Forced Mechanical Vibrations

Transcription

1 Forced Mechanical Vibrations Today we use methods for solving nonhomogeneous second order linear differential equations to study the behavior of mechanical systems.. Forcing: Transient and Steady State Solutions 2. Forcing: Resonance Transient and Steady State Solutions Let us now consider my + by + ky = g(t) where g(t), a forcing function, is non-zero. The forcing function is just an outside force applied to the system. (It might represent the platform the spring is on being shaken, for example.) We are usually interested in periodic forcing functions. Example: Let us return to the spring-mass system. We had a 2 pound mass, an equilibrium stretch of L = 2 feet, and a resistance which was four times velocity. For the homogeneous problem y + 4y + 6y = 0, we found the general solution to be y(t) = c e 2t cos(2 t) + c 2 e 2t sin(2 t). Let us now introduce an external force g(t) = 0 cos(t) which is applied to the system. If the spring is stretched two feet down and then released, solve for the position of the spring and describe the motion. Our initial value problem is now y + 4y + 6y = 0 cos(t), y(0) = 2, y (0) = 0. This is non-homogeneous, so we will need the complementary solution to the homogeneous equation which we found above.

2 Next, we need a particular solution to the non-homogeneous equation. Since g(t) = 0 cos(t), we can use the method of undetermined coefficients. (Note that g(t) does not duplicate any part of the complementary solution.) We get: y p (t) = A cos(t) + B sin(t) y p (t) = A sin(t) + B cos(t) y p(t) = 9A cos(t) 9B sin(t) We plug this into our non-homogeneous equation and get y p + 4y p + 6y p = [ 9A cos(t) 9B sin(t)] + 4 [ A sin(t) + B cos(t)] + 6 [A cos(t) + B sin(t)] = [ 9A + 2B + 6A] cos(t) + [ 9B 2A + 6B] sin(t) = [7A + 2B]cos(t) + [7B 2A]sin(t) Since we want y p + 4y p + 6y p = 0 cos(t), we have the equations 7A + 2B = 0 2A + 7B = 0 Solving this system yields A = 70/9, B = 20/9, so our particular solution is and the general solution is y p (t) = cos(t) sin(t) y(t) = c e 2t cos(2 t) + c 2 e 2t sin(2 t) cos(t) sin(t) From our initial conditions, we know y(0) = 2, so since to see that y(0) = c , then c = 6/9. By plugging in y (0) = 0, we determine c 2 = 6 9. Thus the solution is 6 9 e 2t cos(2 t) e 2t sin(2 t) cos(t) sin(t) Notice that the first two terms (which originate in the solution to the homogeneous problem) have a negative exponential and therefore decay to zero. Such a part of a solution is said to be transient. The last two terms (which originate from the forcing function) represent simple harmonic motion, and do not decay with time. The portion of the solution which remains 2

3 as t is referred to as the steady state solution. We can see in this case that the steady state solution has period 2π/ and amplitude ( ) 70 2 ( ) 20 2 R = + = The graphs of the transient and steady state solutions are shown below from t = 0 to t = 0: Transient Solution 6 9 e 2t cos(2 t) Steady State Solution 20 cos(t) + sin(t) 9 9 e 2t sin(2 t)] You can see the effects of the transient solution vanishing as t increases in the plot of the solution: Solution y(t) = y c (t) + y p (t) 2 Forcing Functions: Resonance Again, we will consider the differential equation my + by + ky = g(t) for a spring with mass m, damping coefficient b, and forcing function g(t). Let us return to the case in which there is no damping, and consider the corresponding homogeneous equation: my + ky = 0 We then have characteristic equation mr 2 + k = 0 which has roots r = ±i k/m. Thus, our solution has angular frequency k/m.

4 If we introduce a forcing function, we expect our solution to change. An interesting case occurs when the forcing function has the same frequency as the natural frequency k/m of the system. Example: Suppose we have a mass of 2 pounds on a spring which stretches by /2 foot from the weight. If we have a forcing function g(t) = cos(8t), no damping, and y(0) = y (0) = 0, describe the motion of the system. Our mass is 2/2 = /8, and k = mg/l = 2/(/2) = 24, so our differential equation is 8 y + 24y = cos(8t) or y + 64y = 8 cos(8t) The homogeneous equation has characteristic equation r = 0, so r = ±8i. Thus the complementary solution is y c (t) = c cos(8t) + c 2 sin(8t) Note that the angular frequency here is 8, which is the same as the angular frequency of the forcing function. Now we solve the non-homogeneous part by undetermined coefficients. For the forcing function 8 cos(8t), we would normally try A cos(8t) + B sin(8t), but since cos(8t) and sin(8t) duplicate the homogeneous solution, we must multiply by t: y p (t) = At cos(8t) + Bt sin(8t) y p (t) = A cos(8t) 8At sin(8t) + B sin(8t) + 8Bt cos(8t) y p (t) = 8A sin(8t) 8A sin(8t) 64At cos(8t) + 8B cos(8t) + 8B cos(8t) 64Bt sin(8t) = 6A sin(8t) 64At cos(8t) + 6B cos(8t) 64Bt sin(8t) Plugging into y + 64y = 8 cos(8t) yields 6A sin(8t) 64At cos(8t) + 6B cos(8t) 64Bt sin(8t) + 64At cos(8t) + 64Bt sin(8t) = 8 cos(8t) 6A sin(8t) + 6B cos(8t) = 8 cos(8t) So finally we have A = 0 and B = /6. Thus our particular solution is and the general solution is y p (t) = 6 t sin(8t) y(t) = y c (t) + y p (t) = c cos(8t) + c 2 sin(8t) + 6 t sin(8t) 4

5 Solving so that y(0) = y (0) = 0 gives c = c 2 = 0, so we have y(t) = 6 t sin(8t) Notice the behavior: as t, y(t) oscillates and grows: The above is an example of resonance. When the forcing function has a frequency which matches the natural frequency of the system, we will duplicate the complementary solutions and will require a particular solution of the form At cos(ω 0 t) + Bt sin(ω 0 t) This solution will grow and oscillate as t, just as in the above example. 5