Mathematics 304 – Ordinary Differential Equations Lab/Problem Set 7 – Solutions and Comments November 8, 2004

B) When R = 0, the equation becomes

LCv′′ + v = a sin(ωt)

The natural frequency of the solutions of the corresponding homogeneous equation

LCv′′ + v = 0

comes from the roots of the characteristic polynomial: LCλ^{2} + 1 = 0, so λ = ±i\frac{1}{\sqrt{LC}}. If ω ≠ \frac{1}{\sqrt{LC}}, then as particular solution we take v_{p}(t) = A cos(ωt) + B sin(ωt). Then

LCv′′_{p}(t) + v_{p}(t) = (1 − LCω^{2})A cos(ωt) + (1 − LCω^{2}) sin(ωt)

Setting this equal to a sin(ωt), we get

B = \frac{a}{1 − LCω^{2}} , A = 0.

So

v_{p}(t) = \frac{a}{1 − LCω^{2}} sin(ωt).

If ω = \frac{1}{\sqrt{LC}}, then we must take a different form for the particular solution:

v_{p}(t) = At cos(ωt) + Bt sin(ωt)

Then

v′_{p}(t) = (A + Bωt) cos(ωt) + (B − Aωt) sin(ωt)
v′′_{p}(t) = (2Bω − Aω^{2}t) cos(ωt) + (−2Aω − Bω^{2}t) sin(ωt)

So using ω^{2}LC + 1 = 0, we get

LCv′′_{p} + v_{p} = 2BLCω cos(ωt) − 2ALCω sin(ωt) = a sin(ωt)

which implies:

B = 0, A = \frac{−a}{2LCω} = \frac{−a}{2\sqrt{LC}}

The particular solution in this case looks like

v_{p}(t) = \frac{at}{2\sqrt{LC}} cos(ωt).

Comment: It is the extra factor of t that produces the resonance when ω = \frac{1}{\sqrt{LC}}. The amplitude of the oscillations of v_{p}(t) increases without bound as t → ∞. In a physical system, of course, this would not be possible because the object undergoing this type of oscillation would never be able to withstand it forever; eventually something must “break” and the situation described by the differential equation would not hold any longer. Resonant situations tend to be “very bad” for this reason. With sound vibrations, for instance, it is possible for a very strong singer to create resonance in objects by hitting exactly the right frequency. It is possible to shatter glasses, windows, etc. this way(!)

E) Does resonance also occur in the presence of damping (R ≠ 0), say in the underdamped case? For the damped equation,

LCv′′ + RCv′ + v = a sin(ωt),

the characteristic polynomial is

LCλ^{2} + RCλ + 1 = 0

and

λ = \frac{−R}{2L} ± \frac{\sqrt{(RC)^{2} − 4LC}}{2LC} .

Write α = −R/(2L) and β = ^{\frac{\sqrt{4LC−(RC)^{2}}}{2LC}}. We see that the solutions of the homogenous equation are

v(t) = c_{1}e^{αt} cos(βt) + c_{2}e^{αt} sin(βt)

and sin(ωt) is never a solution (because it does not contain the exponential factor). This means that a particular solution of the form

v_{p}(t) = A cos(ωt) + B sin(ωt)

always exists (without the extra factor of t). Thus, we never get the resonant solutions with unbounded amplitude as t → +∞. On the other hand, the amplitude of the particular solution (the steady state - what remains after the transients decay to close to zero) can be significantly larger than a (the amplitude of the forcing term). The formulas work out like this:

LCv′′_{p} + RCv′_{p} + v_{p} = ((1 − LCω^{2})A + RCωB) cos(ωt) + ((1 − LCω^{2})B − RCωA) sin(ωt)

So

\begin{aligned}(1 − LCω^{2})A + RCωB & = 0 \\ −RCωA + ((1 − LCω^{2})B & = a \end{aligned}

By Cramer’s Rule, we get

A = \frac{−aωRC}{(1 − LCω^{2})^{2} + (RCω)^{2}} ; B = \frac{a(1 − LCω^{2})}{(1 − LCω^{2})^{2} + (RCω)^{2}} .

The amplitude of v_{p} is

\sqrt{A^{2} + B^{2}} = \frac{a}{\sqrt{(1 − LCω^{2})^{2} + (RCω)^{2}}}

which has a maximum at ω = \frac{1}{\sqrt{LC}}.