AP Physics C: Mechanics, Unit 7: Oscillations, Solutions

Q1EASY7.2 Mass-Spring Period7.2 弹簧质量系统周期Calculator

$m = 0.50~\mathrm{kg}$, $k = 200~\mathrm{N/m}$. SHM period?$m = 0.50~\mathrm{kg}$，$k = 200~\mathrm{N/m}$，求简谐运动周期。

(A) $0.10~\mathrm{s}$
(B) $0.16~\mathrm{s}$
(C) $0.31~\mathrm{s}$
(D) $0.63~\mathrm{s}$

Answer:答案： (C)

$$T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.50/200} = 2\pi(0.050) \approx 0.314~\mathrm{s}$$

Q2EASY7.5 Simple Pendulum Period7.5 单摆周期Calculator

Simple pendulum, $L = 1.0~\mathrm{m}$, $g = 9.8~\mathrm{m/s^2}$.单摆，$L = 1.0~\mathrm{m}$，$g = 9.8~\mathrm{m/s^2}$。

(A) $1.0~\mathrm{s}$
(B) $2.0~\mathrm{s}$
(C) $4.0~\mathrm{s}$
(D) $6.3~\mathrm{s}$

Answer:答案： (B)

$$T = 2\pi\sqrt{L/g} = 2\pi\sqrt{1.0/9.8} \approx 2.01~\mathrm{s}$$

The classic 1-m seconds-pendulum result.这是经典的1米秒摆结果。

Q3EASY7.1 Defining SHM7.1 简谐运动定义No Calculator

An object undergoes simple harmonic motion when the net force on it is当作用在物体上的合力满足以下条件时，物体做简谐运动：

(A) constant in magnitude and direction大小和方向均恒定
(B) proportional to its speed and opposite in direction与速度成正比且方向相反
(C) proportional to its displacement from equilibrium and opposite in direction与偏离平衡位置的位移成正比且方向相反
(D) proportional to the square of its displacement from equilibrium与偏离平衡位置的位移的平方成正比

Answer:答案： (C)

SHM ⇔ $F = -kx$ (linear restoring force). Trap (B) is the damping law; (D) gives anharmonic motion.简谐运动 ⇔ $F = -kx$（线性回复力）。选项(B)是阻尼规律；选项(D)给出非简谐运动。

Q4EASY7.4 Energy at Amplitude7.4 振幅处的能量No Calculator

Block on a spring at $x = +A$. Its energy is弹簧上的滑块位于 $x = +A$ 处，其能量为

(A) entirely kinetic全部为动能
(B) entirely potential全部为势能
(C) half kinetic and half potential动能与势能各占一半
(D) zero (the spring is fully relaxed)零（弹簧完全松弛）

Answer:答案： (B)

At $x = \pm A$ the block is momentarily at rest (turning point), so $K = 0$ and $E = U = \tfrac{1}{2}kA^2$. Trap (D) confuses "no motion" with "no energy", the spring is fully stretched here, not relaxed.在 $x = \pm A$ 处，滑块瞬间静止（转折点），故 $K = 0$，$E = U = \tfrac{1}{2}kA^2$。选项(D)将"无运动"误认为"无能量"，此处弹簧是完全伸长的，而非松弛。

Q5MEDIUM7.3 Initial Conditions7.3 初始条件No Calculator

$x(t) = A\cos(\omega t)$. Velocity at $t = 0$?$x(t) = A\cos(\omega t)$，求 $t = 0$ 时的速度。

(A) $0$
(B) $A$
(C) $A\omega$
(D) $-A\omega$

Answer:答案： (A)

$$v(t) = \dot x = -A\omega\sin(\omega t),\quad v(0) = -A\omega\sin 0 = 0$$

At $t = 0$ the particle is at $x = A$ (turning point), naturally at rest.在 $t = 0$ 时，质点位于 $x = A$（转折点），自然处于静止状态。

Q6MEDIUM7.4 Maximum Speed7.4 最大速度Calculator

$m = 0.20~\mathrm{kg}$, $k = 50~\mathrm{N/m}$, $A = 0.10~\mathrm{m}$. Maximum speed?$m = 0.20~\mathrm{kg}$，$k = 50~\mathrm{N/m}$，$A = 0.10~\mathrm{m}$，求最大速度。

(A) $0.5~\mathrm{m/s}$
(B) $1.0~\mathrm{m/s}$
(C) $1.6~\mathrm{m/s}$
(D) $2.5~\mathrm{m/s}$

Answer:答案： (C)

$$v_\max = A\omega = A\sqrt{k/m} = 0.10\sqrt{50/0.20} = 0.10\sqrt{250} \approx 1.58~\mathrm{m/s}$$

Q7MEDIUM7.5 Pendulum Length Scaling7.5 单摆摆长缩放No Calculator

Pendulum length increases by $4$×. New period vs. original?摆长增大为原来的 $4$ 倍，新周期与原周期的关系？

(A) $T/2$
(B) $T$
(C) $2T$
(D) $4T$

Answer:答案： (C)

$T \propto \sqrt{L}$, so $L \to 4L$ gives $T \to 2T$. Trap (D) takes the ratio linearly.$T \propto \sqrt{L}$，所以 $L \to 4L$ 时，$T \to 2T$。选项(D)错误地将比例关系视为线性。

Q8MEDIUM7.2 Mass-Spring Frequency7.2 弹簧质量系统频率Calculator

$m = 0.40~\mathrm{kg}$, $k = 100~\mathrm{N/m}$. Frequency?$m = 0.40~\mathrm{kg}$，$k = 100~\mathrm{N/m}$，求频率。

(A) $0.4~\mathrm{Hz}$
(B) $1.6~\mathrm{Hz}$
(C) $2.5~\mathrm{Hz}$
(D) $6.3~\mathrm{Hz}$

Answer:答案： (C)

$$f = \frac{1}{2\pi}\sqrt{k/m} = \frac{1}{2\pi}\sqrt{250} \approx 2.52~\mathrm{Hz}$$

Trap (D) reports $\omega = \sqrt{k/m}$ instead of $f$.选项(D)给出的是 $\omega = \sqrt{k/m}$ 而非频率 $f$。

Q9MEDIUM7.4 Energy Fraction at $x = A/2$7.4 $x = A/2$ 处的能量比例No Calculator

Fraction of $E$ that is kinetic at $x = A/2$?在 $x = A/2$ 处，$E$ 中动能所占的比例？

(A) $1/4$
(B) $1/2$
(C) $3/4$
(D) $7/8$

Answer:答案： (C)

$$\frac{U(A/2)}{E} = \frac{\tfrac{1}{2}k(A/2)^2}{\tfrac{1}{2}kA^2} = \tfrac{1}{4}$$

So $K/E = 1 - \tfrac{1}{4} = \tfrac{3}{4}$.故 $K/E = 1 - \tfrac{1}{4} = \tfrac{3}{4}$。

Q10MEDIUM7.5 Physical Pendulum (Rod)7.5 实体摆（均匀杆）No Calculator

Uniform rod $M$, $L$, pivoted at one end. Period of small-amplitude swing?均匀杆，质量 $M$，长度 $L$，绕一端转动，求小振幅摆动周期。

(A) $2\pi\sqrt{\dfrac{L}{g}}$
(B) $2\pi\sqrt{\dfrac{L}{2g}}$
(C) $2\pi\sqrt{\dfrac{2L}{3g}}$
(D) $2\pi\sqrt{\dfrac{3L}{2g}}$

Answer:答案： (C)

$T = 2\pi\sqrt{I/(Mgd)}$ with $I = \tfrac{1}{3}ML^2$ and $d = L/2$:$T = 2\pi\sqrt{I/(Mgd)}$，其中 $I = \tfrac{1}{3}ML^2$，$d = L/2$：

$$T = 2\pi\sqrt{\frac{\tfrac{1}{3}ML^2}{Mg(L/2)}} = 2\pi\sqrt{\frac{2L}{3g}}$$

Note this is faster than the simple pendulum of length $L$ (since $2/3 < 1$), the mass is distributed closer to the pivot.注意，这比摆长为 $L$ 的单摆更快（因为 $2/3 < 1$），因为质量分布更靠近转轴。

Q11MEDIUM7.3 Phase Constant7.3 初相位No Calculator

$x(t) = A\cos(\omega t + \phi)$ with $x(0) = 0$ and $\dot x(0) > 0$. Phase $\phi$?$x(t) = A\cos(\omega t + \phi)$，且 $x(0) = 0$，$\dot x(0) > 0$，求初相位 $\phi$。

(A) $0$
(B) $+\pi/2$
(C) $-\pi/2$
(D) $\pi$

Answer:答案： (C)

$x(0) = A\cos\phi = 0 \Rightarrow \phi = \pm\pi/2$. Then $v(0) = -A\omega\sin\phi > 0 \Rightarrow \sin\phi < 0 \Rightarrow \phi = -\pi/2$.$x(0) = A\cos\phi = 0 \Rightarrow \phi = \pm\pi/2$。又 $v(0) = -A\omega\sin\phi > 0 \Rightarrow \sin\phi < 0 \Rightarrow \phi = -\pi/2$。

Q12MEDIUM7.4 Speed at Equilibrium7.4 平衡位置处的速度No Calculator

Speed at the equilibrium position $x = 0$ in SHM (amplitude $A$, angular frequency $\omega$)?简谐运动（振幅 $A$，角频率 $\omega$）在平衡位置 $x = 0$ 处的速度？

(A) $0$
(B) $A\omega$
(C) $kA$
(D) $A/\omega$

Answer:答案： (B)

At $x = 0$ all energy is kinetic: $\tfrac{1}{2}m v_\max^2 = \tfrac{1}{2}kA^2 \Rightarrow v_\max = A\sqrt{k/m} = A\omega$. Trap (C) has wrong units (N, not m/s).在 $x = 0$ 处，所有能量均为动能：$\tfrac{1}{2}m v_\max^2 = \tfrac{1}{2}kA^2 \Rightarrow v_\max = A\sqrt{k/m} = A\omega$。选项(C)单位错误（N，而非 m/s）。

Q13MEDIUM7.5 Pendulum on the Moon7.5 月球上的单摆No Calculator

Pendulum: $T_\mathrm{E}$ on Earth ($g$). Moon's $g_\mathrm{M} = g/6$. New period?单摆在地球上的周期为 $T_\mathrm{E}$（重力加速度 $g$），月球上 $g_\mathrm{M} = g/6$，求新周期。

(A) $T_\mathrm{E}/6$
(B) $T_\mathrm{E}/\sqrt{6}$
(C) $T_\mathrm{E}\sqrt{6}$
(D) $6\,T_\mathrm{E}$

Answer:答案： (C)

$T \propto 1/\sqrt{g}$, so $T_\mathrm{M}/T_\mathrm{E} = \sqrt{g/(g/6)} = \sqrt{6}$. The pendulum swings slower on the Moon.$T \propto 1/\sqrt{g}$，故 $T_\mathrm{M}/T_\mathrm{E} = \sqrt{g/(g/6)} = \sqrt{6}$。单摆在月球上摆动更慢。

Q14HARD7.4 Where KE Equals PE7.4 动能等于势能的位置No Calculator

Position where $K = U$ in SHM (amplitude $A$)?简谐运动（振幅 $A$）中动能等于势能的位置？

(A) $x = A/2$
(B) $x = A/\sqrt{2}$
(C) $x = A\sqrt{3}/2$
(D) $x = A$

Answer:答案： (B)

$K = U \Rightarrow U = E/2 \Rightarrow \tfrac{1}{2}k x^2 = \tfrac{1}{2}\bigl(\tfrac{1}{2}kA^2\bigr) \Rightarrow x^2 = A^2/2 \Rightarrow x = A/\sqrt{2} \approx 0.707\,A$. Trap (A) is where $U = E/4$; (C) is where $U = 3E/4$.$K = U \Rightarrow U = E/2 \Rightarrow \tfrac{1}{2}k x^2 = \tfrac{1}{2}\bigl(\tfrac{1}{2}kA^2\bigr) \Rightarrow x^2 = A^2/2 \Rightarrow x = A/\sqrt{2} \approx 0.707\,A$。选项(A)对应 $U = E/4$ 处；选项(C)对应 $U = 3E/4$ 处。

Q15HARD7.5 Numerical Physical Pendulum7.5 实体摆数值计算Calculator

Rod $L = 1.5~\mathrm{m}$ pivoted at one end; small-amplitude period?长 $L = 1.5~\mathrm{m}$ 的均匀杆绕一端转动，求小振幅周期。

(A) $1.6~\mathrm{s}$
(B) $2.0~\mathrm{s}$
(C) $2.5~\mathrm{s}$
(D) $3.1~\mathrm{s}$

Answer:答案： (B)

$$T = 2\pi\sqrt{\frac{2L}{3g}} = 2\pi\sqrt{\frac{3.0}{29.4}} \approx 2.01~\mathrm{s}$$

Q16HARD7.3 Speed at $x = A/2$7.3 $x = A/2$ 处的速度No Calculator

$v(A/2) / v_\max$?求 $v(A/2) / v_\max$。

(A) $1/2$
(B) $\sqrt{3}/2$
(C) $3/4$
(D) $1$

Answer:答案： (B)

$v(x) = \omega\sqrt{A^2 - x^2}$ from energy conservation. At $x = A/2$:由能量守恒，$v(x) = \omega\sqrt{A^2 - x^2}$。在 $x = A/2$ 处：

$$\frac{v}{v_\max} = \frac{\sqrt{A^2 - A^2/4}}{A} = \sqrt{3/4} = \tfrac{\sqrt{3}}{2} \approx 0.866$$

Reciprocal of the position-to-amplitude ratio in some sense, this is also the cosine identity since $K/E = 3/4$.这在某种意义上是位置振幅比的余角对应值，也与 $K/E = 3/4$ 的余弦恒等式一致。

Q17HARD7.4 Total Energy Expressions7.4 总能量表达式No Calculator

Statements I. $E = \tfrac{1}{2}kA^2$ and II. $E = \tfrac{1}{2}m\omega^2 A^2$. Which are correct?命题 I. $E = \tfrac{1}{2}kA^2$，命题 II. $E = \tfrac{1}{2}m\omega^2 A^2$，哪些正确？

(A) I only仅 I
(B) II only仅 II
(C) Both I and III 和 II 均正确
(D) Neither均不正确

Answer:答案： (C)

Statement I: $E$ at $x = \pm A$ is all PE, $\tfrac{1}{2}kA^2$. Statement II uses $\omega^2 = k/m$:命题 I：在 $x = \pm A$ 处，$E$ 全为势能，$\tfrac{1}{2}kA^2$。命题 II 利用 $\omega^2 = k/m$：

$$\tfrac{1}{2}m\omega^2 A^2 = \tfrac{1}{2}m(k/m)A^2 = \tfrac{1}{2}kA^2$$

Identical expressions.两式等价。

Q18HARD7.1 Differential Equation of SHM7.1 简谐运动微分方程No Calculator

$m\,\ddot x + kx = 0$. Angular frequency?$m\,\ddot x + kx = 0$，求角频率。

(A) $\dfrac{k}{m}$
(B) $\sqrt{\dfrac{k}{m}}$
(C) $2\pi\sqrt{\dfrac{k}{m}}$
(D) $\dfrac{m}{k}$

Answer:答案： (B)

Write as $\ddot x + (k/m) x = 0$, the canonical SHM form $\ddot x + \omega^2 x = 0$, so $\omega = \sqrt{k/m}$. Trap (A) is $\omega^2$; (C) is $\omega\cdot 2\pi$ (wrong units).改写为 $\ddot x + (k/m) x = 0$，即标准简谐运动形式 $\ddot x + \omega^2 x = 0$，故 $\omega = \sqrt{k/m}$。选项(A)是 $\omega^2$；选项(C)是 $\omega\cdot 2\pi$（单位错误）。

FRQ 1MEDIUM7.2 / 7.4 Mass-Spring SHM7.2 / 7.4 弹簧质量简谐运动Calculator

$m = 0.40~\mathrm{kg}$, $k = 100~\mathrm{N/m}$, $A = 0.10~\mathrm{m}$, frictionless horizontal.$m = 0.40~\mathrm{kg}$，$k = 100~\mathrm{N/m}$，$A = 0.10~\mathrm{m}$，水平无摩擦面。

(a) $\omega$, $T$, $f$:求 $\omega$、$T$、$f$：

$$\omega = \sqrt{k/m} = \sqrt{100/0.40} = \sqrt{250} \approx 15.8~\mathrm{rad/s}$$

$$T = \frac{2\pi}{\omega} \approx 0.397~\mathrm{s},\qquad f = \frac{1}{T} \approx 2.52~\mathrm{Hz}$$

(b) Max speed and max acceleration:最大速度和最大加速度：

$$v_\max = A\omega = 0.10(15.8) \approx 1.58~\mathrm{m/s}$$

$$a_\max = A\omega^2 = 0.10(250) = 25~\mathrm{m/s^2}$$

(c) Total energy:总能量：

$$E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}(100)(0.01) = 0.50~\mathrm{J}$$

(d) At $x = 0.05~\mathrm{m}$ (i.e. $x = A/2$):在 $x = 0.05~\mathrm{m}$（即 $x = A/2$）处：

$$U = \tfrac{1}{2}k x^2 = \tfrac{1}{2}(100)(0.0025) = 0.125~\mathrm{J}$$

$$K = E - U = 0.500 - 0.125 = 0.375~\mathrm{J}$$

$K/E = 3/4$, matching Q9's analytic ratio.$K/E = 3/4$，与 Q9 的解析比例一致。

FRQ 2MEDIUM7.3 Phase Analysis7.3 相位分析Calculator

$x(t) = 0.20\cos(4 t + \pi/3)~\mathrm{m}$.$x(t) = 0.20\cos(4 t + \pi/3)~\mathrm{m}$。

(a) Read off parameters: $A = 0.20~\mathrm{m}$, $\omega = 4~\mathrm{rad/s}$, $T = 2\pi/\omega = \pi/2 \approx 1.57~\mathrm{s}$, $\phi = \pi/3$.直接读出各参数：$A = 0.20~\mathrm{m}$，$\omega = 4~\mathrm{rad/s}$，$T = 2\pi/\omega = \pi/2 \approx 1.57~\mathrm{s}$，$\phi = \pi/3$。

(b) Derivatives:求导：

$$v(t) = -A\omega\sin(\omega t + \phi) = -0.80\sin(4 t + \pi/3)~\mathrm{m/s}$$

$$a(t) = -A\omega^2\cos(\omega t + \phi) = -3.2\cos(4 t + \pi/3)~\mathrm{m/s^2}$$

Note $a(t) = -\omega^2 x(t)$.注意 $a(t) = -\omega^2 x(t)$。

(c) At $t = 0$ ($\cos(\pi/3) = 0.5$, $\sin(\pi/3) = \tfrac{\sqrt 3}{2}$):在 $t = 0$ 时（$\cos(\pi/3) = 0.5$，$\sin(\pi/3) = \tfrac{\sqrt 3}{2}$）：

$$x(0) = 0.20(0.5) = 0.10~\mathrm{m}$$

$$v(0) = -0.80(0.866) \approx -0.69~\mathrm{m/s}$$

$$a(0) = -3.2(0.5) = -1.6~\mathrm{m/s^2}$$

(d) First $t > 0$ with $x = 0$: $\cos(4t + \pi/3) = 0 \Rightarrow 4 t + \pi/3 = \pi/2 + n\pi$.第一个 $t > 0$ 时 $x = 0$：$\cos(4t + \pi/3) = 0 \Rightarrow 4 t + \pi/3 = \pi/2 + n\pi$。

$$4 t = \pi/2 - \pi/3 = \pi/6 \;\Longrightarrow\; t = \pi/24 \approx 0.131~\mathrm{s}$$

FRQ 3HARD7.5 Physical Pendulum7.5 实体摆Calculator

Rod $M = 0.50~\mathrm{kg}$, $L = 1.20~\mathrm{m}$ pivoted at one end, released from $\theta_0 = 30^\circ$ off vertical.均匀杆 $M = 0.50~\mathrm{kg}$，$L = 1.20~\mathrm{m}$，绕一端转动，从偏离竖直方向 $\theta_0 = 30^\circ$ 处由静止释放。

(a) Torque about the pivot from gravity (acting at the rod's CM at distance $L/2$): $\tau = -Mg(L/2)\sin\theta$. For small $\theta$, $\sin\theta \approx \theta$:重力对转轴的力矩（重力作用于距轴 $L/2$ 处的质心）：$\tau = -Mg(L/2)\sin\theta$。小角度近似 $\sin\theta \approx \theta$：

$$I\ddot\theta = -M g\,(L/2)\,\theta,\quad I = \tfrac{1}{3}ML^2$$

$$\ddot\theta = -\frac{Mg(L/2)}{(1/3)ML^2}\,\theta = -\frac{3g}{2L}\,\theta$$

Canonical SHM with $\omega^2 = 3g/(2L)$, so这是标准简谐运动，$\omega^2 = 3g/(2L)$，故

$$T = 2\pi\sqrt{\frac{2L}{3g}}$$

(b) Plug numbers:代入数值：

$$T = 2\pi\sqrt{\frac{2(1.20)}{3(9.8)}} = 2\pi\sqrt{0.0816} \approx 1.79~\mathrm{s}$$

(c) Exact energy conservation: the rod's CM falls by $(L/2)(1 - \cos\theta_0)$ as $\theta$ goes from $\theta_0$ to $0$.精确能量守恒：当 $\theta$ 从 $\theta_0$ 变为 $0$ 时，质心下降 $(L/2)(1 - \cos\theta_0)$。

$$M g\,(L/2)(1 - \cos\theta_0) = \tfrac{1}{2}I\omega^2 = \tfrac{1}{6}ML^2\omega^2$$

$$\omega^2 = \frac{3g(1 - \cos\theta_0)}{L} = \frac{3(9.8)(1 - 0.866)}{1.20} \approx 3.28~\mathrm{rad^2/s^2}$$

$$\omega \approx 1.81~\mathrm{rad/s}$$

(d) Small-angle prediction $\omega_\max = \theta_0\sqrt{3g/(2L)}$ with $\theta_0 = \pi/6 \approx 0.524~\mathrm{rad}$:小角度近似预测 $\omega_\max = \theta_0\sqrt{3g/(2L)}$，其中 $\theta_0 = \pi/6 \approx 0.524~\mathrm{rad}$：

$$\omega_\max^\mathrm{SHM} = 0.524\sqrt{3(9.8)/(2 \cdot 1.20)} = 0.524(3.50) \approx 1.83~\mathrm{rad/s}$$

Percent difference vs. exact: $(1.83 - 1.81)/1.81 \approx 1.1\%$. The small-angle approximation is excellent here, the $\theta - \sin\theta$ error at $30^\circ$ is just under 5% in the force, but the integrated motion happens to wash out into a $\sim 1\%$ peak-speed error.与精确值的百分比偏差：$(1.83 - 1.81)/1.81 \approx 1.1\%$。小角度近似在此处表现优异，$30^\circ$ 时 $\theta - \sin\theta$ 误差在力上不足5%，但积分运动后峰值速度误差仅约1%。

FRQ 4HARD7.3 / 7.4 Vertical Spring7.3 / 7.4 竖直弹簧Calculator

$m = 0.30~\mathrm{kg}$, $k = 60~\mathrm{N/m}$ vertical spring; block pulled down $A = 0.10~\mathrm{m}$ below the new equilibrium and released from rest.$m = 0.30~\mathrm{kg}$，$k = 60~\mathrm{N/m}$ 竖直弹簧；滑块从新平衡位置向下拉 $A = 0.10~\mathrm{m}$ 后由静止释放。

(a) New equilibrium below natural length:新平衡位置在自然长度以下：

$$k x_0 = m g \;\Longrightarrow\; x_0 = \frac{m g}{k} = \frac{(0.30)(9.8)}{60} \approx 0.049~\mathrm{m}$$

(b) Let $y$ be displacement from the new equilibrium (downward positive). Net force at displacement $y$:设 $y$ 为从新平衡位置向下的位移。在位移 $y$ 处的合力：

$$F_\text{net} = -k(x_0 + y) + m g = -k y$$

since $k x_0 = m g$. Newton's second law gives $m\ddot y = -k y$, canonical SHM:由于 $k x_0 = m g$。由牛顿第二定律得 $m\ddot y = -k y$，这是标准简谐运动：

$$\omega = \sqrt{k/m} = \sqrt{60/0.30} = \sqrt{200} \approx 14.1~\mathrm{rad/s}$$

$$T = \frac{2\pi}{\omega} \approx 0.444~\mathrm{s}$$

The presence of gravity just shifts the equilibrium; it doesn't change $\omega$.重力的存在只是移动了平衡位置，并不改变 $\omega$。

(c) Max speed and acceleration:最大速度和加速度：

$$v_\max = A\omega = 0.10(14.1) \approx 1.41~\mathrm{m/s}$$

$$a_\max = A\omega^2 = 0.10(200) = 20~\mathrm{m/s^2}$$

(d) $K = 3U \Rightarrow U = E/4$ (since $K + U = E$ and $K = 3U \Rightarrow 4U = E$).$K = 3U \Rightarrow U = E/4$（由 $K + U = E$ 和 $K = 3U \Rightarrow 4U = E$）。

$$\tfrac{1}{2}k y^2 = \tfrac{1}{4}\,\tfrac{1}{2}kA^2 \;\Longrightarrow\; y^2 = A^2/4 \;\Longrightarrow\; y = \pm A/2 = \pm 0.050~\mathrm{m}$$

Both signs are reached during the oscillation.振动过程中正负两个位移都会达到。

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