Suspended Spring Mass System - Lab Report Example

Assignment Report Spring Suspended Mass al Affiliation Topics to be discussed: Solution of Section A MATLAB Simulink Model Explanation for Section A Solution of Section B MATLAB Simulink Model Explanation for Section B Verification of Results Solver used for Simulink Modeling References Section A Solution: Differential equation describing the dynamic behavior of suspended spring mass is given as: Part (1) Solution: Matlab Simulation Matlab Simulink Simulation is attached with the following parameters. M = 0.5 kg M* = 0.5 kg (step change mass) Ks = 25 kg/s2 Transfer function model of Figure 1 is: (1) Figure 1: Spring Mass System Part (2) Solution: Finding amplitude and natural frequency of Figure 1 Required form of the response is: where; A = Amplitude  = Angular frequency Time domain response of Figure 1 is: t where; MT = M + M* By comparison of above two equations we can get ‘A’ and ‘’ in term of system parameters as follows; A = ;  = Assuming zero initial mass zero i.e. M = 0  Taking MT = M* = 0.5 kg Ks = 25 kg/s2 Solution (i) A  0.36 m (Obtained from Simulink Waveform) Solution (ii)  = =  6.97 rad/s Where T = Time period of sin wave in sec obtain from Simulink Waveform MATLAB Simulink Model Explanation for Section A: Simulink model for un-damped model is shown below. Actually this Simulink model implement the following mathematical model of suspend spring mass system. = Figure 2: Simulink Model for Suspended Spring Mass Step by step explanation is given below: First we take the Add1 block from Simulink Math Library, and give two inputs M (a constant initial mass 0.5 i.e. kg) and M* (step change mass i.e. 0.5 kg). Output of this block is MT which is the sum of both M and M*. Now this MT is further used by two blocks, 1/MT is found by taking the reciprocal of MT and Force F is found by multiplying it with gravitational constant g which is required as clearly seen from the mathematical model. Output coming from second add block is F – (Ks) x which is the multiplied by 1/MT to get double derivative of x i.e. which is the displacement of the mass with respective to reference position. To obtain x from, it is passed through two times into the integrator block. Once we got the x, we fed into the Add block after with multiplication of Ks. Scope is attached to see the output response. This completes my model of suspended spring mass system for un-damped case. As there is no damping involved in the system, so I got oscillatory response which is shown below. Figure 3: Reponses of Suspended Spring Mass When M = 0.5 kg Figure 4: Reponses of Suspended Spring Mass When M = 0 kg Section B Solution: Part (1) Solution: Adding the effect of damping coefficient ‘b’ kg/s in the Figure 1 Damping coefficient ‘b’ kg/s can be modeled by adding a damper with in parallel with spring constant Ks. Writing the differential equation of motion using Newton’s Law to sum to zero all of the forces acting on mass MT. = Taking the Laplace transform, assuming zero initial conditions, (2) Part (2) Solution: Find damping present in the system when time constant ‘’ of the system is 5 sec. Here MT = 1 Using the simulation parameter of Section A Part 1 we get,  = = = 5 Given Time constant =  = 5 sec (obtained from experiment) General formula for the time constant of 2nd order system is;  = where  = damping ratio of the system  = = = = 0.16 General 2nd order system has the following transfer function in standard form; = where B = Damping constant B = b = damping coefficient = 2     = 2  0.16  5 = 1.6 kg/s Transfer function of the system now can be written as; Part (3) Solution: Relating damping coefficient with system parameters assuming M = 0 Using equation 2; Putting the following values into above expression to got transfer function model of system with the presence of damping coefficient; MT = 0. 5 kg ; b = 1.6 kg/s ; Ks = 25 kg/s2 = (3) General 2nd order system has the following transfer function in standard form; = where B = Damping constant B = 2b = 2  damping coefficient = 2  1. 6 = 3.2 kg/s Time constant can be related to standard 2nd order transfer function as follows; 2 = B   = B/2 2 = A   =  = Where time constant for general 2nd order system is; = Putting value of  and   = In terms of system parameter; B =  = Part (4) Solution: Observation when damping is included When damping is added to the system, system response changes its behavior from un-damped to under-damped. If we change the system damping, then settling time of the system varies inversely to the damping. If we increase the damping of the system then, peak time and %S reduces and vice versa. MATLAB Simulink Model Explanation for Section B: Now we have included the damping b kg/s in the system, due to this damping system response will be under-damped, and due to this damping system will stop oscillating unlike earlier case and will reach its steady state after sometime. Simulink model for this situation and its corresponding response curves are shown below. Figure 5: Simulink Model When Damping is included Figure 6: Simulink Model When M = 0.5 kg Figure 6: Simulink Model When M = 0 kg Verification of results: There is verification of Simulink results with mathematical formulas derived above. For sec B Part 2: b = 1.6 kg/s ; MT = 1 = = = 5 sec  Verified from Simulink Model Waveforms For sec B Part 3: b = 1.6 kg/s ; MT = 0.5 = = = 2.5 sec  Verified from Simulink Model Waveforms Similarly all others results can be verified from Simulink Model. Solver used for Simulink Modeling: Solver used for Simulink Modeling is ode45. References Friedland, Bernard. 2012. ‘An Introduction to State-Space Methods,’ Control System Design. Niece, Norman. 2010. Control Engineering 6th Edition. Read More