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Simple Nature - Light and Matter

Simple Nature - Light and Matter

Simple Nature - Light and Matter

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end up canceling out, however:Q = ω o∆ω= 2πf o2π∆f= f of≈ 10In other words, once the musician stops blowing, the horn willcontinue sounding for about 10 cycles before its energy falls offby a factor of 535. (Blues <strong>and</strong> jazz saxophone players will typicallychoose a mouthpiece that gives a low Q, so that they canproduce the bluesy pitch-slides typical of their style. “Legit,” i.e.,classically oriented players, use a higher-Q setup because theirstyle only calls for enough pitch variation to produce a vibrato,<strong>and</strong> the higher Q makes it easier to play in tune.)Q of a radio receiver example 50⊲ A radio receiver used in the FM b<strong>and</strong> needs to be tuned in towithin about 0.1 MHz for signals at about 100 MHz. What is itsQ?⊲ As in the last example, we’re given data in terms of f s, not ωs,but the factors of 2π cancel. The resulting Q is about 1000, whichis extremely high compared to the Q values of most mechanicalsystems.TransientsWhat about the motion before the steady state is achieved?When we computed the undriven motion numerically on page 172,the program had to initialize the position <strong>and</strong> velocity. By changingthese two variables, we could have gotten any of an infinite numberof simulations. 13 The same is true when we have an equation of motionwith a driving term, ma+bv +kx = F m sin ωt (p. 176, equation[1]). The steady-state solutions, however, have no adjustable parametersat all — A <strong>and</strong> δ are uniquely determined by the parametersof the driving force <strong>and</strong> the oscillator itself. If the oscillator isn’tinitially in the steady state, then it will not have the steady-statemotion at first. What kind of motion will it have?The answer comes from realizing that if we start with the solutionto the driven equation of motion, <strong>and</strong> then add to it anysolution to the free equation of motion, the result,x = A sin(ωt + δ) + A ′ e −ct sin(ω f t + δ ′ ) ,13 If you’ve learned about differential equations, you’ll know that any secondorderdifferential equation requires the specification of two boundary conditionsin order to specify solution uniquely.184 Chapter 3 Conservation of Momentum

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