Forced Oscillations and Resonance — Question 1

Question 1

Amplitude and lag come from two force components

A driven oscillator obeys $m\ddot x+b\dot x+kx=F_0\cos\Omega t$, with $b>0$. After transients decay, write $x=X\cos(\Omega t-\delta)$, with $X>0$ and $0<\delta<\pi$.

1. Derive $X$ and the correct quadrant of the phase lag.

2. For $m=1\,\mathrm{kg}$, $k=25\,\mathrm{N/m}$, $b=3\,\mathrm{kg/s}$, $F_0=2\,\mathrm{N}$ and $\Omega=4\,\mathrm{rad/s}$, find $X,\delta$. What happens to the lag when $\Omega$ exceeds $\sqrt{k/m}$?

Question 1 - Solution

Amplitude and lag come from two force components

(a) Derivation and calculation

Let $\psi=\Omega t-\delta$. Matching the cosine and sine coefficients gives$$(k-m\Omega^2)X=F_0\cos\delta,\quad b\Omega X=F_0\sin\delta.$$$$\boxed{X=\frac{F_0}{\sqrt{(k-m\Omega^2)^2+(b\Omega)^2}}},$$$$\boxed{\delta=\operatorname{atan2}(b\Omega,k-m\Omega^2)}.$$

(b) Result and interpretation

$$k-m\Omega^2=25-16=9,\qquad b\Omega=3(4)=12\,\mathrm{N/m}.$$$$\boxed{X=2/\sqrt{9^2+12^2}=.133333\,\mathrm m},\quad\boxed{\delta=.92730\,\mathrm{rad}}.$$$$\Omega>\omega_0:\quad\cos\delta<0,\quad\sin\delta>0\Rightarrow\pi/2<\delta<\pi.$$A one-argument principal arctangent loses the required quadrant.

Physical check

The lag passes through $\pi/2$ when the elastic and inertial components cancel.