Undetermined Coefficients — Question 6

Question 6 – Solution

Strategy. Solve the two coefficient equations, then minimize the amplitude denominator instead of differentiating a complicated square root.

Step 1: Match the harmonic components. The equations are $$(5-\omega^2)A+2\omega B=F,\qquad -2\omega A+(5-\omega^2)B=0.$$ Writing $\Delta=(5-\omega^2)^2+4\omega^2>0$, we obtain $$\boxed{A=F(5-\omega^2)/\Delta,\qquad B=2F\omega/\Delta.}$$ At $\omega=0$, use the constant particular solution $F/5$.

Step 2: Maximize the amplitude. For all $\omega\ge 0$, including this constant-response convention, $$R=\frac{F}{\sqrt{\omega^4-6\omega^2+25}} =\frac{F}{\sqrt{(\omega^2-3)^2+16}}.$$ The denominator is minimized exactly at $\omega=\sqrt 3$, giving $\boxed{R_{\max}=F/4}$.

Step 3: Locate the quarter-cycle phase lag. At $\omega=\sqrt 5$, $A=0$ and $B=F/(2\sqrt 5)$, so $y_p=[F/(2\sqrt 5)]\sin(\sqrt 5t)$. This is a cosine delayed by phase $\pi/2$. Its amplitude $F/(2\sqrt 5)$ is smaller than $F/4$; the quarter-cycle phase frequency is not the amplitude-maximizing frequency.

Step 4: Prove the steady response is unique. Every difference from $y_p$ is $e^{-t}(C\cos 2t+D\sin 2t)$ and tends to zero forward. If a second solution had the same period $T=2\pi/\omega$, this difference would be periodic too. For any fixed $t$, its value equals its value at $t+nT$, whose limit is zero. Hence the difference is identically zero, proving uniqueness of the periodic response.