LAB1_411_RENATO_FERNANDEZ_FALL_2023
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Fernandez 1 The city College of New York LAB 1 EXPERIMENT 1B: TIME-DOMAIN DYNAMICS OF RECTILINEAR MECHANICAL SYSTEMS Renato Fernandez ME 411- System dynamics and control
Fernandez 2 Abstract: The topic of time domain of rectilinear mechanical systems was discussed and applied for this experiment. Four cases from three different spring and mass configuration were performed for the results to be analyzed and used to find certain system parameters such as undamped natural frequency, damped natural frequency, and damping ratio. Additionally, depending on the type of response, which was determined from the computer during the acquisition of data process, certain specifications were calculated such as maximum overshoot, peak time, rise time, settling time, delay time, and steady state error. Pre-experimental questions: 1.
What are the major components in this configuration accounting for most of m? The 500g weights that were added to the carriages and the weight of the devices themselves. 2.
What components in this configuration, other than the spring, contribute to k? Other than the spring, the carriages can contribute to k. 3.
Although there is no damping element included in this configuration, the term b in the above equation is called the incidental damping constant, which accounts for all damping/frictional effect involved in this experiment. What are the major components in this configuration accounting for most of b? The impact between carriages which may send the first carriage back to a position close to its initial one and any form of friction between the devices and the table. 4.
Although minor, what other components in this configuration may contribute to b? A minor component that contributes to damping is the energy dissipation and air resistance. 5.
Will the incidental damping obey the law of linear viscous damping? Justify your reason. The incidental damping should indeed obey the law of linear damping, this is because this law states that the velocity of an element is directly proportional to its damping. In this case, the incidental damping produced by energy dissipation and air resistance could increase if the carriages were to move faster.
Fernandez 3 Case 1 (IB2) For case 1, the free response of a second order undamped system of a linear system was investigated. The driving function for case 1 was : With initial conditions: These initial conditions are typical for a free response. The input for this case, f(t), was a step function with a step value of 0 since this is a free response where the carriage is held initially manually and not moved by an external force from the device. The configuration for input was as follows: Post experiment analyses: 1)
Use the free-response experimental data to determine the system parameters: undamped natural frequency W
n, damping ratio ζ, damped natural frequency Wd and 2% settling time ts(2%) using the theories you learned in Topic 5: Time-Domain Analysis of the course (e.g., logarithmic decrement and free response of 2nd-order underdamped systems).
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Fernandez 4 The required values shown below were found using the plot of position vs data for free response which is shown above. •
T(period) = 1.895-1.43= 0.465 •
damped natural frequency ➔
2pi/Wd = 0.465 Wd = 13.51 •
Logarithmic decrement ln(11.9401/7.00132) = 0.5338 •
n = 2 •
ζ
= 0.5338/sqrt((4pi^2 + (0.5338)^2) ➔
ζ=
0.08465 •
Undamped natural frequency ➔
2pi / (0.465* sqrt(1-0.08465^2)) Wn = 13.56 •
Settling time = 4/(0.08465*13.56) •
ts = 3.48476s 2)
With m
theoretical known determine also the corresponding k
experimental
and b
incidental
. Compare k
experimental
with k
theoretical
. Discuss the reason for difference if exists. m
theoretical = 1kg Experiment stiffness (K
experimental
) = Wn^2 * m
theoretical
= 183.874 N/m B
incidental = 2 * 0.08465*sqrt(183.87*1) = 2.2957 K
experimental
(450N/m) is way larger than K
theoretical
(183.87 N/m) and it
’
s mostly due to the friction between the equipment and the table which is not accounted for during this experiment in addition to the air resistance that the carriages may have experienced. Case 2 (IB3) For this case, the forced response for a second order undamped rectilinear dynamic system was analyzed. This time, as the name suggests, an external force was exerted on carriage 1 in order for movement to take place. The driving function for case 2 was as follows.
Fernandez 5 With initial conditions: There were three types of input involved for this case: step, impulse, and ramp. Configuration for step response was as follows: It can be observed that the maximum value for this step response was 5 N. Configuration for impulse response was as follows: It can be noticed that maximum value of the input is 40 N. Configuration for ramp response was as follows: It can be noticed that the maximum value for input to ramp response was 6 N. A)
Step response
Fernandez 6 B)
Impulse response C)
Ramp response
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Fernandez 7 1)
Use the experimentally-obtained step-response data to determine the time-domain step-response specifications: maximum overshoot Mp in %, the corresponding peak time tp, 0-100% rise time tr(0-00%), 10-90% rise time tr(10-90%), ±2% settling time ts(±2%), delay time td and steady-state step response error in %. Determine also the system parameters: undamped natural frequency Wn, damping ratio ζ and damped natural frequency Wd
. Using step response plot •
T = 0.425 •
N=2 •
Logarithmic decrement ln(A1/An) = ln(25.4/20.84) = 0.1979 •
Damping ratio: 0.1979/(sqrt(4pi^2 + (0.1979)^2)) ζ = 0.03148
•
Damped natural frequency(Wd): 2pi / 0.425 Wd= 14.784 •
Undamped natural frequency (Wn): 2pi / (0.425 * sqrt(1-0.03148^2)) Wn= 14.79 •
Mp = exp(-0.03148*pi/(sqrt(1-0.03148^2)) *100% Mp = 90.579% •
Tp = pi/14.784 Tp= 0.2125s •
Tr (0-100) = (pi –
tan^-1 (sqrt(1-0.03148^2)/0.03148))/(wd) Tr(0-100) = 0.1084 s •
Tr (10-90) = (0.8 + 2.5*0.03148)/(14.79) Tr(10-90) = 0.0594 s •
Settling time Ts = 4/(0.03148*14.784) Ts= 8.59475 s •
Delay time Td = (1 + 0.7*0.03148)/14.79 Td= 0.0691s •
STEADY STATE RESPONSE ERROR = 5.5% 2)
Based on the acquired impulse-response data, determine again the system parameters: undamped natural frequency ωn, damping ratio ζ, damped natural frequency ωd and 2%
settling time ts(2%). Compare the values of ωn, ζ, and ωd obtained from free-, step- and impulse responses. If differences exist, discuss the possible sources, which should be accounted for these differences. Compare also ts(2%) obtained from free- and impulse-responses. Again, discuss the possible sources for the difference. Using Impulse response plot •
T = 0.456 •
N=3
Fernandez 8 •
Logarithmic decrement ln(A1/An) = ln(10.8272/5.8782) = 0.6108 •
Damping ratio: ((1/2)*0.6108)/(sqrt(4*pi^2+ (1/2 * 0.6108)^2)) ζ =
0.0485 •
Damped natural frequency (Wd) : 2pi / 0.456 Wd = 13.7789 •
Undamped natural frequency (Wn): Wn =13.7952 •
Ts = 5.9724 •
Steady state impulse response error = 77.9% Comparing results: Table 1. Analytical results for free response, step response, and impulse response. Type of response Wn Damping ratio Wd Ts Free response 13.56 0.08465 13.51 3.48476 Forced step response 14.79 0.03148 14.784 8.59475 Forced impulse response 13.7952 0.0485 13.7789 5.9724 It can be observed that there are several differences between the values obtained for the three types of responses shown in the table above. One of the reasons could be an error in either the calculation of the system parameters or an error when acquiring the raw data from the experiment. 3)
Based on the ramp-response experimental data, make a judgment if the two key time-
domain ramp-response specifications: steady-state ramp-response error and steady-
state ramp-response time-delay exist or not. If they exist, compare them with the theoretical values calculated using the above-mentioned system parameters: undamped natural frequency ωn and damping ratio ζ. If they do NOT exist, discuss the reasons why. Based on the data obtained and plotted for ramp response, it can be concluded that it is not possible to determine a steady state. Finding time domain ramp response specifications is not possible as they do not exist. It was not possible to find period, undamped natural frequency, and damped natural frequency since the plot does not present a sinusoidal shape.
Fernandez 9 Case 3 (IB-4) For this case, a step response was analyzed again but this time, for a first order rectilinear dynamic system. The driving function was: With initial conditions: Configuration for step input was as follows: 1)
Position plot
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Fernandez 10 2)
Velocity plot 3)
Acceleration plot
Fernandez 11 Pre-experimental question •
Why the acceleration result is much noisier than the velocity result, which in turn is much noisier than the displacement result? This is probably because there were more factors interfering with the readings of velocity and acceleration than there were for position. 1)
Use the experimental step-response data to determine the system parameters: time constant τ, 98% settling time ts(98%) and steady
-state step response error in %. •
Time constant = 1/a = 0.54 •
a = 1.85 •
Settling time(98) = 4/a Ts(98%) =2.162 •
Steady state response error = 36.54% 2)
With the mtheoretical evaluated in Section 1B-2, determine the corresponding bincidental. Compare this bincidental value with the one obtained in Section 1B-2. Discuss the reason for difference if exists. •
m= 1kg •
From transfer function of first order systems = 1/(s+a) •
Taking laplace transform of governing equation msX(s) + bX(s) = F(s) X(s)/F(s) = 1/(ms+b) Considering m = 1kg b = 1.85 Compared to the incidental b of case 1, 1.85 is smaller. This could be a consequence of an error in calculation or in the set up that led to the acquisition of data. Case 4 (IB-5) For this case, a step response for a second order rectilinear dynamic system due to base excitation was analyzed. Unlike the previous two cases, this case does require the use of a second carrier which movement will be analyzed for this report. The driving function for this experiment was:
Fernandez 12 With initial conditions: Configuration for step input was as follows: It can be observed now that force is not used as input now but the position of carriage 1 which will oscillate while exciting moving for carriage 2. 1)
Use the acquired experimental data of base-excited step-response to determine the time-domain step-response specifications: maximum overshoot Mp in %, the corresponding peak time tp, 0-100% rise time tr(0-00%), 10-90% rise time tr(10-
90%), ±2% settling time ts(±2%), delay time td and steady-state step response error in %. Determine also the system parameters: undamped natural frequency Wn, damping ratio ζ and damped natural frequency Wd. •
T = 0.42s •
Damping ratio: ζ
= 0.0486 •
Wd = 14.96 •
Wn = 13.7952 •
Ts (±2%) = 5.9692 •
Mp = 85.8314% •
Tp = 0.21s
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Fernandez 13 •
Tr (0-100)= 0.1082 •
Tr (10-90) = (0.8 + 2.5*0.0486)/13.7952 = 0.0668 •
Td = (1+0.7*0.0486)/13.7952= 0.07496 •
Steady state response error = 3.3% 2)
Compare the values of ωn, ζ, and ωd
obtained here with those obtained in from free-
, step- and impulse responses in Sections 1B-2 and 1B-3. If differences exist, discuss the possible sources. Table 2. Analytical results of system parameters. Response Wn Wd Damping ratio Free (IB-2) 13.56 13.51 0.08465 Step (IB-3) 14.79 14.784 0.03148 Impulse (IB-3) 13.7952 13.7789 0.0485 IB5 response 13.7952 14.96 0.0486 It can be observed that the damping ratio for Impulse response of case 2 and free response for case 4 are very similar if not almost the same. However, the damping ratio for the free response of case 1 is noticeably very different (greater) compared to the damping ratio of the rest of the cases. The reason for this is probably the different damping conditions that were utilized during the configurations. In the case of the undamped and damped natural frequencies, it can be noticed that almost all the results range from 13 to a value close to 15 being each of them fairly close. 3)
Compare the values of Mp, tp, tr(0-00%), tr(10-90%), ts(±2%) and td obtained here with those obtained in from the step response in Section 1B-3. If differences exist, discuss the possible sources. It can be noticed that the values for Mp, Tp, and Tr(0-100) are very close for both cases, however, the same can not be stated about tr (10-90), ts, and td which show some differences. Possible sources of error are the different kind of systems that were analyzed ( first and second order systems). Table 3. Analytical resultsof MP, tp, tr, ts, and td . Response Mp (%) Tp Tr(0-100) Tr(10-90) ts td Step (IB3) 90.58 0.2125 0.1084 0.0594 8.59475 0.0691 Step (IB4) 85.83 0.21 0.1082 0.0668 5.9692 0.07496
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- Needs Complete typed solution with 100 % accuracy. Otherwise skip if you can't give complete solution don't use chat gpt or ai i definitely upvote you.arrow_forwardFind the differential equation of the mechanical system in Figure 1(a) To obtain the differential equation of motion of the mass and spring system given in Fig. 1. (a) one may utilize the Newton's law for mass and spring relations defined as shown in Fig. 1. (b) and (c) use f = cv for viscous friction, where v is the velocity of the motion and c is a constant. Z/////// k M F. F, F F F, F, k EF=ma F = k(x, - x,) = kx (b) (c) Figure 1: Mass-spring system (a), Force relations of mass (b) and spring (c)arrow_forwardYou are requested to design an automotive suspension or shock absorber system. In order to simplify the problem to one dimensional multiple mass-spring-damper system, a quarter vehicle model is used. The system parameters and free-body diagram of such system is shown below. M₁: Automobile body mass M₂: Wheel and suspension mass K₁: Spring constant of suspension system K2: Spring constant of wheel and tire B: Damping constant of shock absorber (a) Obtain the transfer function of X₁ (s) F(s) T₁(s) = = 2500 kg = 320 kg and T₂(s) = = 80,000 N/m = 500,000 N/m = 350 N-s/m Automobile- Suspension system Wheel- M₁ M₂ X₁ (s) - X₂ (S) F(s) in terms of the parameters of mass, damper and elastance (M, B and K). (b) Express the T₁ (s) and T₂ (s) with numerical values. c) Plot the x₁ (t) and x₁ (t) = x₂(t) outputs of this passive suspension system for the input torque f(t) = 2,000 N. fit) K₂ x₂(t) -Tirearrow_forward
- attached is a past paper question in which we werent given the solution. a solution with clear steps and justification would be massively appreciated thankyou.arrow_forwardSubject name (Analytical Mechanics) Peace be upon you. I want all the laws that help me solve mathematical problems in the mentioned topics.arrow_forwardVibration Topic. Please help with the solution with neat answer and correct units. Will upvote to answer. Thanksarrow_forward
- Forcing Function Spring Constant (2)4 Mass k Friction Constant b m Mass Displacement y(t) 3. (20 pts.) Consider the following spring-mass-damper mechanical system (it is placed sideways, so that you won't need to consider gravity). The input is given by f(t), and the output is y(t). Find an equation in frequency domain (s domain) that defines the relationship between the input F(s) and the output Y(s).arrow_forwardGiven the vibrating system below: K1 K3 K4 Find the following 1. Keq 2. Ceq 3. Natural angular velocity K2 m C1 C3 C5 C2 C4 4. Damped angular velocity 5. Type of Damping 6. Equation of motion x(t). Assume Initial conditions for displacement and velocity Graph 2 cycles of the vibrating system. You can use third party app for this. 7. M = 10 kg K1=100 N/m K2= 80 N/m K3=75 N/m K4= 120 N/m C1 = 20Ns/m C2= 40 Ns/m C3= 35Ns/m C4= 15 Ns/m C5= 10 Ns/marrow_forwarda) A two-degree-of-freedom coupled pendulum model is shown in Figure Q1a. The parameters of the model are: k=100 N/m; m=10 kg and L=1 m. The equations of motion of the system are given below. J₁ä + (mgL + kĽ² ) 8₁ - kľ² 0₁₂ = 0 J₁₂+(mgL+kĽ²)₂-kĽ² 0₁ = 0 i) Calculate the natural frequencies. ii) Draw the mode shapes. m Figure Q1a. b) In the above system discuss the implications of the following changes in coupling stiffness. i) k is reduced considerably to 0.1 N/m. ii) k is increased considerably to 1x105 N/m. c) Draw a schematic of modal test arrangement of a beam using instrumented impact hammer. d) Why is it necessary to measure force in a modal test?arrow_forward
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