The damped driven oscillator, which is most of linear physics

Idea. One quadratic, four times. Compare \(\gamma\) with \(2\omega_0 = 16\) before touching the square root.

Working. \(\gamma = 400/(R+5)\).

Bases of \(\ker L\), real-valued:

• \(\{e^{-0.2t}\cos 7.9975t,\ e^{-0.2t}\sin 7.9975t\}\);
• \(\{e^{-2.5t}\cos 7.5993t,\ e^{-2.5t}\sin 7.5993t\}\);
• \(\{e^{-8t},\ te^{-8t}\}\);
• \(\{e^{-0.8082t},\ e^{-79.19t}\}\).

For the underdamped rows the decay is set by the envelope \(e^{-\gamma t/2}\), time constant \(2/\gamma\); for the overdamped row by the slow root, and \(\sqrt{6144} = 78.3837\) gives \(\lambda_+ = (-80+78.3837)/2 = -0.80816\). The critical setting, \(R = 20\,\Omega\), brings the needle to rest soonest: its time constant \(0.125\) s beats every other row, including the heavily damped one.

Check. \(\lambda_+\lambda_- = \omega_0^2 = 64\): \(0.2^2 + 7.9975^2 = 0.04 + 63.96 = 64.00\) ✓; \(2.5^2+7.5993^2 = 6.25+57.75 = 64.00\) ✓; \((-8)^2 = 64\) ✓; \(0.80816\times79.1918 = 64.00\) ✓. Sums: \(0.8082+79.1918 = 80 = \gamma\) ✓. Limit: as \(R \to \infty\), \(\gamma \to 0\), \(\lambda_\pm \to \pm 8i\), a free oscillator — as an open circuit must be.

Idea. Everything follows from \(\exp(z+w) = \exp z\exp w\) and Euler.

Working.

1. \(\cos(a+b) + i\sin(a+b) = e^{i(a+b)} = e^{ia}e^{ib} = (\cos a + i\sin a)(\cos b + i\sin b) = (\cos a\cos b - \sin a\sin b) + i(\sin a\cos b + \cos a\sin b)\). Both \(a, b\) real, so the real and imaginary parts may be equated, giving both formulas at once.
2. \((1+i)^2 = 2i\), so \((1+i)^8 = (2i)^4 = 16\,i^4 = 16\). \(e^{i\pi/3} = \cos60^\circ + i\sin60^\circ = \tfrac12 + \tfrac{\sqrt3}{2}i\). \(|e^{2-3i}| = e^2 = 7.38906\).

3. Proof. Write \(z = u + iv\) with \(u, v\) real-valued. Since differentiation acts componentwise and \(\gamma, \omega_0^2\) are real, \(Lz = (\ddot u + \gamma\dot u + \omega_0^2u) + i(\ddot v + \gamma\dot v + \omega_0^2v) = Lu + i\,Lv\), and both \(Lu\) and \(Lv\) are real-valued. A complex number vanishes exactly when its real and imaginary parts do, so \(Lz = 0 \Rightarrow Lu = Lv = 0\). \(\square\)

   With \(\lambda_+ = -\gamma/2 + i\omega_\mathrm{d}\), \(e^{\lambda_+t} = e^{-\gamma t/2}(\cos\omega_\mathrm{d}t + i\sin\omega_\mathrm{d}t)\), so \(x_1 = e^{-\gamma t/2}\cos\omega_\mathrm{d}t\) and \(x_2 = e^{-\gamma t/2}\sin\omega_\mathrm{d}t\). At \(t = 0\): \(x_1 = 1\), \(\dot x_1 = -\gamma/2\), \(x_2 = 0\), \(\dot x_2 = \omega_\mathrm{d}\), so \(W(0) = 1\cdot\omega_\mathrm{d} - 0 = \omega_\mathrm{d} \ne 0\) and by Abel \(W(t) = \omega_\mathrm{d}e^{-\gamma t}\). A basis.

4. The step that fails is "\(Lu\) and \(Lv\) are real". Take \(Lx = \ddot x + 2i\dot x - x\), whose characteristic polynomial is \((\lambda+i)^2\), so \(z = e^{-it} \in \ker L\). But \(\operatorname{Re}z = \cos t\) gives \(L(\cos t) = -\cos t - 2i\sin t - \cos t = -2\cos t - 2i\sin t \ne 0\).

Check. Part 1 at \(a = b = \pi/4\): \(\cos(\pi/2) = \tfrac12-\tfrac12 = 0\) ✓. Part 3 in the limit \(\gamma \to 0\): \(W = \omega_0\), constant — Abel with \(\gamma = 0\) ✓. Number: for the galvanometer at \(R = 995\,\Omega\), \(W(0) = \omega_\mathrm{d} = 7.9975\).

Idea. Differentiate \(W\) once; everything else is bookkeeping. Then use Abel backwards as a first-order equation for the unknown second solution.

Working.

1. \(\dot W = \dot x_1\dot x_2 + x_1\ddot x_2 - \dot x_2\dot x_1 - x_2\ddot x_1 = x_1\ddot x_2 - x_2\ddot x_1\). Here and only here substitute \(\ddot x_i = -\gamma\dot x_i - \omega_0^2x_i\): \[\dot W = x_1(-\gamma\dot x_2-\omega_0^2x_2) - x_2(-\gamma\dot x_1-\omega_0^2x_1) = -\gamma(x_1\dot x_2 - x_2\dot x_1) = -\gamma W,\] the \(\omega_0^2\) terms cancelling identically. Then \((We^{\gamma t})^{\boldsymbol\cdot} = e^{\gamma t}(\dot W + \gamma W) = 0\), so \(We^{\gamma t}\) is constant: \(W(t) = W(t_0)e^{-\gamma(t-t_0)}\). \(\square\)
2. (i) \(\Rightarrow\) (ii): the exponential never vanishes. (ii) \(\Rightarrow\) (iii): \(W(t_0) \ne 0\) makes \(M = \begin{pmatrix}x_1(t_0)&x_2(t_0)\\\dot x_1(t_0)&\dot x_2(t_0)\end{pmatrix}\) invertible, so \(\mathrm{ev}(c_1x_1+c_2x_2) = M(c_1,c_2)^{\!\top}\) covers all of \(\mathbb{R}^2\); since \(\mathrm{ev}\) is a bijection onto \(\mathbb{R}^2\) and \(\dim\ker L = 2\), the span is all of \(\ker L\), and two spanning elements of a two-dimensional space are a basis. (iii) \(\Rightarrow\) (i): contrapositive — if \(W(t_0) = 0\), choose \((c_1,c_2) \ne 0\) in \(\ker M\); then \(x = c_1x_1+c_2x_2\) has \(\mathrm{ev}(x) = 0\), hence \(x \equiv 0\) by injectivity of \(\mathrm{ev}\), so \(x_1, x_2\) are linearly dependent. \(\square\)
3. \(\gamma = 2\omega_0\), \(x_1 = e^{-\omega_0t}\). Normalise \(W(0) = 1\), so \(W = e^{-2\omega_0t}\). Divide \(x_1\dot x_2 - \dot x_1x_2 = W\) by \(x_1^2 = e^{-2\omega_0t}\): \[\Big(\frac{x_2}{x_1}\Big)^{\boldsymbol\cdot} = \frac{W}{x_1^2} = \frac{e^{-2\omega_0t}}{e^{-2\omega_0t}} = 1 \implies \frac{x_2}{x_1} = t + C.\] Take \(C = 0\): \(x_2 = te^{-\omega_0t}\). No guessing was involved, and the factor \(t\) appears because \(W/x_1^2\) happens to be constant — which is exactly the statement \(\lambda_+ = \lambda_-\).
4. \(x_1 = e^{-\gamma t/2}\cos\omega_\mathrm{d}t\), \(W = W(0)e^{-\gamma t}\), so on an interval where \(\cos\omega_\mathrm{d}t \ne 0\), \((x_2/x_1)^{\boldsymbol\cdot} = W(0)e^{-\gamma t}/(e^{-\gamma t}\cos^2\omega_\mathrm{d}t) = W(0)\sec^2\omega_\mathrm{d}t\), whence \(x_2/x_1 = W(0)\tan(\omega_\mathrm{d}t)/\omega_\mathrm{d}\) and \(x_2 = \frac{W(0)}{\omega_\mathrm{d}}e^{-\gamma t/2}\sin\omega_\mathrm{d}t\). Choosing \(W(0) = \omega_\mathrm{d}\) gives the sine solution exactly; it extends to all \(t\) by continuity, or by noting that it solves \(Lx = 0\) everywhere.

Check. Dimensions: \([W] = [x]^2\mathrm{T^{-1}}\) and the exponent \(\gamma(t-t_0)\) is a pure number ✓. Limit \(\gamma \to 0\): \(W\) constant, the undamped flow preserving phase-space area ✓. Number: at \(\gamma = 0.4\), \(W(1\,\mathrm{s})/W(0) = e^{-0.4} = 0.6703\).

Idea. The degenerate root is where the two exponentials collide; the extremal claim is two inequalities about where the roots sit.

Working.

1. \(\gamma = 2\omega_0\) makes the discriminant \(\gamma^2-4\omega_0^2 = 0\), so \(\lambda = -\omega_0\) twice, \(p(\lambda) = p'(\lambda) = 0\), and \(L(te^{\lambda t}) = p(\lambda)te^{\lambda t}+p'(\lambda)e^{\lambda t} = 0\). The Wronskian at \(t = 0\) is \(1\cdot1 - 0\cdot(-\omega_0) = 1 \ne 0\), so they are a basis. With \(x(0) = x_0\), \(\dot x(0) = 0\): \[x(t) = x_0\,e^{-\omega_0t}(1+\omega_0t), \qquad \dot x = -x_0\,\omega_0^2\,t\,e^{-\omega_0t}.\]
2. Proof. For \(\gamma > 2\omega_0\) both roots are real and negative, with \(\lambda_+\lambda_- = \omega_0^2\) and \(|\lambda_-| > |\lambda_+|\) (because \(\lambda_- < \lambda_+ < 0\)). Hence \(|\lambda_+|^2 < |\lambda_+||\lambda_-| = \omega_0^2\), so \(|\lambda_+| < \omega_0\): every overdamped solution decays strictly more slowly than \(e^{-\omega_0t}\). And \(|\lambda_+| = \omega_0^2/|\lambda_-| \le \omega_0^2/(\gamma/2) \to 0\) as \(\gamma \to \infty\). \(\square\)

3. Proof. For \(\gamma < 2\omega_0\), the solution from rest is \[x(t) = x_0e^{-\gamma t/2}\Big(\cos\omega_\mathrm{d}t + \tfrac{\gamma}{2\omega_\mathrm{d}}\sin\omega_\mathrm{d}t\Big) = x_0A\,e^{-\gamma t/2}\cos(\omega_\mathrm{d}t-\psi),\] with \(A = \sqrt{1+\gamma^2/4\omega_\mathrm{d}^2} > 0\) and \(\tan\psi = \gamma/2\omega_\mathrm{d}\). The cosine vanishes at \(t = (\psi + (n+\tfrac12)\pi)/\omega_\mathrm{d}\) for every integer \(n\), and the exponential prefactor never does; so \(x\) changes sign infinitely often. \(\square\)

   Combining: the non-overshooting cases are exactly \(\gamma \ge 2\omega_0\), and among them the eventual decay rate is \(\omega_0\) at \(\gamma = 2\omega_0\) and \(|\lambda_+| < \omega_0\) for every larger \(\gamma\). It is therefore maximised at \(\gamma = 2\omega_0\): critical damping is the unique fastest return without overshoot.

4. \(\gamma = 16\): solve \((1+8t)e^{-8t} = 0.01\). Putting \(u = 8t\), \((1+u)e^{-u} = 0.01\) gives \(u = 6.6384\), so \(t = 0.830\) s. \(\gamma = 80\): \(x/x_0 = (\lambda_-e^{\lambda_+t}-\lambda_+e^{\lambda_-t})/ (\lambda_--\lambda_+) = 1.0103\,e^{-0.80816t} - 0.0103\,e^{-79.19t}\); the fast term is negligible after a few hundredths of a second, and \(1.0103\,e^{-0.80816t} = 0.01\) gives \(t = 5.71\) s. Ratio \(5.711/0.8298 = 6.88\).

Check. Dimensions: \(\omega_0t\) is a pure number ✓. Limit: as \(\gamma \to 2\omega_0^+\), \(|\lambda_+| \to \omega_0\) and the overdamped settling time tends continuously to the critical one ✓. Number: at \(\gamma = 80\) the instrument is \(6.88\) times slower than at \(\gamma = 16\), which is the whole reason a shorted galvanometer is useless as a meter.

Idea. Maximise \(|\chi|^2\) in the variable \(u = \omega^2\); then notice that the velocity and power curves are a different function of \(\omega\), whose maximum is exact.

Working.

1. \(|\chi|^{-2} = D(\omega) = (\omega_0^2-\omega^2)^2 + \gamma^2\omega^2\). With \(u = \omega^2 \ge 0\), \(D = (\omega_0^2-u)^2 + \gamma^2u\) is a quadratic in \(u\) with positive leading coefficient, minimised where \(\mathrm{d}D/\mathrm{d}u = -2(\omega_0^2-u)+\gamma^2 = 0\), i.e. \(u = \omega_0^2-\gamma^2/2\). That is admissible (\(u > 0\)) exactly when \(\gamma^2 < 2\omega_0^2\), i.e. \(Q > 1/\sqrt2\); otherwise the minimum over \(u \ge 0\) is at \(u = 0\) and \(|\chi|\) decreases from the start. At the interior minimum, \[D = \Big(\frac{\gamma^2}{2}\Big)^2 + \gamma^2\Big(\omega_0^2-\frac{\gamma^2}{2}\Big) = \gamma^2\Big(\omega_0^2-\frac{\gamma^2}{4}\Big) = \gamma^2\omega_\mathrm{d}^2,\] so \(|\chi|_{\max} = 1/(\gamma\omega_\mathrm{d})\).
2. The velocity amplitude and the mean absorbed power are \[\omega|\chi| = \Big[\frac{(\omega_0^2-\omega^2)^2}{\omega^2}+\gamma^2\Big]^{-1/2}, \qquad \bar P = \tfrac12\gamma F_0^2\Big[\frac{(\omega_0^2-\omega^2)^2}{\omega^2} + \gamma^2\Big]^{-1}.\] Both are decreasing functions of \(g(\omega)^2\), where \(g(\omega) = \omega_0^2/\omega - \omega\). Since \(g^2 \ge 0\) with equality only at \(\omega = \omega_0\), both are maximal exactly there, for every \(\gamma\) and with no approximation.
3. Half maximum needs \(g^2 = \gamma^2\), i.e. \(|\omega_0^2-\omega^2| = \gamma\omega\), i.e. \(\omega^2 \pm \gamma\omega - \omega_0^2 = 0\). The positive roots are \(\omega_\mp = \mp\tfrac\gamma2 + \sqrt{\omega_0^2+\gamma^2/4}\), so \(\Delta\omega = \omega_+-\omega_- = \gamma\) exactly, and \(\omega_0/\Delta\omega = \omega_0/\gamma = Q\).

4. \(\gamma = 0.4\): \(\omega_0 = 8.00000\), \(\omega_\mathrm{d} = \sqrt{64-0.04} = 7.99750\), \(\omega_{\text{peak}} = \sqrt{64-0.08} = 7.99500\). Half-power points \(\mp0.2+\sqrt{64.04} = 7.8025\) and \(8.2025\), difference \(0.4000\).

   A physicist saying "the resonant frequency" means \(\omega_0\), and is right to: it is where the system absorbs power, it is where the phase is exactly \(\pi/2\), and both statements are exact for every damping. \(\omega_{\text{peak}}\) is a fact about the amplitude curve, and \(\omega_\mathrm{d}\) about the free ring-down; neither is what an absorption measurement returns.

Check. Dimensions: all four frequencies are \(\mathrm{T^{-1}}\), and \(\gamma^2/2\) is too ✓. Limit \(\gamma \to 0\): all three collapse to \(\omega_0\) and \(|\chi|_{\max} \to \infty\) — undamped resonance ✓. Number: \(|\chi|_{\max} = 1/(0.4\times7.99750) = 0.3125977\) against \(|\chi(\omega_0)| = 1/3.2 = 0.3125000\); the peak is \(0.031\%\) higher.

Idea. \(\varphi = \arg(\omega_0^2-\omega^2+i\gamma\omega)\), so every question is about one complex number's argument.

Working.

1. At \(\omega = \omega_0\) the real part vanishes and the imaginary part is \(\gamma\omega_0 > 0\), so the number is on the positive imaginary axis and \(\varphi = \pi/2\) — independent of \(\gamma\). The imaginary part \(\gamma\omega\) is positive for all \(\omega > 0\), so \(\varphi \in (0,\pi)\) always; the real part decreases strictly from \(\omega_0^2\) to \(-\infty\), and \(\cot\varphi = (\omega_0^2-\omega^2)/(\gamma\omega) = \omega_0^2/\gamma\omega - \omega/\gamma\) decreases strictly, so \(\varphi\) increases strictly, from \(0^+\) to \(\pi^-\).
2. \(\varphi = \pi/4\) needs \(\cot\varphi = +1\), i.e. \(\omega_0^2-\omega^2 = \gamma\omega\), i.e. \(\omega = \omega_- = -\tfrac\gamma2 + \sqrt{\omega_0^2+\gamma^2/4}\). \(\varphi = 3\pi/4\) needs \(\cot\varphi = -1\), giving \(\omega = \omega_+ = +\tfrac\gamma2 + \sqrt{\omega_0^2+\gamma^2/4}\). Difference \(\gamma\), exactly. These are the half-power points of P5: the power falls to half precisely where the phase has turned a further quarter-turn either side of \(\pi/2\).
3. Differentiate \(\cot\varphi = \omega_0^2/\gamma\omega - \omega/\gamma\): \(-\csc^2\!\varphi\;\varphi' = -\omega_0^2/(\gamma\omega^2) - 1/\gamma\). At \(\omega = \omega_0\), \(\csc^2\varphi = 1\), so \(\varphi'(\omega_0) = \dfrac{2}{\gamma} = \dfrac{2Q}{\omega_0}\).

4. \(\gamma = 0.4\), \(Q = 20\). A \(1\%\) shift is \(\Delta\omega = 0.08\, \mathrm{rad\,s^{-1}}\). Exactly: at \(\omega = 8.08\), \(\varphi = 111.70^\circ\), a change of \(+21.70^\circ\); at \(\omega = 7.92\), \(\varphi = 68.10^\circ\), a change of \(-21.90^\circ\). (The linearised estimate \(2\Delta\omega/\gamma = 0.4\) rad \(= 22.92^\circ\) is a few per cent high, as it must be: the phase curve flattens on both sides of its steep middle, as an arctangent does.) The amplitude meanwhile changes by \(-8.0\%\) above and \(-6.3\%\) below.

   Phase is the better instrument for two reasons: a \(21.7^\circ\) change is far easier to measure than an \(8\%\) one, and phase is signed — it tells you which way to move — whereas the amplitude falls on both sides and cannot.

Check. Dimensions: \(\varphi' = 2/\gamma\) has dimension \(\mathrm{T}\), as \(\mathrm{d}(\text{angle})/\mathrm{d}\omega\) must ✓. Limit \(\gamma\to0\): \(\varphi'(\omega_0) \to \infty\) — the phase jumps from \(0\) to \(\pi\) at \(\omega_0\), which is the undamped case ✓. Number: \(2Q/\omega_0 = 40/8 = 5\) radians per \(\mathrm{rad\,s^{-1}}\).

Idea. Convolution with a constant is an integral of \(G\), and that integral has a closed form. The piecewise route must give the same thing, and the corner is where the check bites.

Working.

The step response. Let \[H(t) = \int_0^{t}G(u)\,\mathrm{d}u = \frac{1}{\omega_0^2}\Big[1 - e^{-\gamma t/2}\Big(\cos\omega_\mathrm{d}t + \frac{\gamma}{2\omega_\mathrm{d}}\sin\omega_\mathrm{d}t\Big)\Big] \equiv \frac{h(t)}{\omega_0^2}.\] Verification, by differentiating (all trigonometric arguments are \(\omega_\mathrm{d}t\)): \[h'(t) = \tfrac{\gamma}{2} e^{-\gamma t/2}\Big(\cos + \tfrac{\gamma}{2\omega_\mathrm{d}}\sin\Big) - e^{-\gamma t/2}\Big(-\omega_\mathrm{d}\sin + \tfrac{\gamma}{2}\cos\Big) = e^{-\gamma t/2}\sin\,\Big(\tfrac{\gamma^2}{4\omega_\mathrm{d}} + \omega_\mathrm{d}\Big),\] which is \(\frac{\omega_0^2}{\omega_\mathrm{d}}e^{-\gamma t/2}\sin\omega_\mathrm{d}t = \omega_0^2\,G(t)\), using \(\gamma^2/4+\omega_\mathrm{d}^2 = \omega_0^2\). And \(h(0) = 0\). ✓

1. By convolution. \(f(s) = F_0\) on \([0,T]\), so with \(u = t-s\) \[t \le T:\quad \theta = F_0\!\int_0^{t}\! G(u)\,\mathrm{d}u = \frac{F_0}{\omega_0^2}h(t); \qquad t > T:\quad \theta = F_0\!\int_{t-T}^{t}\! G(u)\,\mathrm{d}u = \frac{F_0}{\omega_0^2}\big[h(t)-h(t-T)\big].\]
2. By matching. On \([0,T]\) the constant \(\theta = F_0/\omega_0^2\) is a particular solution, so \(\theta = F_0/\omega_0^2 + e^{-\gamma t/2}(A\cos \omega_\mathrm{d}t + B\sin\omega_\mathrm{d}t)\). \(\theta(0) = 0\) gives \(A = -F_0/\omega_0^2\); \(\dot\theta(0) = -\tfrac\gamma2A + \omega_\mathrm{d}B = 0\) gives \(B = \gamma A/2\omega_\mathrm{d}\). Hence \(\theta = (F_0/\omega_0^2)h(t)\) on \([0,T]\), and \[\theta(T) = \frac{F_0}{\omega_0^2}h(T), \qquad \dot\theta(T) = \frac{F_0}{\omega_0^2}h'(T) = F_0\,G(T).\] For \(t > T\), \(\theta\) solves \(L\theta = 0\) with these data.

3. They agree. Put \(\theta_2(t) = (F_0/\omega_0^2)[h(t)-h(t-T)]\). Since \(L\big((F_0/\omega_0^2)h\big) = F_0\) for \(t > 0\) (it is the step response, and \(L\) applied to \(F_0\!\int_0^tG\) gives \(F_0\) by Duhamel), we get \(L\theta_2 = F_0 - F_0 = 0\) for \(t > T\). At \(t = T\): \(h(0) = 0\) and \(h'(0) = \omega_0^2G(0) = 0\), so \(\theta_2\) and \(\dot\theta_2\) take the values \((F_0/\omega_0^2)h(T)\) and \(F_0G(T)\) from the left as well as from the right. Same equation, same data, so by uniqueness (unit 01) the two solutions coincide.

   \(\ddot\theta = f - \gamma\dot\theta - \omega_0^2\theta\), and \(\theta, \dot\theta\) are continuous while \(f\) drops from \(F_0\) to \(0\); hence \(\ddot\theta\) jumps by \(-F_0 = -20\,\mathrm{rad\,s^{-2}}\). This is not a defect: the drive itself is discontinuous, and the equation says the acceleration must follow it. The solution is \(C^1\) everywhere and \(C^2\) except at \(t = 0, T\), which is the right regularity for a discontinuous forcing.

4. Numbers, with \(F_0 = 20\), \(F_0/\omega_0^2 = 0.3125\) rad, \(\gamma = 0.4\), \(\omega_\mathrm{d} = 7.997500\), \(T = 0.1\) s:

   \(t\) / s	\(\theta\) / rad	\(\theta\) / deg
   \(0.05\)	\(0.02451\)	\(1.404\)
   \(0.10\)	\(0.09354\)	\(5.360\)
   \(0.20\)	\(0.22007\)	\(12.609\)
   \(0.50\)	\(-0.09797\)	\(-5.614\)
   \(1.00\)	\(0.19469\)	\(11.155\)

   The maximum is where \(\dot\theta = 0\), i.e. \(G(t) = G(t-T)\), at \(t = 0.24345\) s, with \(\theta_{\max} = 0.23416\,\mathrm{rad} = 13.416^\circ\).

   The ballistic estimate treats the pulse as an impulse of the same area, \(F_0T = 2\,\mathrm{rad\,s^{-1}}\), predicting \(F_0T\,G(t_\ast) = 2\times0.120260 = 0.24052\) rad — \(2.72\%\) high. It should be: \(T = 0.1\) s is not small compared with \(T_0 = 0.785\) s, and spreading the impulse over a time during which the oscillator moves costs amplitude.

Check. Dimensions: \(F_0/\omega_0^2\) is an angle ✓. Limit \(T \to \infty\): \(\theta \to F_0/\omega_0^2 = 0.3125\) rad, the static deflection for \(0.1\) mA (and indeed \(\alpha i_0/\kappa = 2\times10^{-3}\times10^{-4}/6.4\times10^{-7} = 0.3125\)) ✓. Limit \(T \to 0\) at fixed \(F_0T\): the two-term expression tends to \(F_0T\,G(t)\), the impulse response ✓. Number: \(\theta_{\max} = 13.4^\circ\) is well inside full scale, so the instrument is being used in its linear range.

Idea. Superpose the steady state with the free solution that cancels it at \(t = 0\); the result is a product of sines, and resonance is its limit.

Working.

1. The steady state is \(F_0\cos\omega t/(\omega_0^2-\omega^2)\); adding \(-F_0\cos\omega_0t/(\omega_0^2-\omega^2) \in \ker L\) gives a solution with \(x(0) = 0\), and \(\dot x(0) = 0\) automatically since both cosines have zero slope at the origin. So \[x(t) = \frac{F_0(\cos\omega t-\cos\omega_0t)}{\omega_0^2-\omega^2} = \frac{2F_0}{\omega_0^2-\omega^2}\, \sin\!\Big(\frac{\omega_0+\omega}{2}t\Big)\sin\!\Big(\frac{\omega_0-\omega}{2}t\Big),\] using \(\cos a - \cos b = 2\sin\frac{a+b}{2}\sin\frac{b-a}{2}\). With \(\Delta = \omega_0-\omega\), the fast factor oscillates at \((\omega_0+\omega)/2\) and the envelope is \(|\sin(\Delta t/2)|\), which returns to a maximum every \(2\pi/|\Delta|\), first at \(t = \pi/|\Delta|\). The envelope's maximum is \(2F_0/|\omega_0^2-\omega^2|\) (a ceiling \(|x|\) never exceeds).
2. Fix \(t\) and let \(\omega \to \omega_0\). Both numerator and denominator vanish; differentiating each with respect to \(\omega\), \(\partial_\omega[F_0(\cos\omega t-\cos\omega_0t)] = -F_0t\sin\omega t\) and \(\partial_\omega[\omega_0^2-\omega^2] = -2\omega\), so by l'Hôpital \[x(t) = \frac{F_0\,t\sin\omega_0t}{2\omega_0}.\] Verify: \(\dot x = \frac{F_0}{2\omega_0}(\sin\omega_0t + \omega_0t\cos\omega_0t)\), \(\ddot x = \frac{F_0}{2\omega_0}(2\omega_0\cos\omega_0t - \omega_0^2t\sin\omega_0t)\), so \(\ddot x + \omega_0^2x = F_0\cos\omega_0t\) ✓, and \(x(0) = \dot x(0) = 0\) ✓.
3. \(L(Ae^{i\omega_0t}) = p(i\omega_0)Ae^{i\omega_0t} = 0\), because \(\gamma = 0\) makes \(p(i\omega_0) = \omega_0^2-\omega_0^2 = 0\): the one-dimensional space \(\mathbb{C}e^{i\omega_0t}\) that the ansatz searches is mapped by \(L\) to \(\{0\}\), so it cannot contain a preimage of \(F_0e^{i\omega_0t}\). The general rule: an exponential ansatz fails exactly when the drive already lies in \(\ker L\), and the cure is to multiply by \(t\) — \(L(te^{\lambda t}) = p'(\lambda)e^{\lambda t}\) when \(p(\lambda) = 0\), and \(p'(i\omega_0) = 2i\omega_0 \ne 0\).

4. \(F_0 = 0.2\,\mathrm{rad\,s^{-2}}\), \(\omega_0 = 8\). At exact resonance the envelope is \(F_0t/2\omega_0 = 0.0125\,t\) rad, reaching full scale \(1.5625\) rad at \(t = 1.5625/0.0125 = \mathbf{125}\) s — over two minutes, which is what "infinite response" looks like in practice.

   Off resonance, the beat maximum is \(2F_0/|\omega_0^2-\omega^2| = 1.5625\) when \(|\omega_0^2-\omega^2| = 0.4/1.5625 = 0.256\), i.e. \(\omega^2 = 63.744\), \(\omega = 7.98398\). The detuning is \(\Delta = 0.01602\,\mathrm{rad\,s^{-1}}\), a fractional detuning of \(0.20\%\); the first full-scale excursion arrives at \(\pi/\Delta = 196\) s and they recur every \(392\) s.

Check. Dimensions: \([F_0t/\omega_0] = [x]\) ✓. Limit: as \(\Delta \to 0\) at fixed \(t\), \(2F_0\sin(\Delta t/2)/(2\omega_0\Delta) \to F_0t/2\omega_0\), reproducing part 2 ✓. Number: \(2F_0/(\omega_0^2-\omega^2)\) at \(\omega = 7.98398\) is \(0.4/0.256 = 1.5625\) rad, full scale ✓.

Idea. The pulse is a probability density of width \(\tau\); convolving with it is the same as averaging \(G\) over a small window, and averaging shifts a maximum without changing its height at first order.

Working.

1. The drive is \(f(s) = (\alpha q/I)\,w(s)\) with \(w(s) = \tau^{-1}e^{-s/\tau}\) for \(s \ge 0\), so \(\int_0^\infty w = 1\). Duhamel gives \(\theta(t) = (\alpha q/I)\int_0^{t}G(t-s)w(s)\,\mathrm{d}s\). For \(t \gg \tau\) the tail beyond \(s = t\) is \(O(e^{-t/\tau})\) and may be added back; expanding \(G(t-s) = \sum_n G^{(n)}(t)(-s)^n/n!\) and using \(\int_0^\infty s^nw(s)\,\mathrm{d}s = n!\,\tau^n\), \[\theta(t) = \frac{\alpha q}{I}\sum_{n\ge0}\frac{(-1)^n}{n!}G^{(n)}(t)\,n!\,\tau^n = \frac{\alpha q}{I}\Big(G(t) - \tau\dot G(t) + \tau^2\ddot G(t) - \cdots\Big).\]

2. Write \(\Theta(t) = G - \tau\dot G + \tau^2\ddot G\), correct to \(O(\tau^3)\). Its maximum is where \(\Theta' = \dot G - \tau\ddot G + O(\tau^2) = 0\); putting \(t = t_\ast+\delta\) and using \(\dot G(t_\ast) = 0\), \(\delta\ddot G(t_\ast) - \tau\ddot G(t_\ast) + O(\tau^2,\delta^2) = 0\), so \(\delta = \tau + O(\tau^2)\).

   At that point, expanding each term about \(t_\ast\) and keeping \(O(\tau^2)\): \(G(t_\ast+\tau) = G + \tfrac12\tau^2\ddot G\); \(\tau\dot G(t_\ast+\tau) = \tau^2\ddot G\); \(\tau^2\ddot G(t_\ast+\tau) = \tau^2\ddot G\). Summing, \(\Theta_{\max} = G(t_\ast) + \big(\tfrac12 - 1 + 1\big)\tau^2\ddot G(t_\ast) = G(t_\ast) + \tfrac12\tau^2\ddot G(t_\ast)\). Finally \(LG = 0\) at \(t_\ast\) with \(\dot G(t_\ast) = 0\) gives \(\ddot G(t_\ast) = -\omega_0^2G(t_\ast)\), so \[\theta_{\max} = \frac{\alpha q}{I}G(t_\ast)\Big(1-\tfrac12\omega_0^2\tau^2\Big)+O(\tau^3).\] The first-order term cancels because \(\dot G(t_\ast) = 0\): shifting a smooth function's argument near its maximum changes the argument at first order but the value only at second.

3. \(\omega_0\tau = 8\times0.010 = 0.08\), so the fractional error is \(-\tfrac12(0.08)^2 = -0.0032 = \mathbf{-0.32\%}\). (Solving the problem numerically gives \(-0.319\%\), and the peak displaced by \(10.0\) ms — exactly \(\tau\), as predicted.) For an error of \(1\%\), \(\tfrac12\omega_0^2\tau^2 = 0.01\) gives \(\tau = \sqrt{0.02}/8 = 17.7\) ms.

Check. Dimensions: \(\omega_0\tau\) is a pure number ✓. Limit \(\tau \to 0\): the series collapses to \((\alpha q/I)G(t)\), the ballistic answer ✓. Number: with \(q = 1\,\mu\mathrm{C}\) and \(\gamma = 0.4\), the ideal throw is \(0.024052\) rad and the \(10\) ms discharge gives \(0.023975\) rad — the deck's number, reduced by \(0.32\%\).

Idea. Two computations and one uniqueness argument.

Working.

1. Distinct roots. Take \(x_1 = e^{\lambda_+t}\), \(x_2 = e^{\lambda_-t}\). Then \(W(s) = e^{\lambda_+s}\lambda_-e^{\lambda_-s} - e^{\lambda_-s}\lambda_+ e^{\lambda_+s} = (\lambda_--\lambda_+)e^{(\lambda_++\lambda_-)s}\), and \[K(t,s) = \frac{e^{\lambda_+s}e^{\lambda_-t}-e^{\lambda_-s}e^{\lambda_+t}} {(\lambda_--\lambda_+)e^{(\lambda_++\lambda_-)s}} = \frac{e^{\lambda_-(t-s)}-e^{\lambda_+(t-s)}}{\lambda_--\lambda_+} = \frac{e^{\lambda_+(t-s)}-e^{\lambda_-(t-s)}}{\lambda_+-\lambda_-} = G(t-s).\]

   Repeated root \(\lambda = -\omega_0\). Take \(x_1 = e^{\lambda t}\), \(x_2 = te^{\lambda t}\); then \(W(s) = e^{\lambda s}(1+\lambda s)e^{\lambda s} - se^{\lambda s}\lambda e^{\lambda s} = e^{2\lambda s}\), and \(K(t,s) = \big(e^{\lambda s}te^{\lambda t}-se^{\lambda s}e^{\lambda t}\big)/ e^{2\lambda s} = (t-s)e^{\lambda(t-s)} = G(t-s)\).

   Independence of basis. If \((y_1,y_2) = (x_1,x_2)M\) with \(M\) an invertible constant matrix, then the numerator of \(K\) is \(\det\begin{pmatrix}y_1(s)&y_2(s)\\y_1(t)&y_2(t)\end{pmatrix}\), which picks up a factor \(\det M\); and \(W_y = \det M\cdot W_x\) likewise. The factors cancel, so \(K\) is unchanged. (It had to be: \(K\) is determined by the equation and the data, and no basis was mentioned in the statement of the problem \(Lx = f\), \(x(t_0) = \dot x(t_0) = 0\).)

2. Put \(x_h(t) = x_0\big(\dot G(t-t_0)+\gamma G(t-t_0)\big) + v_0G(t-t_0)\). Each of \(G(\cdot-t_0)\) and \(\dot G(\cdot-t_0)\) is in \(\ker L\) on \(t > t_0\): the first because \(LG = 0\) there, the second because \(L\) has constant coefficients and therefore commutes with \(\mathrm{d}/\mathrm{d}t\) (\(L\dot G = \frac{\mathrm{d}}{\mathrm{d}t}LG = 0\)). So \(x_h \in \ker L\). At \(t = t_0^+\): \(G = 0\), \(\dot G = 1\), and \(\ddot G = -\gamma\dot G-\omega_0^2G = -\gamma\), so \[x_h(t_0) = x_0(1+0) + 0 = x_0, \qquad \dot x_h(t_0) = x_0(-\gamma+\gamma) + v_0 = v_0 .\] Meanwhile \((G*f)(t_0) = 0\) and its derivative there is \(G(0)f(t_0) + \int_{t_0}^{t_0}\dot G\,f = 0\). So \(x = x_h + G*f\) satisfies \(Lx = f\) with the stated data, and by unit 01's uniqueness theorem it is the solution. \(\square\)
3. If two systems have the same \(G\) then, by part 2, they produce the same \(x\) for every \(f\) and every \((x_0,v_0)\) — the formula contains no other information about the system. The two numbers hidden in \(G\) are \(\gamma\) and \(\omega_0\). From a measured \(G\): read the ringing period to get \(\omega_\mathrm{d} = 2\pi/T_\mathrm{d}\), read the logarithmic decrement of successive maxima to get \(\gamma = 2\delta/T_\mathrm{d}\), and then \(\omega_0 = \sqrt{\omega_\mathrm{d}^2+\gamma^2/4}\). Two readings off one ring-down and the system is known completely.

Check. Dimensions: \([\dot G] = 1\) and \([\gamma G] = 1\), so \(x_0\) multiplies a pure number ✓. Limit \(f = 0\), \(x_0 = 0\): \(x = v_0G\), which is the definition of the impulse response ✓. Number: for the galvanometer at \(\gamma = 0.4\), \(T_\mathrm{d} = 0.78564\) s and \(\delta = 0.15713\) give \(\gamma = 2\times0.15713/0.78564 = 0.4000\) and \(\omega_0 = \sqrt{7.99750^2+0.04} = 8.0000\) ✓.

Idea. Differentiate \(E\) once; then note that the energy is an exponential times something periodic, which makes the "per cycle" statement exact.

Working.

1. \(\dot E = \dot x\ddot x + \omega_0^2x\dot x = \dot x(\ddot x + \omega_0^2x) = \dot x(-\gamma\dot x) = -\gamma\dot x^2 \le 0\), with equality only where \(\dot x = 0\). This is unit 02's gradient-like flow: \(E\) is a Lyapunov function with \(\dot E \le 0\), non-strict, which is exactly the situation LaSalle's invariance principle was built for — and it gives asymptotic stability of the origin without any linearisation.
2. Write \(x = Ae^{-\gamma t/2}\cos(\omega_\mathrm{d}t-\phi)\); then both \(x\) and \(\dot x\) are \(e^{-\gamma t/2}\) times a \(T_\mathrm{d}\)-periodic function, so \(E(t) = e^{-\gamma t}P(t)\) with \(P(t+T_\mathrm{d}) = P(t)\). Hence \[\int_{t+T_\mathrm{d}}^{t+2T_\mathrm{d}}\!\!E = \int_{t}^{t+T_\mathrm{d}}\!\! e^{-\gamma(s+T_\mathrm{d})}P(s+T_\mathrm{d})\,\mathrm{d}s = e^{-\gamma T_\mathrm{d}}\int_{t}^{t+T_\mathrm{d}}\!\! E,\] substituting \(s \mapsto s+T_\mathrm{d}\). The period-average therefore falls by exactly \(e^{-\gamma T_\mathrm{d}}\) each cycle — no small-\(\gamma\) assumption.
3. Successive maxima of \(x\) are \(T_\mathrm{d}\) apart (the stationarity condition \(\tan(\omega_\mathrm{d}t-\phi) = -\gamma/2\omega_\mathrm{d}\) has solutions spaced by \(\pi/\omega_\mathrm{d}\), alternating maxima and minima), and at each the cosine factor repeats; so the ratio is the envelope's, \(\delta = \gamma T_\mathrm{d}/2 = \pi\gamma/\omega_\mathrm{d}\). Then \[\frac{\pi}{\delta} = \frac{\omega_\mathrm{d}}{\gamma} = \frac{\omega_0}{\gamma}\sqrt{1-\frac{1}{4Q^2}} = Q\Big(1 - \frac{1}{8Q^2}+O(Q^{-4})\Big),\] a relative error \(1/8Q^2\).
4. \(\gamma = 0.4\), \(\omega_\mathrm{d} = 7.997500\): \(T_\mathrm{d} = 2\pi/\omega_\mathrm{d} = 0.785644\) s; \(\delta = \pi\gamma/\omega_\mathrm{d} = 0.157129\); successive-maximum ratio \(e^{-\delta} = 0.854594\); energy lost per cycle \(1-e^{-\gamma T_\mathrm{d}} = 0.269669\); \(2\pi/Q = 0.314159\). The last two disagree by \((0.314159-0.269669)/0.269669 = \mathbf{16.5\%}\). The definition used in this course is \(Q = \omega_0/\gamma\); the "energy-per-cycle" reading is a consequence of it, valid to \(O(1/Q)\), and it is the one to distrust.

Check. Dimensions: \(E\) is \([x]^2\mathrm{T^{-2}}\) throughout (unit mass), and \(\gamma\dot x^2\) is \([x]^2\mathrm{T^{-3}}\) as \(\dot E\) must be ✓. Limit \(\gamma\to0\): \(\delta \to 0\), \(\pi/\delta \to \infty = Q\) ✓. Number: \(\pi/\delta = 19.9937\) against \(Q = 20\), a relative error \(3.1\times10^{-4} = 1/(8\times400)\) ✓ — the predicted \(1/8Q^2\), to four figures.

Idea. \(G\) is a sum of two exponentials, so the integral is elementary; the lower limit \(0\) rather than \(-\infty\) is what makes one half-plane special.

Working.

1. Proof. \(G(\tau) = \big(e^{\lambda_+\tau}-e^{\lambda_-\tau}\big)/ (\lambda_+-\lambda_-)\) for \(\tau \ge 0\), with \(\operatorname{Re}\lambda_\pm < 0\). For real \(\omega\) and either root, \[\int_0^\infty e^{(\lambda-i\omega)\tau}\,\mathrm{d}\tau = \Big[\frac{e^{(\lambda-i\omega)\tau}}{\lambda-i\omega}\Big]_0^\infty = \frac{-1}{\lambda-i\omega},\] the upper limit vanishing because \(|e^{(\lambda-i\omega)\tau}| = e^{\operatorname{Re}\lambda\,\tau} \to 0\). Hence \[\hat G(\omega) = \frac{1}{\lambda_+-\lambda_-} \Big[\frac{-1}{\lambda_+-i\omega}+\frac{1}{\lambda_--i\omega}\Big] = \frac{1}{\lambda_+-\lambda_-}\cdot\frac{\lambda_+-\lambda_-} {(\lambda_+-i\omega)(\lambda_--i\omega)} = \frac{1}{(i\omega-\lambda_+)(i\omega-\lambda_-)},\] the last step using \((-1)(-1) = 1\) on both factors. Since \(p(\mu) = (\mu-\lambda_+)(\mu-\lambda_-)\), this is \(1/p(i\omega) = \chi(\omega)\). \(\square\) (The repeated-root case follows by continuity, or by the same integral with \(G = \tau e^{\lambda\tau}\), which gives \(1/(i\omega-\lambda)^2\).)

2. Proof. Let \(\omega = \omega_r + i\omega_i\). Then \(|e^{-i\omega\tau}| = e^{\omega_i\tau}\) and \(|G(\tau)| \le e^{-\gamma\tau/2}/\omega_\mathrm{d}\), so the integrand is bounded by \(e^{(\omega_i-\gamma/2)\tau}/\omega_\mathrm{d}\), which is integrable on \([0,\infty)\) whenever \(\omega_i < \gamma/2\). \(\square\)

   The hypothesis matters. Overdamped, \(0 \le G(\tau) \le e^{\lambda_+\tau}/(\lambda_+-\lambda_-)\) and the same argument gives only \(\omega_i < |\lambda_+|\), which is smaller than \(\gamma/2\) because \(\lambda_+ > -\gamma/2\); and it cannot do better, since \(\chi\) has a pole at \(\omega = i|\lambda_+|\). Critically damped, \(G = \tau e^{-\omega_0\tau}\) gives \(\omega_i < \omega_0 = \gamma/2\). In every regime the closed lower half-plane is inside.

   The property responsible is \(G(\tau) = 0\) for \(\tau < 0\). Had \(G\) been non-zero on the negative axis, the integral would have run over all of \(\mathbb{R}\), and the factor \(e^{\omega_i\tau}\) would have blown up at one end or the other for every \(\omega_i \ne 0\): no half-plane would be distinguished. One half-plane, not both, is the mathematical shadow of "no response before the kick".

3. \(p(i\omega) = -\omega^2+i\gamma\omega+\omega_0^2 = 0\) gives \(\omega^2 - i\gamma\omega - \omega_0^2 = 0\), so \[\omega = \frac{i\gamma \pm \sqrt{4\omega_0^2-\gamma^2}}{2} = \pm\omega_\mathrm{d} + \frac{i\gamma}{2}.\] For the galvanometer at \(\gamma = 0.4\): \(\omega = \pm7.9975 + 0.2\,i\). The real part is the ringing frequency, the imaginary part the decay rate of the envelope: a pole of the response is a free mode of the system, written as one complex number. They sit in the upper half-plane, at height \(\gamma/2\) — on the far side from the region where \(\hat G\) converges, exactly as causality requires.

   As \(\gamma \to 0\) the poles descend to the real axis at \(\pm\omega_0\). A response function with a singularity at a real frequency has no bounded steady state there, which is undamped resonance seen from the complex plane; and the "no singularity in the lower half-plane" statement becomes marginal, which is the analytic face of the fact that an undamped oscillator never forgets its initial conditions.

4. Still to be proved: (i) that an integral of this form defines a holomorphic function of \(\omega\) on the open half-plane, not merely a convergent one — Morera's theorem, or differentiation under the integral sign; (ii) Cauchy's theorem, to say that the contour integral of \(\chi(\omega')/(\omega'-\omega)\) round a closed path in that half-plane vanishes; (iii) an estimate showing the large semicircle contributes nothing, which needs \(\chi(\omega) \to 0\) as \(|\omega|\to\infty\) — true here, since \(\chi \sim -1/\omega^2\); and (iv) the Sokhotski–Plemelj limit, which is where the principal value and the factor \(1/\pi\) come from. All four are unit 20. Attempting them here would be using what has not arrived.

Check. Dimensions: \([\hat G] = \mathrm{T}\times\mathrm{T} = \mathrm{T^2} = [\chi]\) ✓. Limit: at \(\omega = 0\), \(\hat G(0) = \int_0^\infty G = 1/\omega_0^2 = \chi(0)\) ✓ — and that integral is the step response at \(t = \infty\), the static deflection, which is the same statement in the time domain. Number: at \(\gamma = 0.4\), \(\omega_0 = 8\), \(\omega = 5\), numerical integration of \(\int_0^\infty G(\tau)e^{-5i\tau}\mathrm{d}\tau\) gives \(0.0255738 - 0.0013115\,i\), and \(1/(39+2i) = 0.0255738 - 0.0013115\,i\) ✓.