Physics 430: Lecture 13 Driven Oscillations and Resonance Dale E. Gary NJIT Physics Department 5.5 Driven Damped Oscillations Last time we solved the homogeneous equation for damped oscillations: mx bx kx 0. We now wish to consider the case when there is another forcing that depends on time, i.e. driven oscillations: mx bx kx F (t ). Again writing 2b = b/m, and now writing f(t) = F(t)/m, we have the driven damped oscillator equation of motion: x 2 b x o2 x f (t ). The left side of this equation is linear in derivatives of x, in which case it is always possible to define a differential operator d2 d D 2 2b o2 , dt dt where you can think of the operator D as shorthand for the entire right side. Our equation can then be written simply as Dx = f. However, the concept of differential operators is much more than a simple shorthand notation. It is a powerful mathematical tool with applications throughout physics. October 15, 2009 Linear Differential Operators d2 d Let’s look at some operations with the operator D 2 2b o2 , that we dt dt have defined. Because it is a linear operator, we have D(ax) aDx and D( x1 x2 ) Dx1 Dx2 . where a is a constant. Likewise, for a and b constant, we can write D(ax1 bx2 ) aDx1 bDx2 for any two functions x1(t) and x2(t). Any operator that satisfies this equation is called a linear operator. Back to our original, now very simple appearing equation: Dx = f. We can write the homogeneous equation (the damped oscillator one we already solved) as Dx = 0. Let’s call the solution to this the homogeneous solution xh =C1er1t + C2er2t. Let’s also call the solution to the inhomogeneous equation the particular solution xp. Then the sum of these two solutions is also a solution to the inhomogeneous equation, because: D( xh x p ) Dx h Dx p 0 f f . October 15, 2009 Sinusoidal Driving Force d2 d The general solution to Dx = f, with the operator D 2 2b o2 , that dt dt we have defined, is therefore xh x p C1er t C2er t x p , 1 2 where the homogeneous solution xh already carries the required two arbitrary constants. Again, the driving function f(t) = F(t)/m is so far an arbitrary forcing term. Let’s look at the special (although quite common) case that the forcing term is sinusoidal, i.e. f(t) = focos(t)., where fo is the amplitude of the driving force. The equation of motion then becomes: x 2b x o2 x f o cos(t ). Note that the two omegas in this equation are different— is the driving frequency, and o is the “resonant frequency” of the equivalent undamped oscillator without forcing. As before, we can consider this the real part of a complex solution of an equation for z = x + iy, z 2b z o2 z f o eit . October 15, 2009 Sinusoidal Driving Force-2 With the equation in this form, we seek a solution of the form z Ceit , which after substitution into z 2 b z o2 z f o eit , yields 2 it foe . d o2 2 2ib 2 o 2 Ae . Now we have to find expressions for A and d. To do that, first note that CC*=A2, where C* denotes the complex conjugate of C. Thus A 2 it We can now solve for the constant C, and rewrite the constant as a complex number fo id C 2ib Ce 2 o 2b 2 o f o2 4b 2 2 2 2 . Also, d is the arctangent of the imaginary/real part of o2 2 2ib (arctan(y/x)), 2b d arctan 2 . o 2 October 15, 2009 Sinusoidal Driving Force-3 With these substitutions, the solution looks very familiar: z (t ) Aei t d , and the real part is the solution we want: x(t ) A cost d . Remember, this is the particular solution, so to this we have to add the general solution to the homogeneous equation: x(t ) xh x p C1er1t C2er2t A cost d . You may think that this equation has four arbitrary constants, but in fact both A and d are already determined, so only C1 and C2 are arbitrary. There are a couple of amazing things about this solution. The first is that the initial conditions set the value of C1 and C2, but these both die out exponentially with time. Thus, the behavior at long times is just set by the driving force—initial conditions do not matter in the long term! Consider the case of weak damping (b < o). We saw that the homogeneous term becomes an exponentially damped sinusoidal oscillation x(t ) xh x p Atr e bt cos1t d tr A cost d . where we distinguish between the A and d of the transient part (defined by the initial conditions) by adding the subscript tr. Applet October 15, 2009 5.6 Resonance The long term solution for the driven oscillator is x(t ) A cost d , A2 tuning a radio for t >> 1/b, where the amplitude, as we have seen, is A 2 2 o f o2 2 4b 2 2 2 . o This amplitude expression is interesting, because it says that as you drive a system at frequency , its amplitude depends on both how far off you are from the resonant frequency o, and also on how big the damping is. If there is no damping at all, and you drive the system at the resonant frequency, then both terms 2 o2 2 0 and 4b 2 2 0. In this case, the amplitude goes to infinity. The way to think about this is that the driving force pumps energy into the oscillator (like pushing a child on a swing), and if there is no dissipation, there is no loss of energy and the energy grows to become infinite. If you tune the driver frequency (variable ) for a given oscillator (fixed o), what is the value of for which A2 is maximum? 2 2b 2 . 2 o October 15, 2009 Keeping Track of Frequencies We have met a number of frequencies in this discussion, so it might be helpful to list them in one place to help keep them straight. b = 0.1o o k / m natural frequency of undamped oscillator A2 b = 0.2o b = 0.3o 1 o2 b 2 frequency of damped oscillator frequency of driving force 2 o2 2b 2 value of at which response is max To find out the maximum amplitude of a particular driven oscillator, just let ~ o in 2 A2 2 o fo 4b 2 2 2 2 . fo . 2 b o From this you can see that the amplitude goes as b1. It also turns out that the width of the resonance increases with b as FWHM ~ 2b. (Prob. 5.41) i.e. Amax o October 15, 2009 Quality Factor and Phase at Resonance There is a quantitative measure of how sharp the resonance is, which you may have heard of before—the quality factor, or Q. It is just the ratio of the resonance peak frequency o to the peak width 2b. Q o . 2b In the case of a pendulum, which you might think is a good resonator (keeps good time), the Q might be about 100. For a quartz clock, on the other hand, Q may reach 10000. 2 high Q An alternative way of thinking about the Q of a resonator A is to consider it as low Q o decay time 1/ b o Q . period 2 / o 2 b What about the phase? Remember our expression for d: d arctan 2b . 2 2 o d high Q /2 0 low Q October 15, 2009 o