Kernel: SageMath 9.7

C17 More Resonance

Damped Harmonic Oscillator

We found last class that all solutions to

mu(t)+cu(t)+ku(t)=Csin(ωt)mu''(t) + cu'(t) + ku(t) = C\sin(\omega t)

with c>0c > 0 converge to the particular solution

u(t)=C(cω)(kmω2)2+(cω)2cos(ωt)+C(kmω2)(kmω2)2+(cω)2sin(ωt).u(t) = \frac{-C(c\omega)}{(k - m\omega^2)^2 + (c\omega)^2}\cos(\omega t) + \frac{C(k - m\omega^2)}{(k - m\omega^2)^2 + (c\omega)^2}\sin(\omega t).

We also saw that this particular solution could be rewritten

u(t)=CG(ω)sin(ωtΦ(ω))u(t) = CG(\omega)\sin(\omega t - \Phi(\omega))

where the gain is

G(ω)=1(kmω2)2+(cω)2G(\omega) = \frac{1}{\sqrt{(k - m\omega^2)^2 + (c\omega)^2}}

and the phase shift is

Φ(ω)=tan1(cωk+mω2).\Phi(\omega) = \tan^{-1}\left(\frac{c\omega}{k + m\omega^2} \right).

Finally, we found that the gain GG is maximized at the resonance angular frequency

ωmax=km12(cm)2\omega_{\max} = \sqrt{\frac{k}{m} - \frac{1}{2}\left(\frac{c}{m}\right)^2}

which is close to the unforced angular frequency

ωnat=km(c2m)2.\omega_{\text{nat}} = \sqrt{\frac{k}{m} - \left(\frac{c}{2m}\right)^2}.

Of course, the last two definitions make sense only if cc is sufficiently small. Otherwise, the gain is maximized at zero and the unforced ODE does not have an oscillatory solution. What do these results say about the motion of a forced linear oscillator? Show the MIT video.

Example.

Consider u(t)+0.1u(t)+10u(t)=sin(ωt)u''(t) + 0.1u'(t) + 10u(t) = \sin(\omega t). (To get the Greek symbol ω\omega in an input cell, type \omega and the tab key.)

a. Find the particular solution for arbitrary ω\omega. Compare with our earlier work.

b. Plot the gain and phase shift on the interval 0ω50 \leq \omega \leq 5. Compare with our earlier work.

c. Find the resonant radial frequency. Compare with the previous graph.

d. Plot the forcing term sin(ωt)\sin(\omega t) and the particular solution u(t)u(t) for values of ω\omega less than, equal to, and more than the resonant radial frequency Compare with the earlier parts.

var('t ω'); u = function('u') (t); m = 1; c = 0.1; k = 10 de = m*diff(u, t, t) + c*diff(u, t) + k*u == sin(ω*t) usol = desolve(de, dvar = u, ivar = t).subs(_K1 = 0, _K2 = 0) show(usol) G(ω) = 1 / sqrt((k-m*ω^2)^2 + (c*ω)^2) Φ(ω) = arctan2(c*ω, (k - m*ω^2)) plot([G, Φ], (0, 5), figsize = 4).show() ωmax = sqrt((k/m) - (c/m)^2/2); show(ωmax.n())
plot([sin(1*t), usol.subs(ω = 1)], (t, 0, 2*pi), figsize = 4)
plot([sin(ωmax*t), usol.subs(ω = ωmax)], (t, 0, 2*pi), figsize = 4)
plot([sin(5*t), usol.subs(ω = 5)], (t, 0, 2*pi), figsize = 4)

Undamped Harmonic Oscillator

Find the solution to mu(t)+ku(t)=cos(ωt)mu''(t) + ku(t) = \cos(\omega t) with initial conditions u(0)=0u(0) = 0 and u(0)=0u'(0) = 0.. Interpret.

# Find specific solution var('t m k ω'); u = function('u')(t) forget(); assume(m > 0, k > 0) de = m*diff(u, t, t) + k*u == cos(ω*t) usol = desolve(de, dvar = u, ivar = t, ics = (0, 0, 0)) show(usol)
# natural frequency 20 hertz and forced frequency 19 hertz plot(usol.subs(m = 1, k = (2*pi*20)^2, ω = 2*pi*19), (t, 0, 2), figsize = 4)

Open two copies of the tone generator. Play one at 440 hertz and the other at 439 hertz to hear concert A with 1 second frequency beats. Decrease the second to hear the beat frequency increasing until two separate tones are perceived.

Memorex comercial. Play as a preview to an exercise in A06.