26 Jun 2025
Forced Oscillations, Resonance, and Quality Factor
Steady forced response, displacement and power resonance, bandwidth, power dissipation, and quality factor.
A damped oscillator driven by the harmonic force $F_0\cos\omega t$ obeys
\[m\ddot x+b\dot x+kx=F_0\cos\omega t, \qquad m>0,\ b>0,\ k>0.\]The complete motion is the sum of a decaying transient and a steady forced oscillation. After the transient has died away, write
\[x=A\cos(\omega t-\delta), \qquad A>0.\]Then
\[\dot x=-\omega A\sin(\omega t-\delta), \qquad \ddot x=-\omega^2A\cos(\omega t-\delta).\]Substitution in the equation of motion gives
\[A(k-m\omega^2)\cos(\omega t-\delta) -Ab\omega\sin(\omega t-\delta) =F_0\cos\omega t.\]Using
\[\cos(\omega t-\delta) =\cos\omega t\cos\delta+\sin\omega t\sin\delta,\] \[\sin(\omega t-\delta) =\sin\omega t\cos\delta-\cos\omega t\sin\delta,\]the cosine and sine coefficients give
\[A\left[(k-m\omega^2)\cos\delta+b\omega\sin\delta\right]=F_0,\] \[A\left[(k-m\omega^2)\sin\delta-b\omega\cos\delta\right]=0.\]The second equation yields
\[\tan\delta=\frac{b\omega}{k-m\omega^2}.\]The tangent alone does not fix the quadrant. For $\omega>0$, choose
\[\boxed{\delta=\operatorname{atan2}(b\omega,k-m\omega^2)}, \qquad 0<\delta<\pi.\]Squaring and adding the two coefficient equations gives
\[A^2\left[(k-m\omega^2)^2+b^2\omega^2\right]=F_0^2.\]Therefore the steady displacement amplitude is
\[\boxed{A(\omega)= \frac{F_0}{\sqrt{(k-m\omega^2)^2+b^2\omega^2}}}.\]Displacement resonance
The amplitude is largest when its squared denominator
\[D(\omega)=(k-m\omega^2)^2+b^2\omega^2\]is smallest. Differentiation gives
\[\frac{dD}{d\omega} =2(k-m\omega^2)(-2m\omega)+2b^2\omega.\]For a nonzero stationary frequency,
\[-2m(k-m\omega_r^2)+b^2=0.\]Using $\omega_0^2=k/m$ and $\beta=b/(2m)$,
\[\boxed{\omega_r=\sqrt{\omega_0^2-2\beta^2}}.\]A nonzero displacement-resonance peak exists only when
\[\boxed{\beta<\frac{\omega_0}{\sqrt2}}.\]As the damping becomes smaller, the amplitude peak becomes higher and narrower, and $\omega_r\to\omega_0$. If $b=0$ exactly at $\omega=\omega_0$, there is no finite steady-state amplitude; the resonant response grows with time.
Average power dissipation
The positive instantaneous rate at which the damper removes mechanical energy is
\[P_{\rm diss}=b\dot x^2.\]Since $\langle\sin^2(\omega t-\delta)\rangle=1/2$ over a cycle,
\[\langle P\rangle =b\omega^2A^2\left\langle\sin^2(\omega t-\delta)\right\rangle =\frac12b\omega^2A^2.\]Thus
\[\boxed{\langle P\rangle= \frac{F_0^2b\omega^2} {2\left[(k-m\omega^2)^2+b^2\omega^2\right]}}.\]In steady motion, this equals the average input power. Indeed,
\[\langle F\dot x\rangle =\frac12F_0A\omega\sin\delta =\frac12b\omega^2A^2.\]To locate the power maximum, put $y=\omega^2$. Apart from a positive constant,
\[\langle P\rangle\propto \frac{y}{m^2(\omega_0^2-y)^2+b^2y}.\]Differentiating with respect to $y$, the numerator of the derivative reduces to
\[m^2(\omega_0^4-y^2).\]Hence the power is maximum at
\[\boxed{\omega=\omega_0}, \qquad \boxed{\langle P\rangle_{\max}=\frac{F_0^2}{2b}}.\]At this frequency $k-m\omega_0^2=0$, so $\delta=\pi/2$: the velocity is in phase with the driving force.
Sharpness, bandwidth, and quality factor
Let $\omega_1<\omega_0<\omega_2$ be the two frequencies at which the average power is half its maximum. The half-power condition is
\[\frac{F_0^2b\omega^2}{2D(\omega)} =\frac12\frac{F_0^2}{2b}.\]After cancellation,
\[(k-m\omega^2)^2=b^2\omega^2.\]The positive roots on the two sides of $\omega_0$ are
\[\boxed{\omega_1=\sqrt{\omega_0^2+\beta^2}-\beta},\] \[\boxed{\omega_2=\sqrt{\omega_0^2+\beta^2}+\beta}.\]Therefore the full width at half maximum of the power curve is exactly
\[\boxed{\Delta\omega=\omega_2-\omega_1=2\beta=\frac bm}.\]The quality factor is
\[\boxed{Q=\frac{\omega_0}{\Delta\omega} =\frac{m\omega_0}{b} =\frac{\omega_0}{2\beta}}.\]A sharp resonance has $\Delta\omega\ll\omega_0$, equivalently $Q\gg1$.
At $\omega=\omega_0$, the stored mechanical energy is $E=\tfrac12m\omega_0^2A^2$, while the energy dissipated in one period is $\langle P\rangle(2\pi/\omega_0)=\pi b\omega_0A^2$. Hence
\[\boxed{Q=2\pi\frac{\text{energy stored}} {\text{energy dissipated per cycle}}}.\]
The steady response, displacement-resonance condition, power maximum, half-power roots, bandwidth, and quality factor are verified in the Maxima worksheet; every printed residual is zero.
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