Damped Harmonic Oscillator

Transcription

1 Damped Harmonic Oscillator Note: We use Newton s 2 nd Law instead of Conservation of Energy since we will have energy transferred into heat. F spring = -kx; F resistance = -bv. Note also: We use F ar = bv instead of bv 2 for two reasons: 1. usually, because the motion will go back and forth, the speeds don t reach very high values, and so F = -bv is more appropriate; and 2. if we use F = -bv, Newton s 2 nd law gives a linear differential equation which will probably be easier than a non-linear one. If there are no other forces then Newton's Second Law becomes: -kx -bv = ma, or md 2 x/dt 2 + b dx/dt + kx = 0. Note that this is a linear, homogenous, second order differential equation. Because it is second order, there should be two constants of integration (usually initial conditions). Because it is linear, if x 1 (t) is a solution, then C 1 x 1 (t) is also a solution (where C 1 is some constant). Since we need two constants, we would expect a full solution to include the addition of two solutions: x(t) = C 1 x 1 (t) + C 2 x 2 (t). Since the equation is linear (that is, we only have x 1 in any term), and since we know that d(ae at )/dt = aae at (that is, the differential of the exponential function gives itself back again), we might guess that x(t) = Ae pt. To see if and when this works, we simply substitute it back into our differential equation to get: get: mp 2 Ae pt + bpae pt + kae pt mp 2 + bp + k = 0. = 0. We can cancel out the Ae pt in each term to This is the condition that must be satisfied if Ae pt is a solution. Solving for p gives: p = [-b +/- (b 2-4mk) 1/2 ] / 2m = -b/2m +/- [(b/2m) 2 -(k/m)] 1/2. To save writing time, let's define (b/2m) and let's use o = (k/m) 1/2 since (k/m) 1/2 is the frequency for an undamped harmonic oscillator. Note that there are two values of p (unless b 2 =4mk). There are three distinct cases:

2 x(t) = C 1 x 1 (t) + C 2 x 2 (t), x(t) = Ae pt, p = -b/2m +/- [(b/2m) 2 -(k/m)] 1/2 a) (b/2m) 2 > k/m In this case the quantity [(b/2m) 2 -(k/m)] 1/2 = [ 2 - o ] 1/2 (we define to be this quantity) is real, and so the solution becomes: x(t) = C 1 e (- + )t + C 2 e (- - )t. Since > this is obviously a combination of two dying exponentials. This case is called the overdamped case. Homework Problem: Problem #11: Find x(t) and v(t) for a particle of mass, m, subject to a spring force of kx and a damping force of bv if it is displaced an initial distance of x o from the equilibrium position. Do this for the OVERDAMPED case. (This problem amounts to finding C 1 and C 2 in terms of the initial conditions: x(t=0) = x o and v(t=0) = v o.) b) (b/2m) 2 = k/m This is the special case between case a) above: the dying exponential, and case c) below which we will soon see to be the dying but oscillating while it dies case. In this case, the two p values, and hence the two exponentials, become the same. But we must have two, rather than just one, functions. Let's try the function x(t) = C 1 t e pt in addition to the normal x(t) = C 2 e pt. Differentiating the trial solutions give: dx/dt = C 1 e pt + C 1 t p e pt + C 2 p e pt, and d 2 x/dt 2 = C 1 p e pt + C 1 p e pt + C 1 t p 2 e pt + C 2 p 2 e pt. Plugging them into the differential equation (Newton's Second Law) gives: 2mC 1 p e pt + mc 1 tp 2 e pt + mc 2 p 2 e pt + bc 1 e pt + bc 1 tp e pt + bc 2 p e pt + kc 1 t e pt + kc 2 e pt = 0. We can cancel out the e pt in every term. 2mC 1 p + mc 1 tp 2 + mc 2 p 2 + bc 1 + bc 1 tp + bc 2 p + kc 1 t + kc 2 = 0. For this to be true for all time, all terms with t 1 must cancel and all terms with t 0 must also cancel: t 1 terms: mc 1 p 2 + bc 1 p + kc 1 = 0, or mp 2 + bp + k = 0 (but this is our original condition); t 0 terms: 2mC 1 p + mc 2 p 2 + bc 1 + bc 2 p + kc 2 = 0.

3 This 2 nd condition says that: C 1 (2mp + b) + C 2 (mp 2 + bp + k) = 0. Note that the coefficient of the C 2 term is zero (from the first condition); this means that the coefficient of the C 1 term must also be zero: 2mp + b = 0, or p = -b/2m. But this is true for the C 2 coefficient only if (b/2m) 2 = k/m, which was our condition for case b! Hence our solution for this case is: x(t) = C 1 t e -bt/2m + C 2 e -bt/2m. This solution also dies out, since the exponential will die out quicker than the t will go to infinity. At first it looks like there is no k in the solution, but when you realize that the condition for this solution is: (b/2m) 2 = k/m, you see that by knowing b and m, you also determine k! This is called the critically damped case, since x will approach zero quicker than either case a above or c below. This is shown at the end of this section. c) (b/2m) 2 < k/m In this case the quantity [(b/2m) 2 -(k/m)] 1/2 is imaginary, so let's define [(k/m)-(b/2m) 2 ] 1/2 = [ o 2-2 ] 1/2 so that the solution in this case is: x(t) = C 1 e - t e +i t + C 2 e - t e -i t = e - t [C 1 e +i t + C 2 e -i t ]. To see what this is, we recall that: e i = cos( ) + i sin( ) and e -i = cos( ) - i sin( ). [You can show that this is true by expanding each term in a power series, and seeing that you get the same series on both sides of the equality sign.] Thus the above solution indicates some type of exponentially dying oscillation. But how do we work with imaginary solutions to real physical problems? We start by saying that x(t) is a real quantity, so [C 1 e +i t + C 2 e -i t ] must also be a real quantity. But C 1 and C 2 will be complex quantities as well! Let's call C 1 = C 1R + ic 1i where both C 1R and C 1i are real, and C 2 = C 2R + ic 2i where both C 2R and C 2i are real. Working this out gives: (C 1R +ic 1i )*[cos( t) + i sin( t)] + (C 2R +ic 2i )*[cos( t) - i sin( t)] = x R + ix i. x R = C 1R cos( t) - C 1i sin( t) + C 2R cos( t) + C 2i sin( t) = real number, and x i = C 1i cos( t) + C 1R sin( t) + C 2i cos( t) - C 2R sin( t) = 0. For the x i = 0 equation to be true for all t s, we need: C 1i = -C 2i and C 1R = C 2R. [Note that these conditions require that C 2 =C 1 * where C 1 * is the complex conjugate of C 1.]

4 Note that we are back to having two (real) constants for our second order differential equation: C 1R and C 1i. To simplify the notation, let's rename them C 1R = a, and C 1i = b. Our solution is now: x(t) = e - t [C 1 e +i t + C 2 e -i t ] = e - t [(a+ib) e +i t + (a-ib) e -i t ], where a and b are real constants normally determined from the initial conditions (x o and v o ). We can make our answer look even "nicer" if we recognize that a complex number (a+ib) can be expressed in "polar" form: (a+ib) = re i = r cos( ) + i r sin( ). This implies that a = r cos( ) and b = r sin( ), or inversely, r = [a 2 +b 2 ] 1/2 and = tan -1 (b/a). Let's now apply this to our solution for x(t): x(t) = e - t [(a+ib) e +i t + (a-ib) e -i t ] = e - t [r e +i e +i t + r e -i e -i t ] = r e - t [ e +i( t+ ) + e -i( t+ ) ] = r e - t [cos( t+ ) + i sin( t+ ) + cos( t+ ) - i sin( t+ )] = 2 r e - t cos( t+ ) = A e - t cos( t+ ), where (A=2r) and are the two constants that are normally determined from initial conditions. It is easy to see from this last expression that the motion is an oscillation that has an amplitude that dies off exponentially. This is called the underdamped case. Note that the frequency of this oscillation is where [(k/m)-(b/2m) 2 ] 1/2 = [ 2 o - 2 ] 1/2, and that this is less than the o which is the natural frequency of oscillation without damping [ o = (k/m) 1/2 ]. If we started from a physically inspired guess of x(t) = A e - t cos( t+ ), can we show that this is indeed a solution of the differential equation with the conditions that = b/2m and that = [ o 2-2 ] 1/2 = [(k/m) - (b 2 /4m 2 )] 1/2? dx/dt = - A e - t cos( t+ ) + - A e - t sin( t+ ) d 2 x/dt 2 = + 2 A e - t cos( t+ ) + A e - t sin( t+ ) + A e - t sin( t+ ) - 2 A e - t cos( t+ ) = ( 2-2 ) A e - t cos( t+ ) + 2 A e - t sin( t+ ) Substituting these into Newton's Second Law: md 2 x/dt 2 + b dx/dt + kx = 0 gives: m( 2-2 ) A e - t cos( t+ ) + 2m A e - t sin( t+ ) - b A e - t cos( t+ ) + -b A e - t sin( t+ ) + k A e - t cos( t+ ) = 0. There is an A e - t in every term that can cancel out. But we have two different functions of time in this equation: cos( t+ ) and sin( t+ ). For the equation to be satisfied for all times, then the coefficient of each function must be zero. This leads to two equations: m( 2-2 ) + -b +k = 0, and 2m -b = 0.

5 The second equation is easy and gives: = b/2m. The first equation gives 2 = (k/m) (b/m) = (k/m) = (k/m) - 2 = o 2-2. This is exactly what we need! Comparisons: All three cases result in solutions that have a dying exponential, so all three do die out (due to losing energy to air resistance). Which dies out the quickest? The longest lived term is the one with the smallest coefficient of t in the exponent. The quickest to die out has the biggest coefficient of t in the exponential. In all cases below, remember that = (b/2m). For case a) (b/2m) 2 > k/m and x(t) = C 1 e (- + )t + C 2 e (- - )t. The smallest coefficient of t in the exponent in this case is - where = [(b/2m) 2 -(k/m)] 1/2 = (b/2m)[1- ] 1/2, where we define {(k/m)/(b/2m)} 2 so that (k/m) 1/2 = 1/2 * (b/2m) or (b/2m) = (k/m) 1/2 / 1/2 ; note that < 1, so - = (b/2m){1 - [1- ] 1/2 } a) for small, [1- ] 1/2 1 - (½) so - (b/2m)*( /2) = (1/2)*(k/m)/(b/2m) = (½)*{(k/m)/(b/2m) 2 } 1/2 *(k/m) 1/2 = (½)* * (k/m) 1/2 < (k/m) 1/2. b) for =.99, {1 - [1- ] 1/2 } =.9, so - =.9 * (b/2m) =.9 * (k/m) 1/2 /.995 < (k/m) 1/2 For case b) (b/2m) 2 = k/m and x(t) = C 1 t e -bt/2m + C 2 e -bt/2m. The coefficients of t in the exponent in this case are both equal to (b/2m) = (k/m) 1/2. For case c) (b/2m) 2 < k/m and x(t) = A e - t cos( t+ ) where = [ o 2-2 ] 1/2 = b/2m, and o = (k/m) 1/2. The coefficient of t in the exponent in this case is ( = b/2m), but (b/2m) < (k/m) 1/2. From this, we see that case b) has the largest coefficient of t in the exponential and so dies out the quickest. This completes the homogeneous case. The next thing to consider is what happens when we have an applied force on this damped harmonic oscillator.