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(2.62)
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In other words, the swing oscillates 90 degrees out of phase with our pushing. Our push is stronger when the swing is at its point of minimum height and weaker when it is at its point of maximum height. This is exactly what you experience when pushing a child on a swing. As the swing approaches you, you must decrease the intensity of the force so that it vanishes when the swing reaches its point of maximum height. Far away from resonance (i.e., Iw VA), q,(t) :::::: 0 and the swing does not move. Figure 2.6 plots the average of the square of (2.61) over one oscillation period against w for a value of / 100 times smaller than the natural frequency. Notice that it decreases quickly as the frequency becomes different from the natural frequency. The sharp peak in Figure 2.6 is typical of resonant phenomena. To solve (2.56) for any driving force, it is convenient to solve it first for an impulse force c5(t - t'). This solution is called the Green's function, G(tlt'), after George Green, a nineteenth-century mathematician from Nottingham, England. The equation for the Green's function is
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18For weak damping the natural frequency is approximately the same as the natural frequency with no damping.
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Figure 2.6 This is a plot of the average over one period of oscillation of the square of (2.61) as a function of the fr~quency w of the driving force F(t) = Asinwt. All quantities involved have been made dimensionless by expressing them in units of key parameters. The driving frequency w is in units of the natural frequency of oscillation of the swing Wo = y97i. The angle was scaled by the dimensionless factor w5/A and , was taken to be 100 times smaller than woo The sharp peak centered at the natural frequency is a well-known signature of resonance.
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Once we solve (2.63) and find G(tlt'), the solution (t) of (2.56) for an arbitrary driving force F( t) can be obtained by doing the integral
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dt' G(tlt')F(t')
(2.64)
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(2.65)
= F(t).
Let us use a Fourier transform to solve (2.63):
G(tlt') =
m.v eiwtG(w, t').
(2.66)
Substituting (2.66) in (2.63), mUltiplying by (1/27r) exp( -iwt), integrating over t from -00 to 00, and using (F.5), we find that
G(w, t')
1 e- iwt' 2 - hw - g/l' 27rw
(2.67)
The denominator of (2.67) can also be written as
hw -
= {w -
[ii -VI - (i/]}
{w -
[ii + VI - (i/]}
(2.68)
FIAT LUX!
Let us suppose that the friction is small (i.e., r
w 2 W
Vifl). Then (2.68) becomes
[.rfil}{ [.r + VI J . "'2 - VI J "'2 .fil}
t') /fe-<t-t /h /2 sin ([t -
(2.69)
Substituting (2.69) in (2.67) and then back in (2.66), we find that (see Problem 2.1)
G(tlt')
~ 9(t -
t')/f) ,
(2.70)
where 8( u) is the Heaviside step function, which vanishes for negative u and is unity for positive u. Now suppose that we are driving our swing with a force of frequency W from t = 0 until t = T; for example,
F(t) = 9(t)9(T - t) sinwt.
(2.71)
I leave it for the reader to work out integral (2.64) in this case (Problem 2.2). As we have this unrealistic sudden switching on and off of the force, the general solution is not so informative. But the following particular cases are. Assuming that w Vi7l, the solution is given by
<f>(t) ~ { JI/(w 2g) sinwt,
for 0 < t < T forO < T < t.
(2.72)
Assuming that w
= Vi7l, the solution is given by
[1/(4g)] (1 <f>(t)
+ e-l't/2) sin ( Vi7lt)
forO < t < T
+JI/(r2g) (1 - e-l't/2) cos ( Vi7lt) e-l't/2 ([I/(4g)] [eI'T/2 sin ([2T - tlVi7l) +sin( Vi7lt)] + JI/b 2g)
x [el'T /2 cos ( Viflt) - cos (
J97lt) ])
forO < T < t. (2.73)
2.3.2 Confinement is the key: How resonator features arise whenever we attempt to trap dynamic fields
In Section 2.3.1 we introduced the phenomenon of resonance. But as we mentioned earlier to have resonances is only one of the essential features of a resonator. The other essential feature is that the fields acquire particular shapes in space when they vibrate at a resonator's resonance frequency (i.e., the mode patterns). If the reader wishes, he or she can even regard the simple physical example given in Section 2.3.1 as a zero-dimensional resonator: a pathological case that has a resonance, yet lacks