Forced Oscillations and Resonance — Question 3

Question 3

Maximum absorbed power occurs at a different resonance

A damped oscillator is driven by a fixed force amplitude $F_0$. In steady state, displacement amplitude is $X$, velocity amplitude is $V=\Omega X$, and the damping force is $-bv$. Assume $m,k,b>0$.

1. Show that the mean absorbed power is maximized at $\Omega=\omega_0=\sqrt{k/m}$ even when the displacement peak occurs elsewhere.

2. For $m=1\,\mathrm{kg}$, $k=25\,\mathrm{N/m}$, $b=2\,\mathrm{kg/s}$ and $F_0=4\,\mathrm{N}$, find the resonant velocity amplitude, displacement amplitude and mean power.

Question 3 - Solution

Maximum absorbed power occurs at a different resonance

(a) Derivation and calculation

In steady state the mean stored energy does not change:$$\overline P=b\langle v^2\rangle=\frac b2V^2.$$$$V=\Omega X=\frac{F_0}{\sqrt{(k/\Omega-m\Omega)^2+b^2}}.$$$$k/\Omega-m\Omega=0\Rightarrow\boxed{\Omega_{P}=\sqrt{k/m}=\omega_0}.$$

(b) Result and interpretation

$$\omega_0=5.00\,\mathrm{rad/s},\quad\boxed{V_{\max}=F_0/b=4/2=2.00\,\mathrm{m/s}}.$$$$\boxed{X=V/\omega_0=.400\,\mathrm m},\qquad\boxed{\overline P_{\max}=F_0^2/(2b)=4.00\,\mathrm W}.$$$$\delta=\pi/2\Rightarrow x=X\sin\omega_0t,\quad v=V\cos\omega_0t.$$Velocity is in phase with force even though displacement lags by a quarter cycle.

Physical check

The velocity denominator reaches $b$ for any positive damping, so this power peak survives the loss of a displacement peak.