From springs to pendulums, master the elegant mathematics of simple harmonic motion — the language nature uses to describe everything from heartbeats to skyscrapers.
10–15% Exam Weight考试占分 10–15%~12–17 Class Periods约 12–17 课时5 Topics5 个专题
Topic 7.1专题 7.1
Defining Simple Harmonic Motion (SHM)简谐运动（SHM）的定义
Simple harmonic motion is one of the most pervasive phenomena in physics. Whenever a system is disturbed slightly from a stable equilibrium, and the restoring force is proportional to the displacement, the resulting motion is sinusoidal — it repeats itself in a perfectly predictable pattern. SHM is a special case of periodic motion, one where the restoring force has a very specific mathematical form.
The key idea is this: if you pull a mass on a spring a distance $x$ from equilibrium, the spring pulls it back with a force $F = -kx$. The negative sign means the force always opposes the displacement, pointing back toward equilibrium. This is the hallmark of SHM — a linear restoring force.
Big Idea — The SHM Test
To determine whether a system undergoes SHM, check two things: (1) there is a stable equilibrium position, and (2) the net restoring force is directly proportional to displacement from that equilibrium. If $F_{\text{net}} = -k\Delta x$ (where $k$ is a positive constant), the system exhibits SHM.
A restoring force is any force directed opposite to the object's displacement from equilibrium. When you stretch a spring rightward, the spring pulls leftward. When you push a pendulum to the left, gravity's tangential component pulls it back to the right.
Any system that reduces to this form undergoes SHM
任何能化为这种形式的系统都做 SHM
Exam Trap
Not all periodic motion is SHM. A ball bouncing on the floor is periodic but the restoring force is not proportional to displacement. Only systems with a linear restoring force qualify as SHM.
A ball rolls back and forth inside a hemispherical bowl. Is this SHM for small displacements?
小球在半球形碗内来回滚动。 小位移时是否为 SHM？
Step 1第 1 步
Restoring force at displacement $\theta$:角位移 $\theta$ 处的回复力：
$$F = -mg\sin\theta$$
Step 2第 2 步
Use the small-angle approximation $\sin\theta \approx \theta$.小角度近似 $\sin\theta \approx \theta$。
$$F \approx -mg\theta$$
Step 3第 3 步
Express $\theta$ in terms of arc length $s$ and radius $R$:用弧长 $s$ 与半径 $R$ 表示 $\theta$：
$$\theta = \frac{s}{R},\quad F \approx -\left(\frac{mg}{R}\right)s$$
Step 4第 4 步
This has the form $F = -k \cdot (\text{displacement})$ with effective constant $k = mg/R$.该式为 $F = -k\cdot(\text{位移})$，有效"劲度系数" $k = mg/R$。
✓ Yes — for small oscillations, this is SHM.✓ 小幅振荡下确实是 SHM。
A mass on a spring is displaced 0.2 m from equilibrium and released. Which statement best describes the restoring force at maximum displacement?弹簧上的物体被拉离平衡位置 0.2 m 后释放。最大位移处回复力的特点是？
The restoring force is zero at maximum displacement.最大位移处回复力为零。
The restoring force is at its maximum magnitude and directed toward equilibrium.回复力大小最大，方向指向平衡位置。
The restoring force is at its maximum magnitude and directed away from equilibrium.回复力大小最大，方向背离平衡位置。
The restoring force equals the weight of the object.回复力等于物体的重力。
Correct! At maximum displacement (the amplitude), $|F| = kA$ is at its greatest, and the force is directed back toward equilibrium.正确！最大位移（振幅 amplitude）处 $|F| = kA$ 达到最大，方向指回平衡。
The restoring force $F = -kx$ is largest when $|x|$ is largest (at the amplitude). It is zero at equilibrium where $x = 0$, not at the extremes.回复力 $F = -kx$ 在 $|x|$ 最大（振幅处）时最大；$x = 0$（平衡处）才为零。
Topic 7.2专题 7.2
Frequency and Period of SHMSHM 的频率与周期
Every oscillating system has a natural rhythm: it takes a fixed amount of time to complete one full cycle. This time is the period $T$ (seconds). The frequency $f$ is cycles per second (Hz). Angular frequency $\omega$ is radians per second.
Period of a Simple Pendulum (Small Angle)单摆的周期（小角度）
$$T_p = 2\pi\sqrt{\frac{\ell}{g}}$$
Valid when $\theta_{\max} \lesssim 15°$ · Independent of mass and amplitude
在 $\theta_{\max} \lesssim 15°$ 时成立 · 与质量和振幅无关
Key Insight — What Affects the Period?
For a spring oscillator, period depends on $m$ and $k$. For a simple pendulum, period depends on $\ell$ and $g$. In neither case does the amplitude affect the period (for ideal springs and small-angle pendulums). This is the isochronism of SHM.
Derivation Sketch — Spring Period
Starting from $ma = -kx$, write $a = d^2x/dt^2$ to get $\frac{d^2x}{dt^2} = -\frac{k}{m}x$. The solution is $x(t) = A\cos(\omega t + \phi)$ where $\omega^2 = k/m$. Since $T = 2\pi/\omega$, we get $T = 2\pi\sqrt{m/k}$.
Note: amplitude was not given because it doesn't matter!注意：题目没给振幅——因为振幅根本不影响周期！
If the mass in a mass–spring system is quadrupled while the spring constant remains the same, the new period is:若弹簧振子的质量变为原来的 4 倍、劲度系数不变，新周期是：
The same as the original period与原周期相同
Half the original period原周期的一半
Twice the original period原周期的 2 倍
Four times the original period原周期的 4 倍
Correct! $T = 2\pi\sqrt{m/k}$. If $m \to 4m$, then $T' = 2\pi\sqrt{4m/k} = 2T$. The period doubles.正确！$T = 2\pi\sqrt{m/k}$。$m \to 4m$ 时 $T' = 2\pi\sqrt{4m/k} = 2T$，周期翻倍。
Since $T \propto \sqrt{m}$, quadrupling $m$ multiplies $T$ by $\sqrt{4} = 2$. The period doubles.由于 $T \propto \sqrt{m}$，$m$ 变 4 倍则 $T$ 乘以 $\sqrt{4} = 2$，周期翻倍。
Topic 7.3专题 7.3
Representing and Analyzing SHMSHM 的表示与分析
The mathematical backbone of SHM is the sinusoidal function. If an object's position is described by a cosine or sine function of time, and its acceleration is proportional to the negative of its displacement, the motion is simple harmonic.
Why This Is the Solution — Plug-and-Check in Three Lines
The AP exam sometimes asks you to verify that $x(t) = A\cos(\omega t + \phi)$ satisfies the SHM differential equation $\ddot{x} = -\omega^2 x$. Just differentiate twice and compare.
Start with the proposed solution:
$$x(t) = A\cos(\omega t + \phi)$$
First derivative — velocity:
$$\dot{x}(t) = -A\omega\sin(\omega t + \phi)$$
Second derivative — acceleration:
$$\ddot{x}(t) = -A\omega^2\cos(\omega t + \phi) = -\omega^2\,x(t)\;\checkmark$$
The last line matches the ODE for any amplitude $A$ and any phase constant $\phi$.
The constants $A$ and $\phi$ are fixed by initial conditions: if the mass starts at $x_0$ with velocity $v_0$, then $A = \sqrt{x_0^2 + (v_0/\omega)^2}$ and $\tan\phi = -v_0/(\omega x_0)$. Released from rest at $x = A$ gives $\phi = 0$ — the simplest case.
为什么它是解 —— 三行带入验证
AP 考试有时要求你验证 $x(t) = A\cos(\omega t + \phi)$ 满足 SHM 微分方程 $\ddot{x} = -\omega^2 x$——求两次导再对比即可。
假设解：
$$x(t) = A\cos(\omega t + \phi)$$
一阶导数 —— 速度：
$$\dot{x}(t) = -A\omega\sin(\omega t + \phi)$$
二阶导数 —— 加速度：
$$\ddot{x}(t) = -A\omega^2\cos(\omega t + \phi) = -\omega^2\,x(t)\;\checkmark$$
$$v(t) = \frac{dx}{dt} = -A\omega\sin(\omega t + \phi)$$
Maximum speed: $v_{\max} = A\omega$
最大速率：$v_{\max} = A\omega$
Acceleration in SHMSHM 中的加速度
$$a(t) = -A\omega^2\cos(\omega t + \phi) = -\omega^2 x$$
Maximum acceleration: $a_{\max} = A\omega^2$
最大加速度：$a_{\max} = A\omega^2$
Key Insight — Phase Relationships
Velocity leads position by $\pi/2$ (90°): when position is at a maximum, velocity is zero. Acceleration is exactly $\pi$ (180°) out of phase with position — when displacement is maximum positive, acceleration is maximum negative.
When an oscillating system is driven at its natural frequency $f_0 = \omega_0/(2\pi)$, the amplitude grows dramatically. This is resonance. It is why a child on a swing is pushed "in rhythm," and why soldiers break step crossing a bridge.
Exam Trap — Amplitude and Period
Changing the amplitude does not change the period or frequency. A mass on a spring oscillating with amplitude 5 cm has the same period as one with amplitude 50 cm (same $m$ and $k$).
应试陷阱 —— 振幅与周期
改变振幅不会改变周期或频率。$m$、$k$ 相同时，5 cm 与 50 cm 振幅的弹簧振子周期相同。
SHM Explorer — Mass on a SpringSHM 探究器 —— 弹簧振子
Adjust amplitude, spring constant, and mass. Watch position, velocity, and acceleration graphs update in real time with the energy bar.调整振幅、劲度系数与质量；位置、速度、加速度图与能量条会同步更新。
Phase Space: x vs v相空间：x 对 v
ω
20.0
rad/s
Period T周期 T
0.314
s
v_max
2.00
m/s
a_max
40.0
m/s²
E_total
1.00
J
Worked Example — Kinematics of SHM例题 —— SHM 的运动学
A 0.50 kg block on a spring (k = 200 N/m) is pulled 0.10 m from equilibrium and released from rest.
0.50 kg 的滑块系于弹簧（k = 200 N/m）， 从平衡位置拉离 0.10 m 后静止释放。
Position equation (released from x = A at t = 0)位置方程（在 $t = 0$、$x = A$ 处释放）
$$x(t) = 0.10\cos(20t)\;\text{m}$$
$$v(t) = -2.0\sin(20t)\;\text{m/s}$$
$$a(t) = -40\cos(20t)\;\text{m/s}^2$$
Worked Example — Energy Method: Vertical Spring-Mass (FRQ Style)例题 —— 能量方法：竖直弹簧振子（FRQ 风格）
A block of mass $m$ hangs from a vertical spring with constant $k$. Show that gravity shifts the equilibrium but does not change the oscillation frequency from the horizontal case.
$\omega = \sqrt{k/m}$ is identical to the horizontal case. Gravity shifts the equilibrium by $mg/k$ but does not appear in the frequency. ✓$\omega = \sqrt{k/m}$ 与水平情形完全相同。重力只是把平衡位置下移 $mg/k$，不进入频率表达。✓
This explains why you can use $\omega = \sqrt{k/m}$ for any spring orientation — a common AP simplification. ✓这正是 AP 常用简化的依据：弹簧无论取何方向都可直接用 $\omega = \sqrt{k/m}$。✓
An object in SHM has position $x(t) = 0.3\cos(4\pi t)$ (meters). What is the object's maximum speed?做 SHM 的物体位置为 $x(t) = 0.3\cos(4\pi t)$（米）。最大速率是多少？
$v_{\max} = A\omega$. Here $A = 0.3$ m and $\omega = 4\pi$ rad/s. Option (D) would be $A\omega^2$, which is the max acceleration.$v_{\max} = A\omega$，此处 $A = 0.3$ m、$\omega = 4\pi$ rad/s。选项 (D) 是 $A\omega^2$，即最大加速度。
Topic 7.4专题 7.4
Energy of Simple Harmonic Oscillators简谐振子的能量
Energy analysis provides a powerful alternative perspective on SHM. In an ideal SHM system (no friction), total mechanical energy is conserved — it sloshes back and forth between kinetic and potential forms.
The result is time-independent — the Pythagorean identity $\sin^2 + \cos^2 = 1$ is doing all the work.
Since $v_\text{max} = A\omega$, you can also write $E = \tfrac{1}{2}mv_\text{max}^2$ — equating these two forms gives the useful identity $\omega = \sqrt{k/m}$ on exam problems where you're handed $v_\text{max}$ and $A$ but not $k$.
推导 —— 为什么 $E = \tfrac{1}{2}kA^2$ 处处适用
两种切入方式都能得到同样答案，任选其一在 FRQ 上都能拿满分。
方法 1 —— 转折点
在 $x = \pm A$ 处物体瞬时静止，$K = 0$，能量全部为弹性势能：
$$E = K + U = 0 + \tfrac{1}{2}k A^2 = \tfrac{1}{2}kA^2$$
At any position $x$, speed can be found from conservation of energy:
在任意位置 $x$ 处，由能量守恒可求出速率：
Speed at Position x位置 $x$ 处的速率
$$v = \omega\sqrt{A^2 - x^2}$$
Key Insight — Energy Graphs
If you plot $K$ and $U$ vs. position $x$, you get two parabolas: $U = \frac{1}{2}kx^2$ (opening upward) and $K = \frac{1}{2}k(A^2 - x^2)$ (inverted). Their sum is the constant $E = \frac{1}{2}kA^2$. As functions of time, both oscillate as $\cos^2$ and $\sin^2$ at twice the oscillation frequency.
Exam Trap — Amplitude and Energy
Doubling the amplitude quadruples the total energy ($E \propto A^2$). Similarly, to double the energy, increase the amplitude by only $\sqrt{2}$.
Worked Example — Speed at a Given Position例题 —— 指定位置处的速率
A $2.0\;\text{kg}$ block oscillates on a spring with $k = 50\;\text{N/m}$ and amplitude $A = 0.40\;\text{m}$. Find the speed when $x = 0.20\;\text{m}$.
A block on a spring has total energy $E$. At what displacement from equilibrium is the kinetic energy equal to the potential energy?弹簧振子总能量为 $E$。在哪个位移处动能等于势能？
$x = A/2$
$x = A/4$
$x = 0$
$x = A/\sqrt{2}$
Correct! When $K = U$, each equals $E/2$. So $\frac{1}{2}kx^2 = \frac{1}{4}kA^2$, giving $x = A/\sqrt{2}$.正确！$K = U$ 时各占 $E/2$，故 $\frac{1}{2}kx^2 = \frac{1}{4}kA^2$，得 $x = A/\sqrt{2}$。
Set $U = K = E/2$: $\frac{1}{2}kx^2 = \frac{1}{4}kA^2$, so $x = A/\sqrt{2} \approx 0.707A$, not $A/2$.令 $U = K = E/2$：$\frac{1}{2}kx^2 = \frac{1}{4}kA^2$，得 $x = A/\sqrt{2} \approx 0.707A$，并非 $A/2$。
Worked Example — Damped Oscillations (Calculus, FRQ Style)例题 —— 阻尼振动（微积分，FRQ 风格）
A $m = 2.0\;\text{kg}$ block on a spring with $k = 50\;\text{N/m}$ experiences linear damping with coefficient $b = 4.0\;\text{kg/s}$. Determine the type of damping and the oscillation frequency.
Principle: the damped harmonic oscillator obeys $m\ddot{x} + b\dot{x} + kx = 0$, with three regimes set by the sign of $b^2 - 4mk$ — underdamped ($b^2 < 4mk$), critically damped ($b^2 = 4mk$), or overdamped ($b^2 > 4mk$).
AP context: Physics C: Mechanics rarely requires you to solve the damped ODE from scratch. Most exam questions are conceptual — classify the regime (underdamped / critical / overdamped) from the coefficients, explain why the amplitude decays, or relate energy loss to the damping force. Memorize the three conditions on $b^2$ vs. $4mk$ and you'll handle almost any damping prompt.
AP 背景：Physics C: Mechanics 很少要求你从头解阻尼微分方程。大多数题是概念题——由系数判断阻尼类型、解释振幅为何衰减、把能量损失与阻尼力联系起来。记住 $b^2$ 与 $4mk$ 的三种关系，几乎所有阻尼题都能搞定。
Topic 7.5专题 7.5
Simple and Physical Pendulums单摆与复摆
Pendulums provide a second major class of SHM systems, connecting oscillatory motion to rotational dynamics. While a mass–spring system involves linear restoring forces, pendulums involve restoring torques.
A simple pendulum (point mass $m$ on massless string of length $\ell$) at small angle $\theta$ has restoring torque $\tau = -mg\ell\sin\theta \approx -mg\ell\,\theta$. With $I = m\ell^2$:
A physical pendulum is any rigid body oscillating about a fixed pivot, with center of mass at distance $d$ from the pivot:
复摆（physical pendulum）指任意绕固定支点摆动的刚体，质心距支点 $d$：
Physical Pendulum — Restoring Torque复摆 —— 回复力矩
$$\tau = -mgd\sin\theta \approx -mgd\,\theta$$
$d$ = distance from pivot to center of mass
$d$ = 支点到质心的距离
Period of a Physical Pendulum复摆的周期
$$T_{\text{phys}} = 2\pi\sqrt{\frac{I}{mgd}}$$
$I$ = moment of inertia about the pivot, $d$ = distance from pivot to CM
$I$ = 对支点的转动惯量；$d$ = 支点到质心距离
Key Insight — Simple as a Special Case
A simple pendulum is a physical pendulum with $I = m\ell^2$ and $d = \ell$. The formula reduces to $T = 2\pi\sqrt{m\ell^2/(mg\ell)} = 2\pi\sqrt{\ell/g}$.
Equivalent to a simple pendulum of effective length $2L/3$.等效于一个有效长度 $2L/3$ 的单摆。
A physical pendulum has moment of inertia $I$ about its pivot, mass $m$, and $d$ is the pivot-to-CM distance. If the pivot is moved closer to the CM (decreasing $d$) while $I$ stays the same, the period will:复摆绕支点的转动惯量为 $I$、质量为 $m$、支点到质心距离为 $d$。若支点向质心靠近（$d$ 减小）而 $I$ 不变，周期会：
Decrease减小
Increase增大
Stay the same保持不变
Depend on the mass取决于质量
Correct! $T = 2\pi\sqrt{I/(mgd)}$. Decreasing $d$ makes $I/(mgd)$ larger, so $T$ increases. Less torque arm means weaker restoring torque and slower oscillation.正确！$T = 2\pi\sqrt{I/(mgd)}$。$d$ 减小使 $I/(mgd)$ 变大，故 $T$ 增大。力臂变小，回复力矩减弱，振动变慢。
Since $T \propto 1/\sqrt{d}$ (with $I$ held constant), reducing $d$ increases the period.在 $I$ 不变时 $T \propto 1/\sqrt{d}$，$d$ 减小 → $T$ 增大。
Quick Reference — Key Formulas公式速查 —— 关键公式
Spring Period:弹簧周期： $T_s = 2\pi\sqrt{m/k}$
Simple Pendulum:单摆： $T_p = 2\pi\sqrt{\ell/g}$
Physical Pendulum:复摆： $T = 2\pi\sqrt{I/(mgd)}$
SHM Diff. Eq.:SHM 微分方程： $\ddot{x} = -\omega^2 x$
Position:位置： $x = A\cos(\omega t + \phi)$
Velocity:速度： $v = -A\omega\sin(\omega t + \phi)$
Max Speed:最大速率： $v_{\max} = A\omega$
Total Energy:总能量： $E = \frac{1}{2}kA^2$
Worked Example — Physical Pendulum: Uniform Rod (FRQ Style)例题 —— 复摆：均匀杆（FRQ 风格）
A uniform thin rod of mass $M$ and length $L$ is pivoted about one end and swings as a pendulum for small angles. Derive the period and compare it to a simple pendulum of the same length.
质量 $M$、长度 $L$ 的均匀细杆以一端为支点作小角度摆动。推导其周期并与同长度的单摆作比较。
Identify明辨
Principle: $\tau_{\text{net}} = I\alpha$ with small-angle approximation $\sin\theta \approx \theta$ to obtain SHM
The rod pendulum swings ~18% faster than a simple pendulum of the same length.同长度下，杆摆比单摆快约 18%。
Evaluate校验
Physical intuition: The rod's mass is distributed along its length, with the center of mass at $L/2$, not at the tip. Effectively, the "average" bob is closer to the pivot → shorter effective length → shorter period. ✓物理直觉：杆的质量沿长度分布，质心位于 $L/2$ 而非末端。等效"平均球"更接近支点 → 等效长度更短 → 周期更短。✓
How Unit 7 Appears on the AP Exam第 7 单元在 AP 考试中的形式
MC
Multiple Choice — Common StylesMultiple Choice —— 常考题型
Ranking periods: Given different $m$, $k$, or $\ell$ values, rank systems by period. Remember: $T$ depends on $m$ and $k$ (spring) or $\ell$ and $g$ (pendulum), never on amplitude.
Experimental design: Measure $g$ with a pendulum or $k$ with a spring. Linearize: $T^2$ vs. $m$ or $T^2$ vs. $\ell$ gives a straight line whose slope yields the unknown.
Top Mistakes That Lose Points1. Confusing period with frequency — they are reciprocals.2. Thinking amplitude affects period — it does not for ideal SHM.3. Forgetting the small-angle condition for pendulums.4. Mixing up $v_{\max} = A\omega$ and $a_{\max} = A\omega^2$.5. Energy: $E \propto A^2$, not $A$. Doubling amplitude quadruples energy.6. Using wrong $I$ for physical pendulums — must be about the pivot, not the CM.7. Sign errors in the restoring force — missing the negative gives exponential, not sinusoidal.
Restoring force proportional to displacement: $F = -kx$ or $\tau = -\kappa\theta$回复力与位移成正比： $F = -kx$ 或 $\tau = -\kappa\theta$
Period of a mass–spring?弹簧振子的周期？
$$T = 2\pi\sqrt{m/k}$$
Independent of amplitude.与振幅无关。
Period of a simple pendulum?单摆的周期？
$$T = 2\pi\sqrt{\ell/g}$$
Small-angle limit; independent of mass.小角度极限；与质量无关。
$v_\text{max}$ in SHM?SHM 中的 $v_\text{max}$？
$$v_\text{max} = A\omega$$
Occurs at equilibrium ($x = 0$).出现在平衡位置（$x = 0$）。
Total energy of a spring oscillator?弹簧振子的总能量？
$$E = \tfrac{1}{2}kA^2$$
Constant; equals $K + U$ at every instant.恒定；任何时刻都等于 $K + U$。
SHM differential equation?SHM 微分方程？
$$\ddot{x} = -\omega^2 x$$
Solution: $x(t) = A\cos(\omega t + \phi)$.解为 $x(t) = A\cos(\omega t + \phi)$。
Physical pendulum period?复摆周期？
$$T = 2\pi\sqrt{\frac{I}{mgd}}$$
$I$ about the pivot, $d$ = pivot to CM.$I$ 绕支点；$d$ 为支点到质心距离。
What is resonance?什么是共振？
Amplitude increases dramatically when driving frequency matches the system's natural frequency.当驱动频率与系统固有频率相同时，振幅会大幅增大。
Assessment测评
Unit 7 — Practice Quiz第 7 单元 —— 练习测验
1. A mass on a spring oscillates with period $T$. If both the mass and the spring constant are doubled, the new period is:弹簧振子周期为 $T$。若质量和劲度系数都加倍，新周期为：
2. An object in SHM has position $x = 0.5\cos(10t)$ m. At time $t = \pi/20$ s, the velocity is:做 SHM 的物体位置为 $x = 0.5\cos(10t)$ m。在 $t = \pi/20$ s 时速度为：
3. A simple pendulum of length $L$ has period $T$ on Earth. On a planet where $g$ is four times Earth's, the period would be:长度 $L$ 的单摆在地球上的周期为 $T$。在 $g$ 为地球四倍的行星上周期为：
4. A block oscillates with amplitude $A$ and total energy $E$. If the amplitude is tripled to $3A$, the new energy is:振子振幅 $A$、总能量 $E$。若振幅变为 $3A$，新能量为：
5. A uniform disk of mass $M$ and radius $R$ is pivoted at its rim. Using parallel-axis theorem ($I = \frac{3}{2}MR^2$) and $d = R$, its period is:质量 $M$、半径 $R$ 的均匀圆盘以盘缘为支点摆动。由平行轴定理 $I = \frac{3}{2}MR^2$、$d = R$，其周期为：