%% 8. Resonance % % \setcounter{page}{88} % %% % Certain systems have one or more natural frequencies at % which they oscillate. % If you force such a system by an input that oscillates % at close to a natural frequency, % the response may be very large in amplitude. % In this chapter we explore this phenomenon. %% % % The basic idea is linear, autonomous, and inhomogeneous. % Following (4.3), consider the ODE of simple harmonic motion, % $$y'' + \omega^2 y = f(t), \eqno (8.1)$$ % where $\omega$ is a real constant. % One application of this equation is to a linear pendulum, where % $y(t)$ represents the (small) angle from the vertical % at time $t$. % Here $\omega^2$ takes the value $g/L$, where % $g$ is the earth's gravitational acceleration % and $L$ is the length of the pendulum. % Let us imagine that the pendulum is a girl % on a playground swing. % The issue to be examined is the response of the swing to % the forcing function $f$, which we can interpret % as the acceleration introduced by the girl's mother, who is % pushing. % %% % % There is really just one bit of % physics in (8.1), and that is that the associated homogeneous equation % $$y'' + \omega^2 y = 0 \eqno (8.2)$$ % has solutions $\sin(\omega\kern .3pt t)$ % and $\cos(\omega\kern .3pt t)$. This is what we mean when we % say that the {\bf natural frequency} of (8.1) is % $\omega$.\footnote{As in Chapter 4, what we call a frequency could more % fully be called an angular frequency.} % To focus on the simplest possible setting, let us % look at solutions $y(t)$ driven by % the sinusoidal forcing function $f(t) = \sin(\nu\kern .3pt t)$ for some % $\nu \ne \omega$, % $$y'' + \omega^2 y = \sin(\nu\kern .3pt t). \eqno (8.3)$$ % As a reminder of this not very standard notation we note % \begin{displaymath} % \framebox[3.3in][c]{\parbox{3.1in}{\vspace{2pt}\sl % $\omega =$ {\it resonant frequency,} $\quad % \nu =$ {\it forcing frequency.} % \vspace{2pt}}} % \end{displaymath} % We can solve (8.3) analytically. Inserting % the trial solution $y(t) = A\sin(\nu\kern .3pt t)$ gives % $$(-\nu^2 + \omega^2)y = {y\over A},$$ % which implies that a particular solution is % $$y_{\rm p}^{} (t) = {\sin(\nu\kern .3pt t)\over \omega^2 - \nu^2} .$$ % The general solution is accordingly % $$y(t) = {\sin(\nu\kern .3pt t)\over \omega^2 - \nu^2} % + B\sin(\omega\kern .3pt t) + C\cos(\omega\kern .3pt t) \eqno (8.4)$$ % for constants $B$ and $C$. % Note that this is a superposition of sine/cosine waves of % two different frequencies, so in general, $y(t)$ is not periodic. % %% % For example, suppose we take $y(0) = y'(0) = 0$, giving the IVP % $$y'' + \omega^2 y = \sin(\nu\kern .3pt t), \quad t\ge 0, % ~ y(0)=y'(0) = 0. \eqno (8.5)$$ % Choosing $B= -(\nu/\omega )/(\omega^2-\nu^2)$ and % $C=0$ in (8.4) to match the initial conditions, we get the solution % $$y(t) = {\sin(\nu\kern .3pt t) - (\nu/\omega ) \sin(\omega\kern .3pt t) % \over \omega^2 - \nu^2} . \eqno (8.6)$$ % Here is a calculation for $t\in [\kern .3pt 0,200\kern .3pt]$ with % $\omega =1$ and $\nu=0.7$. ODEformats L = chebop(0,200); L.lbc = [0;0]; L.op = @(t,y) diff(y,2) + y; t = chebfun('t',[0 200]); f = sin(0.7*t); subplot(2,1,1), plot(f,'k'), ylim([-2 2]) title('Fig.~8.1.~~Forcing function $f(t) = \sin(0.7t)$',FS,11) y = L\f; subplot(2,1,2), plot(y,CO,ivp) title('Response $y(t)$, resonant frequency $\omega=1$',FS,11) %% % \vskip 1.02em %% % % \noindent % This solution is an incoherent signal of no great interest. % On the other hand, suppose we take $\nu = 0.95$, % a value much closer to $\omega$. % f = sin(0.95*t); subplot(2,1,1), plot(f,'k'), ylim([-2 2]) title(['Fig.~8.2.~~Forcing function $f(t) = \sin(0.95t)$'],FS,11) y = L\f; subplot(2,1,2), plot(y,CO,ivp), ylim([-30,30]) title('Response $y(t)$, resonant frequency $\omega=1$',FS,11) %% % \vskip 1.02em %% % % \noindent % Now the response shows great regularity % and a high amplitude, about $20$, which % builds up over an interval of $60$ time units. This is % {\bf resonance,} which we % can define as the pumping of energy into or out of a system by % a driving function oscillating at close to a natural frequency. % The mother is pushing the swing at close to its natural rate. % The growth in amplitude occurs so long as her % input is in advance of the phase % of the growing oscillation. Eventually, however, after around % time $t=60$, the impulses lag behind the phase of the swing and % she starts extracting energy. % The cycle goes on in the effect known as {\em beating,} % with alternating long % stretches of energy injection and energy extraction. % (No actual mother, of course, would be so mechanical.) % %% % % All of the above is mathematically correct, and an important % base case on which to build % one's understanding of resonance, but it is physically % unrealistic. If an autonomous system is forced by a periodic signal % $f(t) = \sin(\nu\kern .3pt t)$, then surely % we would expect it to have a periodic % solution $y(t)$ with the same frequency, % at least for large $t$, after the effect % of the initial conditions has died away. Why hasn't % this happened? The problem is % that (8.3) includes no damping, so the effect of the % initial conditions never dies away. We can see this % in formulas (8.4) and (8.6), where $B$ and $C$ take values % determined by the initial conditions and the % associated $\sin(\omega\kern .3pt t)$ and $\cos(\omega\kern .3pt t)$ % terms persist forever. % %% % The picture changes fundamentally if % we add a small amount of damping, generalizing (8.3) as in (4.10) to % $$y'' + \varepsilon\kern .2pt y' + \omega^2 y = \sin(\nu\kern .3pt t) \eqno (8.7)$$ % for some small $\varepsilon>0$. Here is the first example % again, but now, with two response curves shown: the first for (8.3) % without damping, the same as before, % and the second for (8.7) with damping coefficient $\varepsilon = 0.04$. L2 = chebop(0,200); L2.lbc = [0;0]; L2.op = @(t,y) diff(y,2) + 0.04*diff(y) + y; f = sin(0.7*t); clf, subplot(3,1,1) plot(f,'k'), ylim([-2 2]), set(gca,XT,0:20:200) set(gca,XTL,{' ',' ',' ',' ',' ',' ',' ',' ',' ',' ',' '}) pp = get(gca,'pos'); pp(2) = pp(2)-.04; set(gca,'pos',pp) title(['Fig.~8.3.~~Undamped and damped responses ' ... 'to $\sin(0.7t)$'],FS,11) y = L\f; subplot(3,1,2) plot(y,CO,ivp), ylim([-5 5]), set(gca,XT,0:20:200) set(gca,XTL,{' ',' ',' ',' ',' ',' ',' ',' ',' ',' ',' '}) pp = get(gca,'pos'); pp(2) = pp(2)-.02; set(gca,'pos',pp) y = L2\f; subplot(3,1,3), plot(y,CO,ivp), ylim([-5 5]) %% % \vskip 1.02em %% % % With damping, it is clear that as $t\to\infty$, the solution % approaches a periodic form with frequency $\nu$. % By appropriate changes in the earlier formulas, we can work % out the details analytically. It is tempting to start with the % trial solution $y(t) = A\sin(\nu\kern .3pt t)$ again, but this will not work % because the $y'$ term will introduce a cosine. % We could get around this with a trial solution containing both % a sine and a cosine, but it is algebraically simpler to % combine the two by introducing complex exponentials. % Thus we replace (8.7) by % $$y'' + \varepsilon y' + \omega^2 y = \exp(i\kern .3pt\nu\kern .3pt t), \eqno (8.8)$$ % and the imaginary part of the solution will correspond to the % solution (8.6). Inserting % $y(t) = A\exp(i\kern .3pt\nu\kern .3pt t)$ gives % $$(-\nu^2 +i\kern .3pt\nu \varepsilon + \omega^2)y = {y\over A},$$ % which implies that a particular solution is % $$y_{\rm p}^{} (t) = {\exp(i\kern .3pt\nu\kern .3pt t)\over \omega^2 % + i\kern .3pt\nu \varepsilon- \nu^2} . \eqno (8.9)$$ % From equation (4.13) in Chapter 4, it can be deduced that the % general solution to (8.8) will be a linear combination of this % function $y_{\rm p}^{}(t)$ % with terms decaying exponentially at the rate $\exp(-\varepsilon\kern .5pt t/2)$. % This confirms that the general solution will be asymptotically periodic % as $t\to\infty$. % %% % Now we add damping to the second experiment, with $\nu=0.95$. % Again the response settles down to a periodic form, with somewhat % smaller amplitude than before. f = sin(0.95*t); subplot(3,1,1) plot(f,'k'), ylim([-2 2]), set(gca,XT,0:20:200); set(gca,XTL,{' ',' ',' ',' ',' ',' ',' ',' ',' ',' ',' '}) pp = get(gca,'pos'); pp(2) = pp(2)-.04; set(gca,'pos',pp) title(['Fig.~8.4.~~Undamped and damped responses ' ... 'to $\sin(0.95t)$'],FS,11) y = L\f; subplot(3,1,2) plot(y,CO,ivp), ylim([-30 30]), set(gca,XT,0:20:200); set(gca,XTL,{' ',' ',' ',' ',' ',' ',' ',' ',' ',' ',' '}) pp = get(gca,'pos'); pp(2) = pp(2)-.02; set(gca,'pos',pp) y = L2\f; subplot(3,1,3), plot(y,CO,ivp), ylim([-20 20]) %% % \vskip 1.02em %% % % The relationship between forcing and response in resonance has an element % of phase as well as amplitude. In the trio of images above, say, % take a look at the forcing and response curves at the final % time $t=200$. % We see that all three curves have a maximum near this point, % reflecting the fact that in (8.4), since $\nu<\omega$, % the denominator is positive, and in (8.9) it is nearly % positive though slightly complex. Thus the input and the % responses are {\em in phase.} Suppose we now change % $\nu$ to $1.05$, a value equally close to $\omega$ but % larger rather than smaller. Now the denominators of (8.4) and % (8.9) are negative and nearly negative, respectively, % and looking near $t=200$ in the figure below confirms that the % forcing and response functions have moved out of phase by an % angle of $\pi$, that is, 180 degrees. This is a basic difference % in the physics of the two kinds of resonance phenomena. % f = sin(1.05*t); clf, subplot(3,1,1) plot(f,'k'), ylim([-2 2]), set(gca,XT,0:20:200) set(gca,XTL,{' ',' ',' ',' ',' ',' ',' ',' ',' ',' ',' '}) pp = get(gca,'pos'); pp(2) = pp(2)-.04; set(gca,'pos',pp) title(['Fig.~8.5.~~Responses to $\sin(1.05t)$, ' ... 'with 180 degree phase shift'],FS,11) y = L\f; subplot(3,1,2) plot(y,CO,ivp), ylim([-25 25]), set(gca,XT,0:20:200) set(gca,XTL,{' ',' ',' ',' ',' ',' ',' ',' ',' ',' ',' '}) pp = get(gca,'pos'); pp(2) = pp(2)-.02; set(gca,'pos',pp) y = L2\f; subplot(3,1,3), plot(y,CO,ivp), ylim([-25 25]) %% % \vskip 1.02em %% % % We can summarize the phase lag situation (for % sufficiently small damping $\varepsilon$) according to % whether the forcing frequency is lower or higher than % the resonant frequency: % \begin{displaymath} % \framebox[3.0in][c]{\parbox{2.8in}{\vspace{2pt}\sl % $\nu < \omega{:}$ {\it no phase lag,} $\quad % \nu >\omega {:} ~180^\circ$ {\it phase lag.} % \vspace{2pt}}} % \end{displaymath} % \noindent For more precision, see Exercise 8.7. % %% % % As $\nu$ gets closer to $\omega$, the maximum amplitude % of $y(t)$ for the undamped equation (8.3) % increases to $\infty$, and so does the % time scale over which it achieves that maximum. % If $\nu = \omega$ exactly, the forcing % and the oscillation are synchronized forever. Energy keeps % getting pumped into the system, and the amplitude grows without limit. % This unbounded growth for undamped oscillation corresponds to a singularity % in the formulas (8.4) and (8.6), but not in the IVP itself or % its solution, which shows simply a linear increase with $t$. % In the damped cases, the growth asymptotes to a fixed % amplitude. % f = sin(t); clf, subplot(3,1,1) plot(f,'k'), ylim([-2 2]), set(gca,XT,0:20:200) set(gca,XTL,{' ',' ',' ',' ',' ',' ',' ',' ',' ',' ',' '}) pp = get(gca,'pos'); pp(2) = pp(2)-.04; set(gca,'pos',pp) title(['Fig.~8.6.~~Responses to $\sin(t)$, ' ... 'with 90 degree phase shift'],FS,11) y = L\f; subplot(3,1,2) plot(y,CO,ivp), ylim([-120 120]), set(gca,XT,0:20:200) set(gca,XTL,{' ',' ',' ',' ',' ',' ',' ',' ',' ',' ',' '}) pp = get(gca,'pos'); pp(2) = pp(2)-.02; set(gca,'pos',pp) y = L2\f; subplot(3,1,3), plot(y,CO,ivp), ylim([-50 50]) %% % \vskip 1.02em %% % % The linearly growing solution to an undamped resonant system % forced at the resonant frequency is called a % {\bf secular solution}. In this case we can derive that % the exact solution of (8.5) is % $$y(t) = {-t \cos (\omega \kern .5pt t)\over 2\kern .5pt \omega^2} % + {\sin(\omega\kern .3pt t)\over 2\kern .5pt\omega^3} ; \eqno (8.10)$$ % the term involving $t\cos(\omega\kern .3pt t)$ is the secular one. % Secular terms arise whenever an ODE has an inhomogeneous % forcing function that is itself a solution to the homogeneous problem. % %% % What about the phase lag in this special situation % $\nu = \omega$? Looking at the figures near % the final time $t=200$, we see that as one might guess, the lag is % now midway between the two cases % we have seen before, namely % one quarter of a wavelength, an angle of $\pi/2$ or % $90^\circ$. We can confirm this algebraically % by noting that in the formula (8.10), % although the forcing function is the sine, % the secular term of the response involves the negative of the cosine. % In a case with damping, the $90^\circ$ lag comes from the % $i$ in the denominator of (8.9). %% % The discussion so far has concerned the response of (8.1) or % its damped cousin to a pure sine % wave, but of course, not every forcing function will be so simple. % More generally, if $f$ is a function that looks approximately % like a sine wave for a certain range of values of $t$, we may % expect to see approximately corresponding behavior for a time. Here is an % experiment to illustrate. Consider % $$f(t) = \sin ((t/100)\cdot t),$$ % a function whose frequency starts at $\nu=0$ for % $t=0$ and then increases linearly to $\nu = 4$ at $t=200$ since % $(t^2/100)' = t/50$. % (In Chapter 17 we will do a number of such experiments % with parameters slowly varying with $t$.) % The response $y(t)$ shown in Figure 8.7 % does nothing very interesting % until near $t=50$, when it grows to a considerable amplitude % due to resonance. Soon the forcing drifts out of tune again, % after which no further significant transfer of energy from % the forcing function takes place. % The undamped response curve shows a permanent record of the brief % moment of resonance near $t=50$, and the damped response curve % dies away slowly on the time scale $\exp(-\varepsilon\kern .5pt t/2)$. f = sin((t/100)*t); clf, subplot(3,1,1) plot(f,'k'), ylim([-2 2]), set(gca,XT,0:20:200) set(gca,XTL,{' ',' ',' ',' ',' ',' ',' ',' ',' ',' ',' '}) pp = get(gca,'pos'); pp(2) = pp(2)-.04; set(gca,'pos',pp) title('Fig.~8.7.~~Responses to $\sin((t/100)t)$',FS,11) y = L\f; subplot(3,1,2) plot(y,CO,ivp), ylim([-15 15]), set(gca,XT,0:20:200) set(gca,XTL,{' ',' ',' ',' ',' ',' ',' ',' ',' ',' ',' '}) pp = get(gca,'pos'); pp(2) = pp(2)-.02; set(gca,'pos',pp) y = L2\f; subplot(3,1,3), plot(y,CO,ivp), ylim([-15 15]) %% % \vskip 1.02em %% % This chapter has focused on linear equations, because the % basic mechanism of resonance is linear. % However, resonance % occurs in nonlinear systems too. For example, generalizing % (8.7) in the case $\omega =1$, % consider the nonlinear pendulum equation % $$y'' + \varepsilon y' + 45\sin(y/45) = \sin(\nu\kern .3pt t), % \eqno (8.11)$$ % where the constant $45$ has been chosen arbitrarily, % corresponding to a rather weak nonlinearity. % Here are the undamped and damped solutions (with $\varepsilon = 0.04$ % as usual) in the same format as before. N = chebop(0,200); N.lbc = [0;0]; N2 = chebop(0,200); N2.lbc = [0;0]; N.op = @(t,y) diff(y,2) + 45*sin(y/45); N2.op = @(t,y) diff(y,2) + .04*diff(y) + 45*sin(y/45); f = sin(0.95*t); clf, subplot(3,1,1) plot(f,'k'), ylim([-2 2]), set(gca,XT,0:20:200); set(gca,XTL,{' ',' ',' ',' ',' ',' ',' ',' ',' ',' ',' '}) pp = get(gca,'pos'); pp(2) = pp(2)-.04; set(gca,'pos',pp) title(['Fig.~8.8.~~Nonlinear responses to $\sin(0.95t)$, ' ... 'eq. (8.11)'],FS,11) y = N\f; subplot(3,1,2) plot(y,CO,ivpnl), ylim([-40 40]), set(gca,XT,0:20:200,YT,-40:40:40); set(gca,XTL,{' ',' ',' ',' ',' ',' ',' ',' ',' ',' ',' '}) pp = get(gca,'pos'); pp(2) = pp(2)-.02; set(gca,'pos',pp) y = N2\f; subplot(3,1,3), plot(y,CO,ivpnl), ylim([-40 40]) set(gca,YT,-40:40:40) %% % \vskip 1.02em %% % % \noindent % Note that in comparison to Figure 8.4, the amplitude and % the time scale have increased. This is because in this example, % nonlinearity has weakened the restoring force, % decreasing the natural frequency of % the system, thereby bringing the forcing frequency closer to resonance. % %% % % \begin{center} % \hrulefill\\[1pt] % {\sc Application: moon, sun, and tides}\\[-3pt] % \hrulefill % \end{center} % %% % Everybody knows that the moon causes the tides, % with a further contribution from the sun. But there are some % surprises along the way related to the material of this chapter. %% % % The resonant system here consists of the earth's oceans, % and the forcing function consists % of the gravitational force from the moon, whose direction % changes as the earth turns. With respect to a % fixed point on Earth, the moon appears to go around about % once every 24.8 hours. This means the tidal forcing from % the moon has a period of about $T = 12.4$ hours, since, % as many books and web pages will tell you, the moon's % gravity both pulls on the ocean nearest it more than on the earth % below {\em and\/} pulls on the earth below more % than on the ocean on the far side. % So our forcing function, with time measured in hours, has frequency % $\nu = 2\pi / T \approx 0.507$. Here's a picture over % 672 hours, i.e., four weeks. % t = chebfun('t',[0,672]); Tmoon = 12.4; numoon = 2*pi/Tmoon; f = cos(numoon*t); clf, plot(f,'k'), ylim([-2.5 2.5]) xlabel('$t$ in hours over a 4-week period',FS,9) ylabel('forcing',FS,9) title('Fig.~8.9.~~First approximation to tidal forcing',FS,11) %% % \vskip 1.02em %% % % There are two big day-to-day corrections to be added % to this picture. The first is the influence of the sun. % As it happens, the % gravitational force from the sun is about 180 times as big as that % from the moon. This big factor doesn't matter, however, since % tides depend % on gravity being stronger on one side of the earth than the % other: what matters % is the {\em gradient} of the gravitational force. This is % about twice as great for the moon as for the sun. Moreover, the % period associated with the sun's forcing is % of course simply 12 hours, not 12.4. The difference in periods % causes a slow beating in and out of phase, giving a % combined gravitational forcing from the moon and sun % like this. % Tsun = 12; nusun = 2*pi/Tsun; f = cos(numoon*t) + 0.46*cos(nusun*t); plot(f,'k'), ylim([-2.5 2.5]) xlabel('$t$ in hours over a 4-week period',FS,9) ylabel('forcing',FS,9) title(['Fig.~8.10.~~Correction for the sun: ' ... 'bigger forcing every 2 weeks'],FS,11) %% % \vskip 1.02em %% % % \noindent % When the moon is new or full, it is aligned with % the sun (a configuration called {\em syzygy}) and % we we have big {\em spring tides.} A week later it % is $90^\circ$ off and we have smaller {\em neap tides.} % %% % The other big correction is the one that makes % tides alternate big-small-big-small. This is caused % by the fact that whereas the moon's orbit % is close to the ecliptic (the plane of the solar system), the earth's % orbit is tilted by $23\%$. That means that if % you live somewhere between the % equator and a pole, as the earth turns during the day, you may % alternate between close to the ecliptic at one high of the % gravitational forcing function and $90^\circ$ away at the next. % So the forcing function near you looks more like this (the % details will depend on latitude and season of the year). f = .8*cos(numoon*t) + .2*cos(numoon*t/2) ... + .8*.46*cos(nusun*t) + .2*.46*cos(nusun*t/2); plot(f,'k'), ylim([-2.5 2.5]) xlabel('$t$ (hours)',FS,9), ylabel('forcing',FS,9) title('Fig.~8.11.~~Correction for tilting of Earth''s axis',FS,11) %% % \vskip 1.02em %% % % Figure 8.11 sketches half of the tide problem, the % gravitational forcing. (We've omitted longer time scale % annual effects, as % the earth moves around the sun, and also the variation in the % moon's distance from the earth since its orbit % is not a circle.) The other half concerns % the resonant response by the water on Earth. The full details of % this are very complicated, % for the oceans have complicated boundaries. % Nevertheless, we % can see a part of the picture by ignoring the continents % and imagining that the ocean is a uniform cover of the % earth a few miles deep. % What is the natural frequency of this system? The answer is % determined by how fast waves travel across the ocean --- not the % little waves one might surf on, but the so-called {\em shallow-water waves} % associated with tsunamis. It turns out that these travel % about half as fast as the earth rotates. (It would be different % if the ocean were deeper.) % %% % % So Earth's tides are in the regime $\nu > \omega$, where % {\em the forcing frequency exceeds the resonant frequency}. % This means that we can expect the tides to be about half a period % out of phase with the moon. Since the moon's forcing peaks % not once but twice per day, this $180^\circ$ difference in % mathematical phase corresponds to a $90^\circ$ difference in % geometric orientation. We can sketch it like this: % earth = chebfun('exp(1i*x)',[0 2*pi]); fill(real(earth),1.2*imag(earth),[.42 .57 .84]), hold on fill(real(earth),imag(earth),[.8 .8 .8]) moon = 6+.4*earth; fill(real(moon),imag(moon),[.96 .95 .72]), hold on text(0,-1.5,'Earth',FS,10,HA,CT) text(-.8,1,'tidal bulge',FS,10,HA,RT) text(6,-.7,'Moon',FS,10,HA,CT), axis equal, axis off title(['Fig.~8.12.~~Tides on Earth, out of phase ' ... 'with gravitational input'],FS,11) %% % \vskip 1.02em %% % % \noindent % Many of us have seen figures like this, perhaps even % when we were teenagers first learning about the tides. But the pictures % usually show the bulges pointing in line with the moon, not at % right angles! That's correct in a certain static sense, as it conveys % the underlying gravitational force, but it's wrong dynamically. % %% % % We hasten to emphasize that the actual behavior of tides on % Earth is far more complex than Figure 8.12 and the discussion % above recognize. For example, the phase lag is diminished % by damping, which is not negligible, and at high latitudes by % the smaller distance around the globe, which brings the % natural frequency of oscillation close to the % frequency of gravitational input. Most importantly, % we have ignored the continents completely, whose presence % changes the details in % a manner that varies from place to place. % See for example E. I. Butikov, A dynamical picture of the % oceanic tides, {\em American Journal of Physics} 70 (2002), % pp.~1001--1011. Because of these effects, if you actually % look at tide data to see whether reality better fits Fig.~8.12 or % its opposite, you are likely to find that the data are % all over the place. See Exercise 8.4. % %% % % \smallskip % {\sc History.} It was Newton who figured out the fundamental dynamics of % Earth's tides, in his {\em Principia,} as a consequence of the same % law of gravity that explains orbits of planets around the sun. % \smallskip % %% % % {\sc Our favorite reference.} For a delightful tour of resonance % and many other effects, check out % E. J. Heller, {\em Why You Hear What You Hear: An Experiential % Approach to Sound, Music, and Psychoacoustics,} Princeton % University Press, 2013. It is hard to read even a single page % of Heller's book without learning something interesting. % \smallskip % %% % % \begin{displaymath} % \framebox[4.7in][c]{\parbox{4.5in}{\vspace{2pt}\sl % {\sc Summary of Chapter 8.} Resonance is the % interaction between a system with a natural frequency of % oscillation and external forcing at a nearby frequency. % Over a period of oscillations, a great deal of energy % can be injected into an oscillating system by this effect. % If the damping is small, % forcing at the resonant frequency introduces % a phase shift of\/ $90$ degrees % and forcing above the resonant frequency results in % a phase shift of\/ $180$ degrees. In systems with % damping, forcing by a periodic function of % frequency\/ $\nu$ leads to a solution that approaches % periodic form as\/ $t\to\infty$. % \vspace{2pt}}} % \end{displaymath} % %% % % \small\parindent=0pt\parskip=1pt % {\em \underline{Exercise $8.1$}. Exploiting resonance to increase amplitude.} % For a given frequency $\nu$, consider the solution % to $y'' + y = 1 - \cos(\nu\kern .3pt t)$, $y(0) = 1$, $y'(0) = 0$. % {\em (a)} If $\nu$ is set equal to the resonant frequency $\omega$ % for this equation, compute the time $t_c^{}$ at which $y(t)$ first % reaches the value $10$. % {\em (\kern .7pt b)} What % is the smallest value of\/ $\nu$ for which $y(t)$ reaches the value % $10$ at some time $t\in [\kern .3pt 0,100\kern .3pt]\kern .5pt ?$ % \par % {\em Exercise $8.2$. RLC circuits and AM radio.} % An AM radio station broadcasts radio waves consisting of a high % frequency carrier (say, $10^6\kern -1.5pt$ Hz) times a low-frequency oscillation % (a few thousand Hz). This is equivalent to saying the signal is contained % in a band of a few thousand hertz about $10^6\kern -1.5pt$ Hz. The radio % is tuned by means of a resonant circuit that % selects energy in this band. The simplest such circuit consists of % a resistor of resistance~$R$ (in ohms), an inductor of % inductance $L$ (in henries), and a capacitor of capacitance~$C$ % (in farads) in series. % If $E(t)$ is the applied voltage (in volts) and % $I(t)$ is the current (in amps) as functions of time, % then $I$ satisfies the ODE $LI'' + R\kern .4pt I' % + C^{-1} \kern -.8pt I = E\kern .5pt{}'$. % {\em (a)} Show that the natural frequency of oscillation (corresponding % to $R$ small enough to be negligible) is $(LC\kern .5pt )^{-1/2}$. % If $L = % 1 \hbox{ henry}$, what value of $C$ is needed to tune in a station % at 680 KHz? % {\em (\kern .7pt b)} Suppose it is desired to make the half-width of the resonance % $1000$ Hz in the sense that signals at $679$ or $681$ KHz generate % responses of half the amplitude of a signal at $680$ KHz. What's the % right choice of $R\kern .7pt$? Is this subcritical, critical, or supercritical % damping? % \par % {\em \underline{Exercise $8.3$}. 1D analogue of Chladni patterns.} % {\em (a)} Plot % the solution of $y'' + 1000\kern .3pt y = e^x$, $y(0) = y(\pi) = 0$. % Why does it have such a regular shape, and why is the number of % maxima what it is? Why does the shape become even more regular if % 1000 is changed to 1020\kern .3pt? % {\em (\kern .7pt b)} Answer the same questions for % $y'''' - 10^6y = e^x$, $y(0) =y''(0) = 0$, % $y(\pi) = y''(\pi) = 0$, with $10^6$ then changing to $1.05\times 10^6$. % \par % {\em Exercise $8.4$. High tides in coastal cities.} % {\em (a)} Find out the date of the next full or new moon in Honolulu, % Lisbon, Sydney, and a fourth coastal city of your choosing. % {\em (\kern .7pt b)} Find out the times of the two high tides in these cities % on this date. For cities observing daylight saving time, be sure to correct % appropriately so as to get high tides in standard local time. % {\em (c)} Figure 8.12 suggests the high tides should be at % approximately 6am and 6pm, whereas if the tidal bulges were aligned with % the moon, it would be noon and midnight. Which of these scenarios % do your data come closer to? How close?\footnote{In the final weeks % of writing this book, author LNT visited the Holy Island of Lindisfarne % just before the solar eclipse of August 21, 2017. When % there is a solar eclipse, there must be a new moon syzygy, % so Fig.~8.12 suggests high tides at 6am and 6pm. Tides matter a great % deal on Lindisfarne, because the causeway is underwater for ten % hours each day! The actual high tide in the afternoon of % August 21 was at 14:24, alas --- not such a good match with the figure.} % \par % {\em \underline{Exercise $8.5$}. Random forcing.} % Repeat Figure 8.7 but with the forcing function replaced % by \verb|f = randnfun([0,200])|, a smooth random function containing % a wide range of wave numbers (see Chapter 12). Run the code % three times so as to see responses to three random functions, and % discuss the shape and size of the outputs. % \par % {\em Exercise $8.6$. Beating.} % Figures 8.2 and 8.10 show beating effects that % arise when two waves of nearby frequencies are added. % {\em (a)} Use trigonometric identities to derive the formula % $\cos(\omega t) + \cos((\omega + 2\kern .5pt\varepsilon) t) = % 2\cos((\omega + \varepsilon)\kern .5pt t)\cos(\varepsilon \kern .3pt t)$ for % arbitrary constants $\omega$ and $\varepsilon$. % {\em (\kern .7pt b)} Relate this formula to one of these % figures, and discuss what adaptation % would be needed to relate it to the other. % \par % {\em \underline{Exercise $8.7$}. Damping and phase lag.} % Let $\omega=1$ and $\varepsilon = 0.05$ and define a chebfun $f$ % for the denominator of (8.9) over the range $0.5\le \nu \le 1.5$. % Plot {\tt angle(1/f)}. Repeat for $\varepsilon = 0.005$ and % relate the plots to the discussion about phase lag as a function of % $\kern .8pt\nu$. %