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

April 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

April 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

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

April 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

April 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”

April 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 )

April Solution Changes With Damping underdamped case critically damped case undamped case overdamped case

April 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”

April 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.

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

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

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

April  What are some parameters? Second Order Equation

April  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

April Second Order Responses

April wnt y/Kx damping ratio= Second Order Responses ntnt

April Second Order System (Continued) A Pressure Transducer

April  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

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

April nd 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.

April Pressure Transducer - An Example of 2nd-Order System

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

April 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.

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

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

April The Solution  Only Concerned with the underdamped response.

April The Solution

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