Standing Waves and Resonance — Question 9

Question 9

Distributed damping sets a finite driven-mode amplitude

Approximate the response by the fundamental mode alone: $y=B\sin(\pi x/L)\sin\omega_1t$. A viscous transverse force density $-b y_t$ acts along the string, where $b$ has units $\mathrm{kg/(m\,s)}$. A midpoint force $F_d\cos\omega_1t$ drives weakly damped steady resonance.

1. Equate mean input power to integrated mean damping loss and find $B$.

2. For $F_d=.0020\,\mathrm{N}$, $b=.0040\,\mathrm{kg/(m\,s)}$, $L=1.0\,\mathrm{m}$ and $\omega_1=100\,\mathrm{rad/s}$, find amplitude and mean input power. What fails as $b\to0$?

Question 9 - Solution

Distributed damping sets a finite driven-mode amplitude

Method and mathematical working

Average force times midpoint velocity for input power. For loss, integrate the average squared velocity over the whole mode before equating the two powers.$$\begin{aligned}\overline P_{\mathrm{in}}&=\frac12F_dB\omega_1,\\\overline P_{\mathrm{loss}}&=\frac12bB^2\omega_1^2\int_0^L\sin^2(\pi x/L)\,dx\\&=\frac14bLB^2\omega_1^2.\\[4pt]B&=\frac{2(0.0020)}{(0.0040)(1.0)(100)}=0.010\,\mathrm m,\\\overline P_{\mathrm{in}}&=\tfrac12(0.0020)(0.010)(100)=0.0010\,\mathrm W.\end{aligned}$$

(a) Calculation and conclusion

Midpoint velocity is $B\omega_1\cos\omega_1t$, so mean input is $F_dB\omega_1/2$. Mean loss is $\int_0^L b\langle y_t^2\rangle dx=bLB^2\omega_1^2/4$. Hence $B=2F_d/(bL\omega_1)$.

(b) Interpretation and limits

$B=.010\,\mathrm{m}$ and mean input is $.0010\,\mathrm{W}$. As $b\to0$, this steady resonant amplitude diverges; an ideal undamped driven mode has growing motion rather than a bounded steady state. Finite-amplitude nonlinearities or other losses would eventually invalidate the ideal model.

Physical check

Equating these positive quantities gives $B=2F_d/(bL\omega_1)$ with units of length. The divergence as damping tends to zero signals loss of a bounded steady resonant solution, not finite power-free motion.