Differential Equations Question of the Day

1. $4\sqrt{5}$
2. $8$
3. $2\sqrt{5}$
4. $2\sqrt{10}$ (correct answer)

Explanation: This problem tests your understanding of resonance in damped harmonic oscillators. When you see a differential equation of the form $my'' + \gamma y' + ky = F(t)$ with a driving force, you're dealing with forced oscillations where amplitude depends on the driving frequency. For maximum steady-state amplitude in a damped system, resonance occurs at the frequency $\omega_{\text{res}} = \sqrt{\omega_0^2 - \frac{\gamma^2}{2m^2}}$, where $\omega_0 = \sqrt{\frac{k}{m}}$ is the natural frequency. From your equation $2y'' + \gamma y' + 18y = F(t)$, we have $m = 2$ and $k = 18$, so $\omega_0 = \sqrt{\frac{18}{2}} = 3$ rad/s. Since resonance occurs at $\omega = 2$ rad/s, we can solve: $2 = \sqrt{9 - \frac{\gamma^2}{8}}$ Squaring both sides: $4 = 9 - \frac{\gamma^2}{8}$ Rearranging: $\frac{\gamma^2}{8} = 5$ Therefore: $\gamma^2 = 40$, so $\gamma = 2\sqrt{10}$ Answer choice A ($4\sqrt{5}$) would give $\gamma^2 = 80$, making the resonant frequency too low. Choice B (8) gives $\gamma^2 = 64$, also incorrect. Choice C ($2\sqrt{5}$) gives $\gamma^2 = 20$, which would place resonance at a higher frequency than 2 rad/s. The correct answer is D. Study tip: Always identify the natural frequency first, then use the resonance condition. Remember that damping shifts the resonant frequency below the natural frequency, and heavier damping means a larger shift.