F = ma, F R + F S = mx.

Transcription

1 Mechanical Vibrations As we mentioned in Section 3.1, linear equations with constant coefficients come up in many applications; in this section, we will specifically study spring and shock absorber systems (your book refers to them as spring and dashpot systems). However, much of the information we learn can be used to describe the motions of other simple mechanical objects, such as pendulums. Recall that in a spring and dashpot system, an object with mass m is attached to a spring; the spring asserts a force F S on the object, while a dashpot on the opposite side provides a resisting force F R : We can describe the position of the object as a function of time, x(t), so that the objects velocity is x (t), and its acceleration is x (t). The object has an equilibrium position, which we think of as x = 0. To determine the position function x, we use Newton s second law F = ma, where F is the total force acting on the object, m is the objects mass, and a = x is its acceleration. Assuming that F S and F R are the only two forces acting on the object, we write F R + F S = mx. To describe F S explicitly, we use Hooke s Law, which says that F S is proportional to the displacement (or position x) of the mass. In addition, we know that when the object is on the right side of the equilibrium position, F S acts to pull it back to equilibrium; so when x > 0, F S < 0. On the other hand, when the object is on the left side of its equilibrium position, F S acts to push it back to equilibrium; so when x < 0, F S > 0. Putting this information together, we write F S = kx, where k > 0. The constant k is the spring constant, and is given by k = force of spring distance stretched/compressed ; 1

2 k is measured in N/m. As for F R, we may assume that the force supplied by the dashpot is proportional to the velocity x (t) of the object. In addition, the dashpot always acts in the direction opposite of the motion x (t) of the object, so we may write F R = cx (t), where c > 0. The constant c is called the damping constant, and is measured in N/(m/s). Now that we have described F R and F S, we arrive at a differential equation relating x, x, x, and t: cx kx = mx, or mx + cx + kx = 0. Since each of m, c, and k are constants, this is a linear homogeneous differential equation with constant coefficients. If F S and F R are not the only forces acting on the object i.e., there is some other force F (t) involved then we must account for this force in Newton s law, writing F R + F S + F (t) = mx. With the same assumptions on F R and F S, this means that mx + cx + kx = F (t). Notice that this equation is a linear non-homogeneous equation with constant coefficients in particular, the equation mx + cx + kx = 0 above is the homogeneous equation associated to mx + cx + kx = F (t). If there is an outside force F (t) acting on the object, we say that the motion is forced; otherwise, we say that the motion is free. Since free motion corresponds to homogeneous equations, we will focus on solving problems with no extra forces involved in this section; in 3.6, once we have more practice solving nonhomogeneous equations, we will investigate problems involving forced motion. The equation relating the object s position x, velocity x, and acceleration x is given by mx + cx + kx = 0. Since the dasphot damps the motion of the spring, we say that the spring s motion is damped. Notice that, if the spring is the only force acting on the mass (the system has no dashpot), then the damping constant c = 0. If this is the case, the equation becomes mx + kx = 0, and we say that the motion of the mass is undamped. We will look at the two cases above separately, starting with the simplest version, free undamped motion. 2

3 Free Undamped Motion In the case of free undamped motion, our differential equation is mx + kx = 0, where m > 0 and k > 0. This is an LHCC equation, so we know exactly how to solve it: replace with its characteristic equation mx + kx = 0 mr 2 + k = 0. Solving the equation, we have k r = ± m. Again, k and m are both positive constants, so the quantity k is a complex number. Setting k ω = m, we rewrite r as k r = ± m = ±ωi. Specifically thinking of these roots as complex numbers a ± bi, we have Finally, the general solution to the equation is m a ± bi = ±ωi so that a = 0, b = ω. x = A cos ωx + B sin ωx. In the case of free undamped motion, there is no dashpot to damp the force of the spring, so we will see continued oscillations; we refer to this as simple harmonic motion. Example. A 5/4 kg weight is attached to a spring. The spring is stretched.5 m by a force of N. At time t = 0, the weight is released from 1 m to the right of its equilibrium position with a velocity of 3 m/s to the left (towards the spring). Find (a) the function modeling the weight s position x(t), (b) the amplitude of the motion, (c) the period of the motion, and (d) the frequency of the motion. 3

4 (a) Since the motion is free and undamped, the equation mx + kx = 0 models the weight s position. We know that m = 5/4, and we can find k easily; since we know that the spring is stretched.5 m by a force of N, k =.5 = 20. Our equation is whose characteristic equation is Solving for r, we see that 5 4 x + 20x = 0, 5 4 r = 0. r = ±4i, so that our solution function is x(t) = c 1 cos 4t + c 2 sin 4t. To find c 1 and c 2, we ll need to use the initial conditions: we know that x(0) = 1 and x (0) = 3. Using the first condition, we see that thus we know that x(t) has form 1 = c 1 cos 0 + c 2 sin 0 = c 1 ; x(t) = cos 4t + c 2 sin 4t. Using x (0) = 3 with x (t) = 4 sin 4t + 4c 2 cos 4t, we have 3 = 4 sin 0 + 4c 2 cos 0 = 4c 2, so that c 2 = 3/4. Thus the equation for the position x(t) of the weight is x(t) = cos 4t 3 sin 4t; 4 the function is graphed below, 0 t : 4

5 (b) In order to continue to analyze the motion of the weight, it would be helpful to rewrite the equation for x(t) so that it only involves one trig function. We can do so using the following identities: a cos ωt + b sin ωt = a 2 + b 2 sin(ωt + ϕ), where sin ϕ = a b and cos ϕ = a 2 + b2 a 2 + b 2. Since x(t) = cos 4t 3 sin 4t, 4 we can set a = 1, b = 3/4, and ω = 4, so that a 2 + b 2 = = = = In addition, we can write so that Finally, we rewrite using cos ϕ = = 3 5, ϕ = arccos 3 x(t) = cos 4t 3 sin 4t 4 a cos ωt + b sin ωt = a 2 + b 2 sin(ωt + ϕ) 5

6 as x(t) = 5 4 sin(4t + arccos 3 5 ). The advantage of rewriting the equation in this form becomes immediately apparent; if then the function has f(t) = A sin(ωt + ϕ), Amplitude A, period 2π ω, and frequency ω 2π. For our function x(t) = 5 3 sin(4t + arccos 4 5 ), we see that the amplitude is A = 5/4 m. (c) The period of the motion is T = π/2 seconds. (d) The frequency of the motion is ν = 2/π hertz. One quick note here we must be extremely careful about calculating ϕ when using the trigonometric identities from the previous example. Due to the restrictions on the range of the arctrig functions, the calculation for ϕ depends upon the signs on a and b, as indicated below: a > 0 b > 0 ϕ = arcsin a a 2 +b 2 = arccos b a 2 +b 2 a > 0 b < 0 ϕ = π arcsin a a 2 +b 2 = arccos b a 2 +b 2 a < 0 b > 0 ϕ = π arcsin a a 2 +b 2 = 2π arccos b a 2 +b 2 a < 0 b < 0 ϕ = arcsin a a 2 +b 2 = 2π arccos b a 2 +b 2 π Free Damped Motion If the motion of the mass is damped, i.e. c > 0, then our differential equation is mx + cx + kx = 0, where each of m, c, and k are positive constants. Since this is an LHCC equation, solving comes down to finding the roots of the corresponding characteristic equation Using the quadratic equation, we see that mr 2 + cr + k = 0. r = c ± c 2 4mk. 6

7 Note that the form of the roots will depend on whether the input for the radical above, c 2 4mk, is negative, positive, or 0. Since we know that m > 0 and k > 0, we can be certain that 4mk < 0, so we have a great deal of control over the situation; we write the cases as 1. c 2 > 4mk 2. c 2 = 4mk 3. c 2 < 4mk. Let us consider the three cases separately: 1. c 2 > 4mk (overdamped): if c 2 > 4mk, then c 2 4mk is a real number, and both roots r = c ± c 2 4mk of the original equation mr 2 + cr + k = 0 are real (and are different). Setting ω 1 = c + c 2 4mk we see that the general solution is and ω 2 = c c 2 4mk, x = c 1 e ω 1t + c 2 e ω 2t. In the case of overdamping, the damping constant is relatively large (in comparison with the mass and spring constant); the dashpot does not allow much movement of the mass, and very quickly damps the motion. 2. c 2 = 4mk (critically damped): if c 2 = 4mk, then c 2 4mk = 0, so that both roots r = c ± c 2 4mk = c = γ of the original equation mr 2 + cr + k = 0 are real (and are equal). Thus the solution is x = c 1 e γt + c 2 te γt. 3. c 2 < 4mk (underdamped): if c 2 < 4mk, then c 2 4mk is complex. Thus both roots r = c ± c 2 4mk 7

8 of the original equation mr 2 + cr + k = 0 are complex. Setting γ = c 4mk c and τ = 2, we see that the roots are the complex numbers so that the general solution is γ ± τi, x = e γt (c 1 cos τt + c 2 sin τt). In the underdamped case, the damping constant is relatively small. Thus the spring is able to cause more motion in the mass, and while the dashpot will eventually return the the mass to equilibrium, it will initially allow a great deal more motion than in the overdamped case. Example. A 5 kg mass is attached to a spring and dashpot. The spring is stretched 1 m by a force of 2 N. The dashpot produces Newtons of resistance for each meter/second of velocity. The mass is released when it is 2 meters to the left of its equilibrium position (i.e., 2 meters closer to the spring), with an initial velocity of 5 meters/second to the right. Find the function modeling the mass s position. Starting with the equation mx + cx + kx = 0, we have m = 5 and c =. The value for the spring constant k is k = 2/1 = 2, so the equation 5x + x + 2x = 0 is our starting point. The characteristic equation is whose roots are simplifying, we have r 1 = r 1 = 1 + 5r 2 + r + 2 = 0, 15 5 and r 2 = 0 40 ; and r 2 = 1 15 Since c 2 > 4mk, we see that the system is overdamped, and the roots r 1 and r 2 are real. The general solution is x(t) = c 1 e r 1t + c 2 e r 2t ; now we must find the particular solution that fits our equation. Since the mass was originally released at 2 meters to the left of equilibrium, with initial velocity of 5 meters/second, we have x(0) = 2 and x (0) = 5. 8

9 Using x(0) = 2, we have 2 = c 1 e 0 + c 2 e 0 = c 1 + c 2, or c 1 = 2 c 2. Since we see that x (t) = r 1 c 1 e r 1t + r 2 c 2 e r 2t and x (0) = 5, 5 = r 1 c 1 e 0 + r 2 c 2 e 0 = r 1 c 1 + r 2 c 2 = r 1 ( 2 c 2 ) + r 2 c 2 = 2r 1 + (r 2 r 1 )c 2 so that Finally, we have c 2 = 5 + 2r 1 r 2 r 1, and c 1 = r 1 r 2 r 1. x(t) = ( r 1 r 2 r 1 )e r 1t r 1 r 2 r 1 e r 2t, with 15 r and r 2 = 1 5 The graph of the position function, 0 t 30, is below: 15 Notice that, since the damping constant is fairly large, the motion of the weight is damped back to equilibrium very quickly. 9

10 Example. The same mass from the previous example is now attached to a dashpot that produces 3 Newtons of resistance for each meter/second of velocity. The spring, initial position, and initial velocity are the same as above. Find the function modeling the mass s position. Again, we have m = 5, k = 2, x(0) = 2, and x (0) = 5, but this time c = 3. The equation modeling the system is 5x + 3x + 2x = 0, whose characteristic equation is The roots are 5r 2 + 3r + 2 = 0. r 1 = and r 2 = ; since c 2 < 4mk, this is a case of underdamping, and the characteristic equation has complex roots. Setting γ = 3 31 and τ =, we see that the roots of the characteristic equation are γ ± τi, so that the general solution to the equation is x(t) = e γx( c 1 cos τt + c 2 sin τt ). To find the particular solution, we must again use the initial conditions x(0) = 2 and x (0) = 5. Starting with x(0) = 2, we see that 2 = e 0( c 1 cos 0 + c 2 sin 0 ) = c 1, so that Using x (0) = 5, with x(t) = e γt( 2 cos τt + c 2 sin τt ). x (t) = γe γt( 2 cos τt + c 2 sin τt ) + e γt( 2τ sin τt + c 2 τ cos τt ), we have 5 = γe 0( 2 cos 0 + c 2 sin 0 ) + e 0( 2τ sin 0 + c 2 τ cos 0 ) = 2γ + c 2 τ so that c 2 = 5 + 2γ. τ Thus the particular solution describing the motion of the mass in our system is with x(t) = e γt( 2 cos τt γ τ γ = 3 and τ = The graph of the function is below: 31. sin τt ),

11 Since the damping constant is so small, the dashpot allows a fair amount of oscillation before damping the motion back to equilibrium. Example. Given the same mass and spring from the previous examples, what resistance must the dashpot provide so that the motion is critically damped? Find the function modeling the mass s position in this case. In order for the motion to be critically damped, we need c 2 = 4mk. Since m = 5 and k = 2, In this case, the differential equation is with associated characteristic equation 4mk = 40 so that c = 40 = 2. 5y + 2 y + 2y = 0, 5r r + 2 = 0. We chose c so that c 2 = 4mk; as a result, we know that c 2 4mk = 0. Thus the characteristic equation has repeated root r = c = 2 Thus the general solution to the equation is = x = c 1 e rt + c 2 te rt, with r = To find the corresponding particular solution, we again return to the initial conditions x(0) = 2, x (0) = 5. Since x(0) = 2, we have 2 = c 1 e 0 + c 2 0 e 0 = c 1, 11

12 so that To find c 2, we ll need to calculate x : Since x (0) = 5, we have so that x = 2e rt + c 2 te rt, with r = x (t) = 2re rt + c 2 e rt + c 2 rte rt. 5 = 2re 0 + c 2 e 0 + c 2 r 0 e 0 = 2r + c 2, c 2 = 5 + 2r. Finally, the particular solution to our equation is The equation is graphed below: x = 2e rt + (5 + 2r)te rt, with r = The term critically damped means that, if we were to reduce the damping constant even a tiny bit, the dashpot would allow the same kind of oscillations we saw in the underdamped case. Below, all three functions discussed above are graphed on the same scale for comparison: 12

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