Each item restates the prompt and choices, marks the correct letter, and gives a brief justification. Numbers use $g = 9.8~\mathrm{m/s^2}$.每道题重述题干和选项，标出正确答案，并给出简要说明。数值计算使用 $g = 9.8~\mathrm{m/s^2}$。
Insight.要点。The mass-spring period is $T=2\pi\sqrt{m/k}$. It increases with mass and decreases with stiffness; amplitude does not appear for ideal SHM.弹簧振子周期为 $T=2\pi\sqrt{m/k}$。周期随质量增大、随劲度系数增大而减小；理想简谐运动中振幅不出现。
The classic 1-m seconds-pendulum result.这是经典的1米秒摆结果。
Insight.要点。A simple pendulum has $T=2\pi\sqrt{L/g}$, independent of mass and amplitude in the small-angle limit. The one-meter pendulum is the classic two-second period.单摆周期为 $T=2\pi\sqrt{L/g}$，小角度极限下与质量和振幅无关。一米单摆是经典的两秒周期。
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
Insight.要点。Simple harmonic motion is defined by a linear restoring force $F=-kx$. The minus sign means the force always points back toward equilibrium, which is what produces oscillation.简谐运动由线性回复力 $F=-kx$ 定义。负号表示力始终指向平衡位置，这正是产生振动的根源。
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。
Insight.要点。At the turning points $x=\pm A$, speed is zero, so all energy is potential: $E=\tfrac12 kA^2$. Zero speed does not mean zero energy; the spring is maximally stretched or compressed.在转折点 $x=\pm A$ 处速度为零，故全部能量都是势能：$E=\tfrac12 kA^2$。速度为零不代表能量为零；弹簧处于最大伸长或压缩状态。
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。
Insight.要点。For $x=A\cos(\omega t)$, velocity is the derivative $-A\omega\sin(\omega t)$. At $t=0$ the particle is at the turning point, so the velocity is naturally zero.对 $x=A\cos(\omega t)$，速度是其导数 $-A\omega\sin(\omega t)$。$t=0$ 时粒子位于转折点，速度自然为零。
Insight.要点。Maximum speed occurs at equilibrium and equals $A\omega=A\sqrt{k/m}$. The mass appears under a square root, and the result must have units of speed.最大速度出现在平衡位置，等于 $A\omega=A\sqrt{k/m}$。质量在根号下，且结果必须具有速度单位。
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。
Insight.要点。Pendulum period scales as $\sqrt{L}$, so quadrupling the length doubles the period. A square-root dependence is not linear; do not multiply by four.单摆周期与 $\sqrt{L}$ 成正比，所以摆长变为四倍时周期变为两倍。这是平方根关系，不是线性关系；不要乘四。
Insight.要点。Frequency and angular frequency are related by $f=\omega/(2\pi)$. Reporting $\omega=\sqrt{k/m}$ as frequency omits the $2\pi$ and gives the wrong units.频率与角频率的关系是 $f=\omega/(2\pi)$。把 $\omega=\sqrt{k/m}$ 当作频率会漏掉 $2\pi$，且单位错误。
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？
Insight.要点。At $x=A/2$, potential energy is one quarter of the total because $U\propto x^2$. Therefore kinetic energy is three quarters of the total, not half.在 $x=A/2$ 处，由于 $U\propto x^2$，势能为总能量的四分之一。因此动能为总能量的四分之三，而不是二分之一。
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。
Insight.要点。A rod pivoted at one end is a physical pendulum: $T=2\pi\sqrt{I/(Mgd)}$. With $I=\tfrac13ML^2$ and $d=L/2$, the period is $2\pi\sqrt{2L/(3g)}$, shorter than a simple pendulum of length $L$.绕一端转动的杆是物理摆：$T=2\pi\sqrt{I/(Mgd)}$。代入 $I=\tfrac13ML^2$、$d=L/2$，得 $T=2\pi\sqrt{2L/(3g)}$，比摆长为 $L$ 的单摆更短。
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 $\phi$。
Insight.要点。The phase constant is fixed by both initial position and the sign of initial velocity. Starting at $x=0$ with positive velocity selects $\phi=-\pi/2$ for the cosine form.初相位由初始位置和初速度的符号共同确定。余弦形式下，从 $x=0$ 出发且初速度为正时，应取 $\phi=-\pi/2$。
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）。
Insight.要点。At equilibrium all energy is kinetic, so $v_{\max}=A\omega$. The answer must be a speed; any expression with force units fails dimensional analysis.在平衡位置全部能量为动能，故 $v_{\max}=A\omega$。答案必须是速度；任何具有力的单位的表达式都通不过量纲分析。
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。
Insight.要点。Pendulum period scales as $1/\sqrt{g}$. With Moon gravity $g/6$, the period grows by $\sqrt6$; the pendulum swings more slowly.单摆周期与 $1/\sqrt{g}$ 成正比。月球重力为 $g/6$ 时，周期变为原来的 $\sqrt6$ 倍；摆得更慢。
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。
Insight.要点。Set $K=U$ by writing $U=E/2$, then solve $\tfrac12 kx^2=\tfrac12(\tfrac12 kA^2)$. The location is $A/\sqrt2$, not $A/2$.令 $K=U$ 等价于 $U=E/2$，再解 $\tfrac12 kx^2=\tfrac12(\tfrac12 kA^2)$。位置是 $A/\sqrt2$，不是 $A/2$。
Insight.要点。A physical pendulum is evaluated with the same formula as any rigid oscillator. Compute the rod's end moment of inertia and the CM distance to the pivot before substituting.物理摆与任何刚体振子使用同一公式。代入前先计算杆绕端点的转动惯量和质心到转轴的距离。
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。
Insight.要点。Energy conservation gives $v(x)=\omega\sqrt{A^2-x^2}$. At $x=A/2$, the ratio to $v_{\max}$ is $\sqrt{3}/2$, consistent with three quarters of the energy being kinetic.能量守恒给出 $v(x)=\omega\sqrt{A^2-x^2}$。在 $x=A/2$ 处，与 $v_{\max}$ 之比为 $\sqrt{3}/2$，与四分之三能量为动能一致。
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 $\omega^2 = k/m$：
Insight.要点。The two total-energy expressions are identical because $\omega^2=k/m$. Evaluating at the amplitude gives $\tfrac12 kA^2$; using the velocity amplitude gives $\tfrac12 m\omega^2A^2$.总能量的两个表达式是等价的，因为 $\omega^2=k/m$。在振幅处计算得 $\tfrac12 kA^2$；用速度振幅则得 $\tfrac12 m\omega^2A^2$。
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）。
Insight.要点。Rewrite $m\ddot x+kx=0$ as $\ddot x+(k/m)x=0$, which is the canonical $\ddot x+\omega^2 x=0$. Therefore $\omega=\sqrt{k/m}$, not $\omega^2$.把 $m\ddot x+kx=0$ 改写为 $\ddot x+(k/m)x=0$，即标准形式 $\ddot x+\omega^2 x=0$。因此 $\omega=\sqrt{k/m}$，而不是 $\omega^2$。
Insight.要点。One mass-spring system yields all the SHM relationships: $\omega$, $T$, $f$, $v_{\max}$, $a_{\max}$, and the energy split at any $x$. Compute them from $k$, $m$, and $A$ in that order.一个弹簧振子系统包含全部简谐运动关系：$\omega$、$T$、$f$、$v_{\max}$、$a_{\max}$ 以及任意 $x$ 处的能量分配。按 $k$、$m$、$A$ 的顺序依次计算。
FRQ 2MEDIUM7.3 Phase Analysis7.3 相位分析Calculator
$x(t) = 0.20\cos(4 t + \pi/3)~\mathrm{m}$.$x(t) = 0.20\cos(4 t + \pi/3)~\mathrm{m}$。
$$4 t = \pi/2 - \pi/3 = \pi/6 \;\Longrightarrow\; t = \pi/24 \approx 0.131~\mathrm{s}$$
Insight.要点。Read the parameters directly from $x=A\cos(\omega t+\phi)$, then differentiate to get velocity and acceleration. The acceleration always satisfies $a=-\omega^2 x$ in SHM.从 $x=A\cos(\omega t+\phi)$ 直接读出参数，再求导得速度和加速度。简谐运动中加速度始终满足 $a=-\omega^2 x$。
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 $\sin\theta \approx \theta$：
$$I\ddot\theta = -M g\,(L/2)\,\theta,\quad I = \tfrac{1}{3}ML^2$$
(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 $(L/2)(1 - \cos\theta_0)$。
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%。
Insight.要点。A physical pendulum is derived by linearizing the torque equation: $\sin\theta\approx\theta$ yields $\omega^2=3g/(2L)$ for a rod about its end. The same energy expression can be evaluated exactly for large angles.物理摆通过线性化力矩方程得到：对绕端点杆有 $\sin\theta\approx\theta$，得 $\omega^2=3g/(2L)$。同一能量表达式也可用于大角度的精确计算。
Both signs are reached during the oscillation.振动过程中正负两个位移都会达到。
Insight.要点。Gravity shifts a vertical spring's equilibrium but does not change its frequency. Measure displacement from the new equilibrium and the equation reduces to the standard horizontal mass-spring form.重力只移动竖直弹簧的平衡位置，不改变频率。以新平衡位置为位移零点，方程就化为标准水平弹簧振子形式。