Companion to the University-Style Practice Set大学风格练习题配套解答
Sections 1 to 7: spring-mass model, free undamped motion, free damped motion, forced vibrations, resonance, the RLC circuit analogy, beats and practical resonance第1至7节：弹簧质量模型、自由无阻尼运动、自由有阻尼运动、受迫振动、共振、RLC电路类比、拍频与实际共振CALC IV
$m = 2\ \text{kg}$, $k = 32\ \text{N/m}$, undamped. IVP: $x(0) = 3$, $x'(0) = -8$. Find $\omega_0$, solve the IVP, and convert to amplitude-phase form $A\cos(\omega_0 t - \phi)$.$m = 2\ \text{kg}$，$k = 32\ \text{N/m}$，无阻尼。初值问题：$x(0) = 3$，$x'(0) = -8$。求 $\omega_0$，解初值问题，并转化为幅相形式 $A\cos(\omega_0 t - \phi)$。
Newton's second law with no damping and no forcing gives $mx'' + kx = 0$, i.e. $2x'' + 32x = 0$, or equivalently $x'' + 16x = 0$. (M1)无阻尼无外力时牛顿第二定律给出 $mx'' + kx = 0$，即 $2x'' + 32x = 0$，等价于 $x'' + 16x = 0$。(M1)
The natural frequency is $\omega_0 = \sqrt{k/m} = \sqrt{32/2} = \sqrt{16} = 4\ \text{rad/s}$. (A1)自然频率为 $\omega_0 = \sqrt{k/m} = \sqrt{32/2} = \sqrt{16} = 4\ \text{rad/s}$。(A1)
The characteristic equation $r^{2} + 16 = 0$ has roots $r = \pm 4i$, so the general solution is (M1)特征方程 $r^{2} + 16 = 0$ 的根为 $r = \pm 4i$，故通解为 (M1)
$$ x(t) = c_1\cos 4t + c_2\sin 4t. $$Applying $x(0) = 3$: $c_1 = 3$. (A1) Differentiating gives $x'(t) = -4c_1\sin 4t + 4c_2\cos 4t$, so $x'(0) = 4c_2 = -8$, hence $c_2 = -2$. (A1)代入 $x(0) = 3$：$c_1 = 3$。(A1) 对 $x$ 求导得 $x'(t) = -4c_1\sin 4t + 4c_2\cos 4t$，故 $x'(0) = 4c_2 = -8$，从而 $c_2 = -2$。(A1)
$$ x(t) = 3\cos 4t - 2\sin 4t. $$With $c_1 = 3$ and $c_2 = -2$, the amplitude is (M1)已知 $c_1 = 3$，$c_2 = -2$，振幅为 (M1)
$$ A = \sqrt{c_1^{2} + c_2^{2}} = \sqrt{9 + 4} = \sqrt{13}. $$The phase satisfies $\cos\phi = c_1/A = 3/\sqrt{13} > 0$ and $\sin\phi = c_2/A = -2/\sqrt{13} < 0$, placing $\phi$ in the fourth quadrant: $\phi = -\arctan(2/3) \approx -0.5880\ \text{rad}$. (A1)相位满足 $\cos\phi = c_1/A = 3/\sqrt{13} > 0$，$\sin\phi = c_2/A = -2/\sqrt{13} < 0$，故 $\phi$ 在第四象限：$\phi = -\arctan(2/3) \approx -0.5880\ \text{rad}$。(A1)
The period is $T = 2\pi/\omega_0 = 2\pi/4 = \pi/2\ \text{s}$, and the solution in amplitude-phase form is $x(t) = \sqrt{13}\,\cos(4t + 0.5880)$. (A1)周期为 $T = 2\pi/\omega_0 = 2\pi/4 = \pi/2\ \text{s}$，幅相形式的解为 $x(t) = \sqrt{13}\,\cos(4t + 0.5880)$。(A1)
(a) $m=1,c=6,k=9$; (b) $m=1,c=4,k=13$; (c) $m=2,c=10,k=8$. Compute $\Delta=c^{2}-4mk$, classify, and write the general solution form.(a) $m=1,c=6,k=9$；(b) $m=1,c=4,k=13$；(c) $m=2,c=10,k=8$。计算 $\Delta=c^{2}-4mk$，判断阻尼类型，并写出通解形式。
$\Delta = 6^{2} - 4(1)(9) = 36 - 36 = 0$. (M1) A repeated characteristic root $r = -c/(2m) = -3$. The general solution is $x = (c_1 + c_2 t)e^{-3t}$. (A1)$\Delta = 6^{2} - 4(1)(9) = 36 - 36 = 0$。(M1) 特征根为重根 $r = -c/(2m) = -3$，通解为 $x = (c_1 + c_2 t)e^{-3t}$。(A1)
$\Delta = 4^{2} - 4(1)(13) = 16 - 52 = -36 < 0$. (M1) The roots are $r = (-4 \pm \sqrt{-36})/2 = -2 \pm 3i$, giving $\beta = 2$ and quasi-frequency $\omega_1 = 3$. General solution: $x = e^{-2t}(c_1\cos 3t + c_2\sin 3t)$. (A1)$\Delta = 4^{2} - 4(1)(13) = 16 - 52 = -36 < 0$。(M1) 特征根为 $r = (-4 \pm \sqrt{-36})/2 = -2 \pm 3i$，阻尼率 $\beta = 2$，准频率 $\omega_1 = 3$，通解：$x = e^{-2t}(c_1\cos 3t + c_2\sin 3t)$。(A1)
$\Delta = 10^{2} - 4(2)(8) = 100 - 64 = 36 > 0$. (M1) The roots are $r = (-10 \pm 6)/(2\cdot 2) = -1$ and $-4$, both real and negative. General solution: $x = c_1 e^{-t} + c_2 e^{-4t}$. (A1)$\Delta = 10^{2} - 4(2)(8) = 100 - 64 = 36 > 0$。(M1) 特征根为 $r = (-10 \pm 6)/(2\cdot 2) = -1$ 和 $-4$，均为负实数，通解：$x = c_1 e^{-t} + c_2 e^{-4t}$。(A1)
Solve $x''+2x'+10x=0$, $x(0)=2$, $x'(0)=-1$. Find $\beta$, $\omega_1$, apply ICs, convert to amplitude-phase form, and explain the relation $\omega_1 < \omega_0$.求解 $x''+2x'+10x=0$，$x(0)=2$，$x'(0)=-1$。求 $\beta$、$\omega_1$，代入初始条件，转化为幅相形式，并解释 $\omega_1 < \omega_0$ 的原因。
The characteristic equation is $r^{2} + 2r + 10 = 0$. Discriminant: $\Delta = 4 - 40 = -36 < 0$, confirming underdamped status. (M1) The roots are特征方程为 $r^{2} + 2r + 10 = 0$。判别式：$\Delta = 4 - 40 = -36 < 0$，确认欠阻尼状态。(M1) 特征根为
$$ r = \frac{-2 \pm \sqrt{-36}}{2} = -1 \pm 3i. $$The damping rate is $\beta = 1$, the quasi-frequency is $\omega_1 = 3\ \text{rad/s}$, and $\omega_0 = \sqrt{10} \approx 3.162$, confirming $\omega_1 < \omega_0$. (A1)阻尼率 $\beta = 1$，准频率 $\omega_1 = 3\ \text{rad/s}$，$\omega_0 = \sqrt{10} \approx 3.162$，确认 $\omega_1 < \omega_0$。(A1)
The general solution is $x(t) = e^{-t}(c_1\cos 3t + c_2\sin 3t)$. Applying $x(0) = 2$: $c_1 = 2$. (M1)通解为 $x(t) = e^{-t}(c_1\cos 3t + c_2\sin 3t)$。代入 $x(0) = 2$：$c_1 = 2$。(M1)
Differentiating: $x'(t) = -e^{-t}(c_1\cos 3t + c_2\sin 3t) + e^{-t}(-3c_1\sin 3t + 3c_2\cos 3t)$. At $t = 0$:对 $x$ 求导：$x'(t) = -e^{-t}(c_1\cos 3t + c_2\sin 3t) + e^{-t}(-3c_1\sin 3t + 3c_2\cos 3t)$。在 $t = 0$ 时：
$$ x'(0) = -c_1 + 3c_2 = -1. $$With $c_1 = 2$: $-2 + 3c_2 = -1$, so $c_2 = \dfrac{1}{3}$. (A1·A1)代入 $c_1 = 2$：$-2 + 3c_2 = -1$，故 $c_2 = \dfrac{1}{3}$。(A1·A1)
The amplitude is $A = \sqrt{c_1^{2}+c_2^{2}} = \sqrt{4 + \tfrac{1}{9}} = \dfrac{\sqrt{37}}{3}$. (M1) Since $c_1 > 0$ and $c_2 > 0$, the phase lies in the first quadrant: $\phi = \arctan(1/6) \approx 0.1651\ \text{rad}$. The solution in amplitude-phase form is振幅为 $A = \sqrt{c_1^{2}+c_2^{2}} = \sqrt{4 + \tfrac{1}{9}} = \dfrac{\sqrt{37}}{3}$。(M1) 因 $c_1 > 0$，$c_2 > 0$，相位在第一象限：$\phi = \arctan(1/6) \approx 0.1651\ \text{rad}$。幅相形式的解为
$$ x(t) = \frac{\sqrt{37}}{3}\,e^{-t}\cos\!\left(3t - \arctan\tfrac{1}{6}\right), \quad T_1 = \frac{2\pi}{3}\ \text{s}. \quad\text{(A1)} $$The quasi-frequency $\omega_1 = \sqrt{\omega_0^{2} - \beta^{2}} < \omega_0$: damping subtracts $\beta^{2}$ from $\omega_0^{2}$ under the root, so the damped system oscillates more slowly than the undamped one because some energy is continuously dissipated by the dashpot. (A1)准频率 $\omega_1 = \sqrt{\omega_0^{2} - \beta^{2}} < \omega_0$：阻尼在根号内从 $\omega_0^{2}$ 中减去 $\beta^{2}$，因此有阻尼系统比无阻尼系统振荡更慢，原因是阻尼器不断耗散能量。(A1)
$x''+9x=6\cos 3t$, $x(0)=0$, $x'(0)=0$. Explain why the standard guess fails, find $x_p$ via the modified guess, apply ICs, and name the unbounded-growth phenomenon.$x''+9x=6\cos 3t$，$x(0)=0$，$x'(0)=0$。解释标准猜测失效的原因，用修正猜测求 $x_p$，代入初始条件，并命名无界增长现象。
The homogeneous equation $x'' + 9x = 0$ has the general solution $c_1\cos 3t + c_2\sin 3t$, so both $\cos 3t$ and $\sin 3t$ already satisfy the homogeneous equation. (M1) Any linear combination of them gives zero on the left side, never $6\cos 3t$; the guess duplicates the complementary solution. The correct modified guess is $x_p = t(A\cos 3t + B\sin 3t)$. (A1)齐次方程 $x'' + 9x = 0$ 的通解为 $c_1\cos 3t + c_2\sin 3t$，故 $\cos 3t$ 和 $\sin 3t$ 均已满足齐次方程。(M1) 它们的任意线性组合代入左端均为零，无法得到 $6\cos 3t$；该猜测与余函数解重复。正确的修正猜测为 $x_p = t(A\cos 3t + B\sin 3t)$。(A1)
Differentiate twice: $x_p'' = -6A\sin 3t + 6B\cos 3t - 9t(A\cos 3t + B\sin 3t)$. (M1) Substituting into $x_p'' + 9x_p$, the $9t(\cdots)$ terms cancel, leaving $-6A\sin 3t + 6B\cos 3t = 6\cos 3t$. Matching: $A = 0$, $B = 1$, so $x_p = t\sin 3t$. (A1)对 $x_p$ 两次求导：$x_p'' = -6A\sin 3t + 6B\cos 3t - 9t(A\cos 3t + B\sin 3t)$。(M1) 代入 $x_p'' + 9x_p$，含 $t$ 的项抵消，剩余 $-6A\sin 3t + 6B\cos 3t = 6\cos 3t$。比较系数：$A = 0$，$B = 1$，故 $x_p = t\sin 3t$。(A1)
General solution: $x = c_1\cos 3t + c_2\sin 3t + t\sin 3t$. $x(0) = 0 \Rightarrow c_1 = 0$. $x'(0) = 3c_2 = 0 \Rightarrow c_2 = 0$. (M1) The complete solution is通解：$x = c_1\cos 3t + c_2\sin 3t + t\sin 3t$。$x(0) = 0 \Rightarrow c_1 = 0$。$x'(0) = 3c_2 = 0 \Rightarrow c_2 = 0$。(M1) 完全解为
$$ x(t) = t\sin 3t. $$The factor of $t$ causes the amplitude to grow without bound ($|x| \le t \to \infty$): this is pure resonance, occurring because the forcing frequency $\omega = 3$ equals the natural frequency $\omega_0 = 3$ of the undamped system. (A1)因子 $t$ 导致振幅无界增长（$|x| \le t \to \infty$）：这就是纯共振，发生条件是激励频率 $\omega = 3$ 恰好等于无阻尼系统的自然频率 $\omega_0 = 3$。(A1)
(a) Prove $c_1\cos\omega t + c_2\sin\omega t = A\cos(\omega t - \phi)$ with $A = \sqrt{c_1^{2}+c_2^{2}}$, $\phi = \arctan(c_2/c_1)$. (b) Convert $3\cos 4t - 4\sin 4t$ explicitly.(a) 证明 $c_1\cos\omega t + c_2\sin\omega t = A\cos(\omega t - \phi)$，其中 $A = \sqrt{c_1^{2}+c_2^{2}}$，$\phi = \arctan(c_2/c_1)$。(b) 对 $3\cos 4t - 4\sin 4t$ 显式完成转换。
Define $A = \sqrt{c_1^{2}+c_2^{2}} > 0$ and $\phi$ by $\cos\phi = c_1/A$ and $\sin\phi = c_2/A$ (the angle is uniquely determined once the quadrant is fixed by the signs of $c_1$ and $c_2$). (M1)定义 $A = \sqrt{c_1^{2}+c_2^{2}} > 0$，并由 $\cos\phi = c_1/A$，$\sin\phi = c_2/A$ 确定 $\phi$（一旦由 $c_1$，$c_2$ 的符号确定象限，角度唯一确定）。(M1)
Expand $A\cos(\omega t - \phi)$ using the cosine subtraction formula $\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$: (M1)利用余弦差角公式 $\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$ 展开 $A\cos(\omega t - \phi)$：(M1)
$$ A\cos(\omega t - \phi) = A\bigl(\cos\omega t\,\cos\phi + \sin\omega t\,\sin\phi\bigr) = A\cdot\frac{c_1}{A}\cos\omega t + A\cdot\frac{c_2}{A}\sin\omega t = c_1\cos\omega t + c_2\sin\omega t. $$(A1) The equality holds for all $t$, so the two expressions are identical. (R1)(A1) 等式对所有 $t$ 成立，故两表达式完全相同。(R1)
Here $c_1 = 3$ and $c_2 = -4$. The amplitude is (M1)此处 $c_1 = 3$，$c_2 = -4$，振幅为 (M1)
$$ A = \sqrt{3^{2} + (-4)^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5. $$(A1) The phase satisfies $\cos\phi = 3/5 > 0$ and $\sin\phi = -4/5 < 0$, placing $\phi$ in the fourth quadrant: (M1)(A1) 相位满足 $\cos\phi = 3/5 > 0$，$\sin\phi = -4/5 < 0$，故 $\phi$ 在第四象限：(M1)
$$ \phi = -\arctan\!\bigl(\tfrac{4}{3}\bigr) \approx -0.9273\ \text{rad}. $$Therefore $x(t) = 5\cos(4t - (-0.9273)) = 5\cos(4t + 0.9273)$. Verification: $5\cos(4t+0.9273) = 5[{}\cos 4t\cos 0.9273 - \sin 4t\sin 0.9273] = 5[0.6\cos 4t - 0.8\sin 4t] = 3\cos 4t - 4\sin 4t$. (A1)故 $x(t) = 5\cos(4t - (-0.9273)) = 5\cos(4t + 0.9273)$。验证：$5\cos(4t+0.9273) = 5[\cos 4t\cos 0.9273 - \sin 4t\sin 0.9273] = 5[0.6\cos 4t - 0.8\sin 4t] = 3\cos 4t - 4\sin 4t$。(A1)
(a) Derive $C(\omega) = (F_0/m)/\sqrt{(\omega_0^{2}-\omega^{2})^{2}+4\beta^{2}\omega^{2}}$ from $mx''+cx'+kx=F_0\cos\omega t$. (b) Show the maximum is at $\omega_r = \sqrt{\omega_0^{2}-2\beta^{2}}$ and describe the behaviour as $\beta\to 0^{+}$.(a) 从 $mx''+cx'+kx=F_0\cos\omega t$ 推导 $C(\omega) = (F_0/m)/\sqrt{(\omega_0^{2}-\omega^{2})^{2}+4\beta^{2}\omega^{2}}$。(b) 证明最大值在 $\omega_r = \sqrt{\omega_0^{2}-2\beta^{2}}$ 处取得，并描述 $\beta\to 0^{+}$ 时的行为。
Try $x_p = P\cos\omega t + Q\sin\omega t$. Compute (M1)令 $x_p = P\cos\omega t + Q\sin\omega t$，计算 (M1)
$$ x_p' = -P\omega\sin\omega t + Q\omega\cos\omega t, \quad x_p'' = -P\omega^{2}\cos\omega t - Q\omega^{2}\sin\omega t. $$Substituting into $mx_p'' + cx_p' + kx_p = F_0\cos\omega t$ and collecting $\cos\omega t$ and $\sin\omega t$ terms: (M1)代入 $mx_p'' + cx_p' + kx_p = F_0\cos\omega t$ 并整理 $\cos\omega t$ 和 $\sin\omega t$ 各项：(M1)
$$ \bigl[(k - m\omega^{2})P + c\omega Q\bigr]\cos\omega t + \bigl[(k-m\omega^{2})Q - c\omega P\bigr]\sin\omega t = F_0\cos\omega t. $$Matching coefficients gives the $2\times 2$ system (M1)比较系数得到 $2\times 2$ 方程组 (M1)
$$ (k-m\omega^{2})P + c\omega Q = F_0, \qquad -c\omega P + (k-m\omega^{2})Q = 0. $$From the second equation $Q = c\omega P/(k-m\omega^{2})$; substituting into the first and using the fact that $P^{2}+Q^{2} = C^{2}$ after solving: the determinant of the system is $D = (k-m\omega^{2})^{2}+(c\omega)^{2}$, so由第二方程得 $Q = c\omega P/(k-m\omega^{2})$，代入第一方程，利用求解后 $P^{2}+Q^{2} = C^{2}$：方程组的行列式为 $D = (k-m\omega^{2})^{2}+(c\omega)^{2}$，故
$$ P = \frac{(k-m\omega^{2})F_0}{D}, \quad Q = \frac{c\omega F_0}{D}, \quad C = \sqrt{P^{2}+Q^{2}} = \frac{F_0}{\sqrt{D}}. $$(A1) Writing $k - m\omega^{2} = m(\omega_0^{2}-\omega^{2})$ and $c = 2m\beta$:(A1) 令 $k - m\omega^{2} = m(\omega_0^{2}-\omega^{2})$，$c = 2m\beta$：
$$ D = m^{2}(\omega_0^{2}-\omega^{2})^{2} + 4m^{2}\beta^{2}\omega^{2}, $$ $$ C(\omega) = \frac{F_0}{m\sqrt{(\omega_0^{2}-\omega^{2})^{2}+4\beta^{2}\omega^{2}}} = \frac{F_0/m}{\sqrt{(\omega_0^{2}-\omega^{2})^{2}+4\beta^{2}\omega^{2}}}. \quad \text{(A1)} $$Maximising $C(\omega)$ is equivalent to minimising $f(u) = (\omega_0^{2}-u)^{2} + 4\beta^{2}u$ where $u = \omega^{2} \ge 0$. (M1) Differentiate with respect to $u$: (M1)最大化 $C(\omega)$ 等价于最小化 $f(u) = (\omega_0^{2}-u)^{2} + 4\beta^{2}u$，其中 $u = \omega^{2} \ge 0$。(M1) 对 $u$ 求导：(M1)
$$ f'(u) = -2(\omega_0^{2}-u) + 4\beta^{2} = 0 \implies \omega_0^{2}-u = 2\beta^{2} \implies u = \omega_0^{2}-2\beta^{2}. $$This gives a minimum of $f$ (hence a maximum of $C$) provided $u > 0$, i.e. $2\beta^{2} < \omega_0^{2}$. (A1) The practical resonance frequency is $\omega_r = \sqrt{\omega_0^{2}-2\beta^{2}}$. Note $\omega_r < \omega_0$ for all $\beta > 0$. (R1)当 $u > 0$（即 $2\beta^{2} < \omega_0^{2}$）时，这给出 $f$ 的极小值（即 $C$ 的极大值）。(A1) 实际共振频率为 $\omega_r = \sqrt{\omega_0^{2}-2\beta^{2}}$，注意对所有 $\beta > 0$ 均有 $\omega_r < \omega_0$。(R1)
As $\beta \to 0^{+}$, the denominator $\sqrt{(\omega_0^{2}-\omega_r^{2})^{2}+4\beta^{2}\omega_r^{2}}$ at $\omega_r \to \omega_0$ approaches $0$, so the peak amplitude $C(\omega_r) \to \infty$: this is the undamped resonance limit. (A1)当 $\beta \to 0^{+}$ 时，$\omega_r \to \omega_0$，分母 $\sqrt{(\omega_0^{2}-\omega_r^{2})^{2}+4\beta^{2}\omega_r^{2}} \to 0$，故峰值振幅 $C(\omega_r) \to \infty$：这是无阻尼共振极限。(A1)
$Lq'' + Rq' + q/C = E(t)$. (a) Write the full mechanical-electrical dictionary. (b) $L=1$, $R=4$, $C=1/5$: classify and find $\omega_1$. (c) Explain why $L$ plays the same role as $m$.$Lq'' + Rq' + q/C = E(t)$。(a) 写出完整的机电对照表。(b) $L=1$，$R=4$，$C=1/5$：判断类型并求 $\omega_1$。(c) 解释 $L$ 与 $m$ 起相同作用的原因。
Comparing $m x'' + c x' + k x = F(t)$ with $L q'' + R q' + (1/C) q = E(t)$ term by term: (A1 for each correct row)逐项比较 $m x'' + c x' + k x = F(t)$ 与 $L q'' + R q' + (1/C) q = E(t)$：（每行正确得 A1）
The electrical discriminant is $R^{2} - 4L/C = 16 - 4(1)(5) = 16 - 20 = -4 < 0$. (M1) The circuit is underdamped. (A1)电学判别式为 $R^{2} - 4L/C = 16 - 4(1)(5) = 16 - 20 = -4 < 0$。(M1) 电路为欠阻尼状态。(A1)
The natural frequency is $\omega_0 = \sqrt{1/(LC)} = \sqrt{5}$ and the damping rate is $\beta = R/(2L) = 2$. The quasi-frequency is $\omega_1 = \sqrt{\omega_0^{2}-\beta^{2}} = \sqrt{5-4} = 1\ \text{rad/s}$. (A1)自然频率为 $\omega_0 = \sqrt{1/(LC)} = \sqrt{5}$，阻尼率为 $\beta = R/(2L) = 2$，准频率为 $\omega_1 = \sqrt{\omega_0^{2}-\beta^{2}} = \sqrt{5-4} = 1\ \text{rad/s}$。(A1)
Inductance $L$ resists changes in current ($v_L = L\,di/dt$) just as mass $m$ resists changes in velocity ($F = m\,a = m\,dv/dt$): both express inertia in their respective domains, and both appear as the coefficient of the second-derivative term in their governing equations. (R1·A1)电感 $L$ 阻碍电流的变化（$v_L = L\,di/dt$），正如质量 $m$ 阻碍速度的变化（$F = m\,a = m\,dv/dt$）：两者均在各自领域体现惯性，且均作为各自控制方程中二阶导数项的系数出现。(R1·A1)
$m=4\ \text{kg}$, $k=100\ \text{N/m}$, $c=40\ \text{N\,s/m}$; $x(0)=0.5\ \text{m}$, $x'(0)=0$. Write the IVP, classify, solve, verify, and describe long-run behaviour.$m=4\ \text{kg}$，$k=100\ \text{N/m}$，$c=40\ \text{N\,s/m}$；$x(0)=0.5\ \text{m}$，$x'(0)=0$。写出初值问题，判断阻尼类型，求解，验证，并描述长期行为。
Newton's second law gives $4x'' + 40x' + 100x = 0$, or dividing by 4: $x'' + 10x' + 25x = 0$. (M1)牛顿第二定律给出 $4x'' + 40x' + 100x = 0$，除以4得 $x'' + 10x' + 25x = 0$。(M1)
Discriminant: $\Delta = c^{2} - 4mk = 40^{2} - 4(4)(100) = 1600 - 1600 = 0$. The system is critically damped. (A1)判别式：$\Delta = c^{2} - 4mk = 40^{2} - 4(4)(100) = 1600 - 1600 = 0$，系统为临界阻尼。(A1)
The characteristic equation $(r+5)^{2} = r^{2}+10r+25 = 0$ has a repeated root $r = -5$. (M1) The general solution is (A1)特征方程 $(r+5)^{2} = r^{2}+10r+25 = 0$ 有重根 $r = -5$。(M1) 通解为 (A1)
$$ x(t) = (c_1 + c_2 t)\,e^{-5t}. $$This is the only form for the critically damped case; the factor of $t$ in the second term distinguishes it from both the underdamped and overdamped solutions. (A1)这是临界阻尼情形唯一的通解形式；第二项中的因子 $t$ 将其与欠阻尼和过阻尼解区分开来。(A1)
From $x(0) = c_1 = 0.5$. (M1) Differentiating: $x'(t) = (c_2 - 5c_1 - 5c_2 t)\,e^{-5t}$, so $x'(0) = c_2 - 5c_1 = 0$, giving $c_2 = 5c_1 = 2.5$. (A1) The specific solution is由 $x(0) = c_1 = 0.5$。(M1) 求导：$x'(t) = (c_2 - 5c_1 - 5c_2 t)\,e^{-5t}$，故 $x'(0) = c_2 - 5c_1 = 0$，得 $c_2 = 5c_1 = 2.5$。(A1) 特定解为
$$ x(t) = (0.5 + 2.5t)\,e^{-5t}. $$Verification: $x(0) = 0.5$ and $x'(0) = 2.5 - 5(0.5) = 0$. (M1) For the ODE: $x'' + 10x' + 25x$: with $x = (c_1+c_2t)e^{-5t}$ and the characteristic polynomial $(r+5)^2$ vanishing at $r=-5$, the substitution gives identically zero, confirming the solution. (A1)验证：$x(0) = 0.5$，$x'(0) = 2.5 - 5(0.5) = 0$。(M1) 对于常微分方程 $x'' + 10x' + 25x$：以 $x = (c_1+c_2t)e^{-5t}$ 代入，特征多项式 $(r+5)^2$ 在 $r=-5$ 处为零，代入后恒等于零，确认解的正确性。(A1)
Since $0.5 + 2.5t > 0$ for all $t \ge 0$, the function $x(t) = (0.5 + 2.5t)e^{-5t}$ is strictly positive and decreasing to zero without ever crossing the equilibrium $x = 0$. The critically damped system returns to equilibrium as quickly as possible without oscillating. (A1)由于对所有 $t \ge 0$ 均有 $0.5 + 2.5t > 0$，函数 $x(t) = (0.5 + 2.5t)e^{-5t}$ 严格正值并单调趋近于零，始终不穿越平衡位置 $x = 0$。临界阻尼系统在不振荡的前提下以最快速度回到平衡位置。(A1)
$x''+2x'+26x=F_0\cos\omega t$: identify $\omega_0$, $\beta$; write $C(\omega)$; find $\omega_r$ and $C(\omega_r)$ for $F_0=1$; compare with undamped resonance.$x''+2x'+26x=F_0\cos\omega t$：确定 $\omega_0$、$\beta$；写出 $C(\omega)$；当 $F_0=1$ 时求 $\omega_r$ 和 $C(\omega_r)$；与无阻尼共振比较。
Comparing $x'' + 2x' + 26x = F_0\cos\omega t$ with the standard form $m=1$, $c=2$, $k=26$: $\omega_0 = \sqrt{k/m} = \sqrt{26}$ and $\beta = c/(2m) = 1$. (M1)将 $x'' + 2x' + 26x = F_0\cos\omega t$ 与标准形式对比，$m=1$，$c=2$，$k=26$：$\omega_0 = \sqrt{k/m} = \sqrt{26}$，$\beta = c/(2m) = 1$。(M1)
Discriminant: $\Delta = c^{2} - 4mk = 4 - 4(1)(26) = 4 - 104 = -100 < 0$, confirming the free system is underdamped. (A1)判别式：$\Delta = c^{2} - 4mk = 4 - 4(1)(26) = 4 - 104 = -100 < 0$，确认自由系统为欠阻尼状态。(A1)
From the formula derived in Q6(a) with $m=1$, $F_0=1$: (M1)利用Q6(a)推导的公式，取 $m=1$，$F_0=1$：(M1)
$$ C(\omega) = \frac{1}{\sqrt{(\omega_0^{2}-\omega^{2})^{2}+4\beta^{2}\omega^{2}}} = \frac{1}{\sqrt{(26-\omega^{2})^{2}+4\omega^{2}}}. $$Expanding: $(26-\omega^{2})^{2}+4\omega^{2} = \omega^{4} - 52\omega^{2} + 676 + 4\omega^{2} = \omega^{4} - 48\omega^{2} + 676$. (A1)展开：$(26-\omega^{2})^{2}+4\omega^{2} = \omega^{4} - 52\omega^{2} + 676 + 4\omega^{2} = \omega^{4} - 48\omega^{2} + 676$。(A1)
Using the result from Q6(b), $\omega_r = \sqrt{\omega_0^{2}-2\beta^{2}} = \sqrt{26-2} = \sqrt{24} = 2\sqrt{6}$. (M1·A1)利用Q6(b)的结论，$\omega_r = \sqrt{\omega_0^{2}-2\beta^{2}} = \sqrt{26-2} = \sqrt{24} = 2\sqrt{6}$。(M1·A1)
Substituting $\omega_r^{2} = 24$ into the radicand: (M1)将 $\omega_r^{2} = 24$ 代入被开方式：(M1)
$$ (26-24)^{2}+4(24) = 4 + 96 = 100. $$ $$ C(\omega_r) = \frac{1}{\sqrt{100}} = \frac{1}{10}. \quad \text{(A1)} $$In the undamped case ($\beta = 0$) the amplitude $C(\omega_0)$ is unbounded: the system resonates to infinity. With positive damping, the peak is finite: $C(\omega_r) = 1/10$ here. (A1) As $\beta \to 0^{+}$, the peak frequency $\omega_r = \sqrt{26-2\beta^{2}} \to \sqrt{26} = \omega_0$, so practical resonance approaches natural resonance in the zero-damping limit. (A1)无阻尼情形（$\beta = 0$）下，振幅 $C(\omega_0)$ 无界：系统共振至无穷大。正阻尼时峰值有限：此处 $C(\omega_r) = 1/10$。(A1) 当 $\beta \to 0^{+}$ 时，峰值频率 $\omega_r = \sqrt{26-2\beta^{2}} \to \sqrt{26} = \omega_0$，故在零阻尼极限下实际共振趋近于自然共振。(A1)
$L=1\ \text{H}$, $R=4\ \Omega$, $C=\tfrac{1}{20}\ \text{F}$, $E(t)=10\cos 2t\ \text{V}$, $q(0)=0$, $q'(0)=0$. Write the ODE, classify the free circuit, find $q_p$, apply ICs to get the full solution, and identify transient and steady-state.$L=1\ \text{H}$，$R=4\ \Omega$，$C=\tfrac{1}{20}\ \text{F}$，$E(t)=10\cos 2t\ \text{V}$，$q(0)=0$，$q'(0)=0$。写出常微分方程，对自由电路分类，求 $q_p$，代入初始条件求完全解，并区分暂态与稳态分量。
Kirchhoff's voltage law gives $Lq'' + Rq' + q/C = E(t)$, i.e. (M1)基尔霍夫电压定律给出 $Lq'' + Rq' + q/C = E(t)$，即 (M1)
$$ q'' + 4q' + 20q = 10\cos 2t. $$For the free circuit ($E=0$): discriminant $R^{2} - 4L/C = 16 - 4(1)(20) = 16 - 80 = -64 < 0$, so the circuit is underdamped. (A1) The damping rate is $\beta = R/(2L) = 2$ and the natural frequency is $\omega_0 = \sqrt{1/(LC)} = \sqrt{20}$. The quasi-frequency is $\omega_1 = \sqrt{\omega_0^{2}-\beta^{2}} = \sqrt{20-4} = \sqrt{16} = 4\ \text{rad/s}$. (A1)对自由电路（$E=0$）：判别式 $R^{2} - 4L/C = 16 - 4(1)(20) = 16 - 80 = -64 < 0$，电路为欠阻尼状态。(A1) 阻尼率 $\beta = R/(2L) = 2$，自然频率 $\omega_0 = \sqrt{1/(LC)} = \sqrt{20}$，准频率 $\omega_1 = \sqrt{\omega_0^{2}-\beta^{2}} = \sqrt{20-4} = \sqrt{16} = 4\ \text{rad/s}$。(A1)
Try $q_p = A\cos 2t + B\sin 2t$. Compute $q_p'' = -4A\cos 2t - 4B\sin 2t$ and $q_p' = -2A\sin 2t + 2B\cos 2t$. Substitute: (M1)令 $q_p = A\cos 2t + B\sin 2t$，计算 $q_p'' = -4A\cos 2t - 4B\sin 2t$，$q_p' = -2A\sin 2t + 2B\cos 2t$，代入：(M1)
$$ (-4A + 8B + 20A)\cos 2t + (-4B - 8A + 20B)\sin 2t = 10\cos 2t. $$ $$ (16A + 8B)\cos 2t + (16B - 8A)\sin 2t = 10\cos 2t. $$(M1) Matching coefficients: $16A + 8B = 10$ and $16B - 8A = 0$. From the second equation $B = A/2$; substituting: $16A + 4A = 20A = 10$, so $A = \tfrac{1}{2}$ and $B = \tfrac{1}{4}$. (A1) The particular solution is(M1) 比较系数：$16A + 8B = 10$，$16B - 8A = 0$。由第二方程得 $B = A/2$；代入：$16A + 4A = 20A = 10$，故 $A = \tfrac{1}{2}$，$B = \tfrac{1}{4}$。(A1) 特解为
$$ q_p(t) = \tfrac{1}{2}\cos 2t + \tfrac{1}{4}\sin 2t. $$The general solution is $q(t) = e^{-2t}(c_1\cos 4t + c_2\sin 4t) + \tfrac{1}{2}\cos 2t + \tfrac{1}{4}\sin 2t$. Applying $q(0) = 0$: $c_1 + \tfrac{1}{2} = 0$, so $c_1 = -\tfrac{1}{2}$. (M1)通解为 $q(t) = e^{-2t}(c_1\cos 4t + c_2\sin 4t) + \tfrac{1}{2}\cos 2t + \tfrac{1}{4}\sin 2t$。代入 $q(0) = 0$：$c_1 + \tfrac{1}{2} = 0$，故 $c_1 = -\tfrac{1}{2}$。(M1)
Differentiating and applying $q'(0) = 0$: $q'(t) = e^{-2t}[(-2c_1+4c_2)\cos 4t + (-2c_2-4c_1)\sin 4t] + (-\sin 2t + \tfrac{1}{2}\cos 2t)$. At $t=0$:求导并代入 $q'(0) = 0$：$q'(t) = e^{-2t}[(-2c_1+4c_2)\cos 4t + (-2c_2-4c_1)\sin 4t] + (-\sin 2t + \tfrac{1}{2}\cos 2t)$。在 $t=0$ 时：
$$ q'(0) = (-2c_1 + 4c_2) + \tfrac{1}{2} = 0. $$With $c_1 = -\tfrac{1}{2}$: $1 + 4c_2 + \tfrac{1}{2} = 0$, so $4c_2 = -\tfrac{3}{2}$, giving $c_2 = -\tfrac{3}{8}$. The complete solution is代入 $c_1 = -\tfrac{1}{2}$：$1 + 4c_2 + \tfrac{1}{2} = 0$，故 $4c_2 = -\tfrac{3}{2}$，得 $c_2 = -\tfrac{3}{8}$。完全解为
$$ q(t) = \underbrace{e^{-2t}\!\left(-\tfrac{1}{2}\cos 4t - \tfrac{3}{8}\sin 4t\right)}_{\text{transient}} + \underbrace{\tfrac{1}{2}\cos 2t + \tfrac{1}{4}\sin 2t}_{\text{steady-state}}. \quad\text{(A1)} $$