_{0}\omega^{2}\cos\omega t = -\omega^{2} x$, consistent with the defining condition, with $$ a_{\max} = \omega^{2} x_{0}\ \text{(at } x = \pm x_{0}). $$ Phase summary. Velocity leads displacement by $90^{\circ}$; acceleration is $180^{\circ}$ out of phase with displacement.
速度。对余弦情形求导得 $v = -x_{0}\omega\sin\omega t$。消去 $t$ 得数据手册中对任意位移都成立的形式： $$ v = \pm\,\omega \sqrt{x_{0}^{2} - x^{2}}, \qquad v_{\max} = \omega x_{0}\ (\text{在 } x = 0). $$ 加速度。再次求导：$a = -x_{0}\omega^{2}\cos\omega t = -\omega^{2} x$，与定义条件一致，且 $$ a_{\max} = \omega^{2} x_{0}\ (\text{在 } x = \pm x_{0}). $$ 相位小结。速度比位移超前 $90^{\circ}$；加速度与位移相位相反（差 $180^{\circ}$）。
A particle performs SHM with amplitude $x_{0} = 0.040\ \mathrm{m}$ and angular frequency $\omega = 15\ \mathrm{rad\,s^{-1}}$, released from rest at maximum displacement at $t = 0$. Find (a) $v_{\max}$, (b) $a_{\max}$, and (c) the speed when the displacement is $x = 0.020\ \mathrm{m}$.某质点作简谐运动，振幅 $x_{0} = 0.040\ \mathrm{m}$、角频率 $\omega = 15\ \mathrm{rad\,s^{-1}}$，在 $t = 0$ 由最大位移处静止释放。求 (a) $v_{\max}$、(b) $a_{\max}$、(c) 位移 $x = 0.020\ \mathrm{m}$ 时的速率。
Identify. Released from rest at the extreme, so $x = x_{0}\cos\omega t$ applies. Known $x_{0} = 0.040\ \mathrm{m}$, $\omega = 15\ \mathrm{rad\,s^{-1}}$.
识别。从端点静止释放，故 $x = x_{0}\cos\omega t$ 适用。已知 $x_{0} = 0.040\ \mathrm{m}$、$\omega = 15\ \mathrm{rad\,s^{-1}}$。
(a) Execute. $v_{\max} = \omega x_{0} = (15)(0.040) = 0.60\ \mathrm{m\,s^{-1}}$ (at the centre, $x = 0$).
(a) 计算。$v_{\max} = \omega x_{0} = (15)(0.040) = 0.60\ \mathrm{m\,s^{-1}}$（在中心 $x = 0$）。
(b) Execute. $a_{\max} = \omega^{2} x_{0} = (15)^{2}(0.040) = 9.0\ \mathrm{m\,s^{-2}}$ (at the extremes, $x = \pm x_{0}$).
(b) 计算。$a_{\max} = \omega^{2} x_{0} = (15)^{2}(0.040) = 9.0\ \mathrm{m\,s^{-2}}$（在端点 $x = \pm x_{0}$）。
(c) Set up & Execute. Use $v = \pm\omega\sqrt{x_{0}^{2} - x^{2}}$:
(c) 列式与计算。用 $v = \pm\omega\sqrt{x_{0}^{2} - x^{2}}$：
$$ v = 15\sqrt{0.040^{2} - 0.020^{2}} = 15\sqrt{1.2\times 10^{-3}} \approx 0.52\ \mathrm{m\,s^{-1}}. $$Evaluate. At half the amplitude the speed is about $87\%$ of $v_{\max}$, not $50\%$, because $v$ depends on $\sqrt{x_{0}^{2} - x^{2}}$, not linearly on $x$. Speed is greatest at the centre and zero at the extremes; acceleration is the reverse.
评估。在半振幅处速率约为 $v_{\max}$ 的 $87\%$，而非 $50\%$，因为 $v$ 取决于 $\sqrt{x_{0}^{2} - x^{2}}$，不是 $x$ 的线性函数。速率在中心最大、端点为零；加速度相反。
Going deeper: deriving $v = \pm\omega\sqrt{x_{0}^{2} - x^{2}}$深入：推导 $v = \pm\omega\sqrt{x_{0}^{2} - x^{2}}$
Start from $x = x_{0}\cos\omega t$. Differentiate to get $v = -x_{0}\omega\sin\omega t$. Now use $\sin^{2}\omega t + \cos^{2}\omega t = 1$:
由 $x = x_{0}\cos\omega t$ 出发，求导得 $v = -x_{0}\omega\sin\omega t$。再用 $\sin^{2}\omega t + \cos^{2}\omega t = 1$：
$$ \sin\omega t = \pm\sqrt{1 - \cos^{2}\omega t} = \pm\sqrt{1 - \left(\frac{x}{x_{0}}\right)^{2}}. $$Substituting back,
代回，
$$ v = \mp\, x_{0}\omega \sqrt{1 - \frac{x^{2}}{x_{0}^{2}}} = \pm\,\omega\sqrt{x_{0}^{2} - x^{2}}. $$The $\pm$ records that the body passes each interior point twice per cycle, once moving each way. At $x = 0$ this gives $v_{\max} = \omega x_{0}$; at $x = \pm x_{0}$ it gives $v = 0$, as expected.
$\pm$ 记录了物体每周期两次经过每个内部位置，往返各一次。$x = 0$ 时得 $v_{\max} = \omega x_{0}$；$x = \pm x_{0}$ 时得 $v = 0$，符合预期。
Energy Interchange in SHM简谐运动中的能量互换 HL only C.1 AHL
Kinetic energy at displacement $x$. Using $v = \pm\omega\sqrt{x_{0}^{2} - x^{2}}$, $$ E_{K} = \tfrac{1}{2} m v^{2} = \tfrac{1}{2} m \omega^{2}\left(x_{0}^{2} - x^{2}\right). $$ Potential energy at displacement $x$. Total minus kinetic: $$ E_{P} = E - E_{K} = \tfrac{1}{2} m \omega^{2} x^{2}. $$ Where each peaks. $E_{K}$ is maximum at $x = 0$; $E_{P}$ is maximum at $x = \pm x_{0}$. Both have the SHM frequency doubled on a time graph (two energy peaks per oscillation).
位移 $x$ 处的动能。用 $v = \pm\omega\sqrt{x_{0}^{2} - x^{2}}$， $$ E_{K} = \tfrac{1}{2} m v^{2} = \tfrac{1}{2} m \omega^{2}\left(x_{0}^{2} - x^{2}\right). $$ 位移 $x$ 处的势能。总能量减动能： $$ E_{P} = E - E_{K} = \tfrac{1}{2} m \omega^{2} x^{2}. $$ 峰值位置。$E_{K}$ 在 $x = 0$ 最大；$E_{P}$ 在 $x = \pm x_{0}$ 最大。在时间图上两者频率为运动频率的两倍（每周期两个能量峰）。
A $0.25\ \mathrm{kg}$ mass undergoes SHM with amplitude $x_{0} = 0.080\ \mathrm{m}$ and angular frequency $\omega = 12\ \mathrm{rad\,s^{-1}}$. Find (a) the total energy, (b) the kinetic and potential energy when $x = 0.040\ \mathrm{m}$.$0.25\ \mathrm{kg}$ 的质量作简谐运动，振幅 $x_{0} = 0.080\ \mathrm{m}$、角频率 $\omega = 12\ \mathrm{rad\,s^{-1}}$。求 (a) 总能量，(b) $x = 0.040\ \mathrm{m}$ 时的动能与势能。
Identify. Known $m = 0.25\ \mathrm{kg}$, $x_{0} = 0.080\ \mathrm{m}$, $\omega = 12\ \mathrm{rad\,s^{-1}}$, $x = 0.040\ \mathrm{m}$.
识别。已知 $m = 0.25\ \mathrm{kg}$、$x_{0} = 0.080\ \mathrm{m}$、$\omega = 12\ \mathrm{rad\,s^{-1}}$、$x = 0.040\ \mathrm{m}$。
(a) Total energy.
(a) 总能量。
$$ E = \tfrac{1}{2} m \omega^{2} x_{0}^{2} = \tfrac{1}{2}(0.25)(12)^{2}(0.080)^{2} = 0.1152\ \mathrm{J} \approx 0.115\ \mathrm{J}. $$(b) Kinetic at $x = 0.040\ \mathrm{m}$.
(b) $x = 0.040\ \mathrm{m}$ 处的动能。
$$ E_{K} = \tfrac{1}{2} m \omega^{2}(x_{0}^{2} - x^{2}) = \tfrac{1}{2}(0.25)(144)(0.080^{2} - 0.040^{2}) = 0.0864\ \mathrm{J}. $$Potential at the same point. $E_{P} = E - E_{K} = 0.1152 - 0.0864 = 0.0288\ \mathrm{J}$.
同一点的势能。$E_{P} = E - E_{K} = 0.1152 - 0.0864 = 0.0288\ \mathrm{J}$。
Evaluate. At $x = x_{0}/2$ the potential is a quarter of the total ($E_{P} \propto x^{2}$), and kinetic is three quarters. Their sum returns the total energy, confirming consistency.
评估。在 $x = x_{0}/2$ 处势能为总能量的四分之一（$E_{P} \propto x^{2}$），动能为四分之三。两者之和回到总能量，验证一致。
Going deeper: the energy "well" picture深入：能量"势阱"图像
Plot $E_{P} = \tfrac{1}{2} m\omega^{2} x^{2}$ against $x$: it is a parabola (a potential well) with minimum at the centre. The horizontal line $E = \tfrac{1}{2} m\omega^{2} x_{0}^{2}$ sits above it. At any $x$, the gap between the line and the parabola is the kinetic energy $E_{K}$.
把 $E_{P} = \tfrac{1}{2} m\omega^{2} x^{2}$ 对 $x$ 作图：是一条抛物线（势阱），中心处取极小。水平线 $E = \tfrac{1}{2} m\omega^{2} x_{0}^{2}$ 位于其上方。在任意 $x$ 处，直线与抛物线之间的差即动能 $E_{K}$。
The turning points (extremes) are where the parabola meets the line: there $E_{K} = 0$ and the body stops momentarily. The well picture makes the amplitude the natural boundary of the motion, the body cannot reach where the parabola exceeds the total energy.
折返点（端点）是抛物线与直线相交处：此处 $E_{K} = 0$，物体瞬时停止。势阱图像使振幅成为运动的自然边界——物体无法到达抛物线高于总能量之处。
SHM Graphs and Phase Relationships简谐运动图像与相位关系 C.1 SL+HL
- $x$-$t$: $x = x_{0}\cos\omega t$, starts at maximum.
- $v$-$t$: $v = -\omega x_{0}\sin\omega t$, starts at zero, peaks a quarter-period later. Velocity leads displacement by $90^{\circ}$ ($T/4$).
- $a$-$t$: $a = -\omega^{2} x_{0}\cos\omega t$, the mirror image of $x$-$t$ (inverted). Acceleration is $180^{\circ}$ out of phase with displacement.
Real-world SHM. A child on a swing (small amplitude), a tuning fork prong, a mass bobbing on a spring, atoms vibrating in a solid lattice, and the balance wheel of a mechanical watch all approximate SHM.
- $x$-$t$：$x = x_{0}\cos\omega t$，从最大值开始。
- $v$-$t$：$v = -\omega x_{0}\sin\omega t$，从零开始，四分之一周期后达峰值。速度比位移超前 $90^{\circ}$（$T/4$）。
- $a$-$t$：$a = -\omega^{2} x_{0}\cos\omega t$，是 $x$-$t$ 的镜像（倒置）。加速度与位移相位相反（差 $180^{\circ}$）。
现实中的简谐运动。荡秋千的孩子（小振幅）、音叉的叉股、弹簧上上下振动的质量、固体晶格中振动的原子、机械表的摆轮，都近似简谐运动。
A displacement-time graph of an oscillator is a cosine starting at its maximum, with peak displacement $0.06\ \mathrm{m}$ and a full cycle every $0.40\ \mathrm{s}$. State the amplitude and period, find $\omega$, and state where on the cycle the velocity is greatest.某振子的位移-时间图为从最大值开始的余弦曲线，峰位移 $0.06\ \mathrm{m}$，每 $0.40\ \mathrm{s}$ 一个完整周期。给出振幅与周期，求 $\omega$，并说明在循环何处速度最大。
Identify. Read amplitude and period straight off the graph: $x_{0} = 0.06\ \mathrm{m}$, $T = 0.40\ \mathrm{s}$.
识别。直接从图上读出振幅与周期：$x_{0} = 0.06\ \mathrm{m}$、$T = 0.40\ \mathrm{s}$。
Execute. $\omega = 2\pi/T = 2\pi/0.40 \approx 15.7\ \mathrm{rad\,s^{-1}}$.
计算。$\omega = 2\pi/T = 2\pi/0.40 \approx 15.7\ \mathrm{rad\,s^{-1}}$。
Where is $v$ greatest? Velocity is the slope of the $x$-$t$ graph, steepest as the curve crosses $x = 0$ (the equilibrium centre). So $v$ is greatest at the centre and zero at the turning points (the peaks).
速度何处最大？速度是 $x$-$t$ 图的斜率，曲线穿过 $x = 0$（平衡中心）时最陡。故 $v$ 在中心最大、在折返点（峰处）为零。
Evaluate. The $v$-$t$ graph is a sine (90° out of phase with the cosine displacement); the $a$-$t$ graph is an inverted cosine (180° out of phase with displacement). Maximum speed is $v_{\max} = \omega x_{0} \approx 0.94\ \mathrm{m\,s^{-1}}$.
评估。$v$-$t$ 图为正弦（与余弦位移相差 90°）；$a$-$t$ 图为倒余弦（与位移相差 180°）。最大速率 $v_{\max} = \omega x_{0} \approx 0.94\ \mathrm{m\,s^{-1}}$。
Going deeper: why velocity leads displacement by a quarter-cycle深入：速度为何比位移超前四分之一周期
Velocity is the time-derivative of displacement. Differentiating a cosine gives a (negative) sine, and differentiating a sine gives a cosine, each shifting the curve by a quarter-period. Concretely, $x = x_{0}\cos\omega t$ leads to $v = -\omega x_{0}\sin\omega t = \omega x_{0}\cos(\omega t + 90^{\circ})$, so $v$ reaches its peak $T/4$ before $x$ would, hence "velocity leads displacement by $90^{\circ}$".
速度是位移对时间的导数。对余弦求导得（负）正弦，对正弦求导得余弦，每次都把曲线平移四分之一周期。具体地，$x = x_{0}\cos\omega t$ 给出 $v = -\omega x_{0}\sin\omega t = \omega x_{0}\cos(\omega t + 90^{\circ})$，故 $v$ 比 $x$ 提前 $T/4$ 达峰，即"速度比位移超前 $90^{\circ}$"。
Differentiating once more flips the sign and shifts another quarter-cycle, accumulating to the $180^{\circ}$ offset between acceleration and displacement: $a = -\omega^{2} x$.
再求导一次会变号并再平移四分之一周期，累计为加速度与位移之间的 $180^{\circ}$ 相位差：$a = -\omega^{2} x$。
Exam Strategy and Common Pitfalls考试策略与常见陷阱
- The defining condition and the SHM equations use $\omega$, never $f$. Convert first: $\omega = 2\pi f$. Forgetting the $2\pi$ is the single most-penalised SHM slip.
- 定义条件与简谐方程用 $\omega$，绝不用 $f$。先换算：$\omega = 2\pi f$。漏掉 $2\pi$ 是简谐运动最常被扣分的失误。
- Read $\omega^{2}$ directly off any $a = -(\text{const})x$ relation. That constant is $\omega^{2}$; take the positive root.
- 从任何 $a = -(\text{常数})x$ 关系直接读出 $\omega^{2}$。该常数即 $\omega^{2}$；取正根。
- $T = 2\pi\sqrt{m/k}$ for a mass–spring; $T = 2\pi\sqrt{L/g}$ for a simple pendulum. Both are amplitude-independent (isochronous).
- 弹簧振子用 $T = 2\pi\sqrt{m/k}$；单摆用 $T = 2\pi\sqrt{L/g}$。两者均与振幅无关（等时）。
- The pendulum formula requires small angles ($\theta \lesssim 10^{\circ}$). State this assumption if a question pushes the amplitude.
- 单摆公式要求小角度（$\theta \lesssim 10^{\circ}$）。若题目加大振幅，要写明该假设。
- Choose $\cos$ if released from rest at an extreme, $\sin$ if launched through equilibrium at $t = 0$. State the choice in your working.
- 从端点静止释放选 $\cos$，在 $t = 0$ 经平衡位置出发选 $\sin$。在解题中写明所选。
- For "speed at displacement $x$", use $v = \pm\omega\sqrt{x_{0}^{2} - x^{2}}$ directly. Do not assume speed is linear in $x$.
- 求"位移 $x$ 处的速率"直接用 $v = \pm\omega\sqrt{x_{0}^{2} - x^{2}}$。不要假定速率与 $x$ 成线性。
- Total energy $E = \tfrac{1}{2} m\omega^{2} x_{0}^{2}$ is constant; $E_{K}$ peaks at the centre, $E_{P}$ at the extremes. $E \propto x_{0}^{2}$.
- 总能量 $E = \tfrac{1}{2} m\omega^{2} x_{0}^{2}$ 恒定；$E_{K}$ 在中心最大，$E_{P}$ 在端点最大。$E \propto x_{0}^{2}$。
- On graphs, velocity leads displacement by $90^{\circ}$; acceleration is anti-phase ($180^{\circ}$). Slope of $x$-$t$ is $v$; slope of $v$-$t$ is $a$.
- 图像上速度比位移超前 $90^{\circ}$；加速度反相（$180^{\circ}$）。$x$-$t$ 斜率为 $v$；$v$-$t$ 斜率为 $a$。
Flashcards闪卡
Unit C.1 Practice Quiz单元 C.1 练习测验
Readiness Checklist备考清单
Tick each item when you can do it cold, without notes, on your first attempt.
每一条都要"裸做"做对（不看笔记、一次过）才打勾。
- State the defining condition $a = -\omega^{2} x$ and explain the proportionality and the restoring sign写出定义条件 $a = -\omega^{2} x$ 并解释比例关系与恢复负号
- Read $\omega^{2}$ off any $a = -(\text{const})x$ relation and take the positive root for $\omega$从任何 $a = -(\text{常数})x$ 关系读出 $\omega^{2}$，并对 $\omega$ 取正根
- Convert freely between $T$, $f$ and $\omega$ using $T = 1/f$ and $\omega = 2\pi f$用 $T = 1/f$ 与 $\omega = 2\pi f$ 在 $T$、$f$、$\omega$ 之间自由换算
- Explain why SHM is isochronous (period independent of amplitude)解释简谐运动为何等时（周期与振幅无关）
- Use $T = 2\pi\sqrt{m/k}$ for a mass–spring system and reason about $m$ and $k$ scaling对弹簧振子用 $T = 2\pi\sqrt{m/k}$ 并分析 $m$、$k$ 的标度关系
- Use $T = 2\pi\sqrt{L/g}$ for a simple pendulum and state the small-angle condition对单摆用 $T = 2\pi\sqrt{L/g}$ 并说明小角度条件
- HL Choose $\cos$ or $\sin$ for $x(t)$ from the initial conditions由初始条件为 $x(t)$ 选择 $\cos$ 或 $\sin$
- HL Find $v$ at any displacement with $v = \pm\omega\sqrt{x_{0}^{2} - x^{2}}$, and $v_{\max} = \omega x_{0}$, $a_{\max} = \omega^{2} x_{0}$用 $v = \pm\omega\sqrt{x_{0}^{2} - x^{2}}$ 求任意位移处的 $v$，并求 $v_{\max} = \omega x_{0}$、$a_{\max} = \omega^{2} x_{0}$
- HL Use $E = \tfrac{1}{2} m\omega^{2} x_{0}^{2}$ and $E_{K} = \tfrac{1}{2} m\omega^{2}(x_{0}^{2} - x^{2})$ to split the energy用 $E = \tfrac{1}{2} m\omega^{2} x_{0}^{2}$ 与 $E_{K} = \tfrac{1}{2} m\omega^{2}(x_{0}^{2} - x^{2})$ 分配能量
- State where $E_{K}$ and $E_{P}$ peak and that total energy $\propto x_{0}^{2}$说明 $E_{K}$ 与 $E_{P}$ 的峰值位置，并知总能量 $\propto x_{0}^{2}$
- Sketch and read $x$-$t$, $v$-$t$ and $a$-$t$ graphs and state their phase offsets画出并读懂 $x$-$t$、$v$-$t$、$a$-$t$ 图并说明相位差
- Give real-world examples of (approximate) SHM and justify the approximation举出（近似）简谐运动的现实例子并论证近似条件
IB Paper-Style PracticeIB 试卷风格练习
C.1 Practice and Solutions are on the roadmap, to ship under Practice Questions/Unit_C1_*.html with the bilingual built-in pattern.
C.1 配套 Practice 与 Solutions 在排期，上线后位于 Practice Questions/Unit_C1_*.html。