1. April 20001 Second Order Systems m Spring force ky F(t) (proportional to velocity) (proportional to displacement)

2. April 20002 Second-Order Systems  Contains the following two elements  Energy storage element  Damper restricts flow of energy  Examples - Ringing a bell - Shock absorber connected to spring in your car - Capacitance-inductance electrical systems - Catheter-transducer

3. April 20003 Basic Equations  General equation for 2nd-order systems  Divide a o throughout the above equation gives us x is forcing function and represents what we’re trying to measure. y is instrumentation output x is forcing function and represents what we’re trying to measure. y is instrumentation output

4. April 20004 Basic Equations  General equation for 2nd-order systems  Define  A general solution is:

5. April 20005 Basic Equations, cont’d If a 1 = 0 ( and therefore  = 0) the damping force is zero and system responds by oscillating with its natural frequency,  n. If 0 <  < 1, system is “under-damped” and responds by ringing down. If  = 1, the system is “critically damped” and responds quickly, w/no oscillations If a 1 is large such that  > 1, then system is “over-damped” and responds slowly, without oscillations

6. April 20006 Equation in terms of natural frequency and damping ratio With our definitions of  n and , our equation becomes If  < 1, our general solution is: ( a decaying sinusoid ) “standard form”

7. April 20007 Particular Solution - Recall that the general solution is for the case where there is no forcing function, that is, when Kx o =0. - To get the total solution, we add the forcing function to the general solution: y total = y gen + y part = y gen + Kx o = - If the forcing function changes abruptly from x = 0 to x= x o at t = 0, an ideal measurement device would read Kx o immediately. But since this is a second order system, we will see damping and ringing (for  < 1) before the measurement system “settles” on the measurand actual value of K x o. For these initial conditions and knowing as t  , y  Kx o, we get (book uses y e = Kx o )

8. April 20008 Solution Changes With Damping underdamped case critically damped case undamped case overdamped case

9. April 20009 Basic 2nd Order Behavior underdamped case (  < 1) critically damped case (  =1) overdamped case (  > 1) t y critically damped case (  =1) underdamped case (  < 1) t y Driving function (Kx o ) is “step up” Driving function (Kx o ) is “step down”

10. April 200010 An Example of the 2 nd -Order Equation  The following equation describes the behavior of a second-order system. Determine the natural frequency and the damping ratio of the system. Covert it to standard form as shown in Eq.(11.21) and find the equilibrium response (response of the system in the absence of the dynamic effect) of the system.

11. April 200011 Solution for  n and   Comparing the coefficients of the above two equations give us underdamped

12. April 200012  Remember Newton’s Second Law?  The Sum of forces: Second Order Equation  Where  m = mass  = damping force coefficient  k = spring constant

13. April 200013  Relate to measured parameters  Where   n = natural frequency   = damping ratio  K = 1/k = 1/spring constant Second Order Equation

14. April 200014  What are some parameters? Second Order Equation

15. April 200015  Consider a step input  Forcing function, F(t)  Change from x = 0 to x = x o at t = 0  The response of the system depends on damping ratio,    < 1, underdamped (damped oscillation)   > 1, overdamped (asymptotic response)  if  = 1, critically damped (response is on the verge of oscillating) Second Order Solution

16. April 200016 Second Order Responses

17. April 200017 0 0.5 1 1.5 2 2.5 0246810 wnt y/Kx damping ratio=0.0 0.25 0.75 1 1.5 2 Second Order Responses ntnt

18. April 200018 Second Order System (Continued) A Pressure Transducer

19. April 200019  x = x o sin  t  forcing function is Kx  Ideal system (without dynamic (d/dt) effects would provide an output y e = Kx o sin  t  Actual response (y) has a damped part that dies out and a long term part that relates to the driving function  Continuing part given by: Sinusoidal Input (Forcing) Function Measurand Measurement System Response

20. April 200020 Second Order Responses Response of a 2nd-order System to a Sinusoidal Input Frequency Ratio (  /  n ) Amplitude Ratio (y/kx o ) Work Examples

21. April 200021 2nd Order System EXAMPLES Pressure Transducer A small tube, 0.5 mm in diameter, is connected to a pressure transducer through a volume of 3.5 m 3. The tube has a length of 7.5 cm. Air at 1 atm and 20 o C is the pressure-transmitting fluid. Calculate the natural frequency for this system.

22. April 200022 Pressure Transducer - An Example of 2nd-Order System

23. April 200023 Pressure Transducer - An Example of 2nd-Order System Calculate the damping ratio and the attenuation of a 100- Hz pressure signal in the system.

24. April 200024 Cantilever Beam - Another Example of 2nd-Order System A 1/16 -in-diameter spring-steel rod is to be used for a vibration frequency measurement as shown in the figure below. The length of the rod may be varied between 1 and 4 in. The density of this material is 489 lbm/ft 3, and the modulus of elasticity is 28.3  10 6 psi. Calculate the range of frequencies that may be measured with this device.

25. April 200025 Cantilever Beam - Another Example of 2nd-Order System =4

26. April 200026 This Week in the Lab   d : the damped frequency   : the damping ratio   n : the undamped natural frequency Measuring Parameters

27. April 200027 The Solution  Only Concerned with the underdamped response.

28. April 200028 The Solution

29. April 200029 Measuring  n  the answer here is a simple calculation