X Determine: k (i) (ii) с ww m E Figure Q1 Figure Q1 shows a forced spring-mass system with damping, where mass m = 1 kg, spring constant k = 0.2 N/m, and damping coefficient c = 0.3 N-s/m. (a) This forced spring-mass system with damping can be described by the following differential equation: F d²x(t) c dx(t) k dt² m dt + -+-x(t)==F(t) m m Laplace Transfer Function of this system, This system's steady state gain, damping ratio and natural frequency.
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- 1. A spring mass system serving as a shock absorber under a car's suspension, supports the M 1000 kg mass of the car. For this shock absorber, k = 1 × 10°N /m and c = 2 × 10° N s/m. The car drives over a corrugated road with force %3| F = 2× 10° sin(@t) N . Use your notes to model the second order differential equation suited to this application. Simplify the equation with the coefficient of x'" as one. Solve x (the general solution) in terms of w using the complimentary and particular solution method. In determining the coefficients of your particular solution, it will be required that you assume w – 1z w or 1 – o z -w. Do not use Matlab as its solution will not be identifiable in the solution entry. Do not determine the value of w. You must indicate in your solution: 1. The simplified differential equation in terms of the displacement x you will be solving 2. The m equation and complimentary solution xe 3. The choice for the particular solution and the actual particular solution x,…An object attached to a spring undergoes simple harmonic motion modeled by the differential equation d²x = 0 where x (t) is the displacement of the mass (relative to equilibrium) at time t, m is the mass of the object, and k is the spring constant. A mass of 3 kilograms stretches the spring 0.2 meters. dt² Use this information to find the spring constant. (Use g = 9.8 meters/second²) m k = + kx The previous mass is detached from the spring and a mass of 17 kilograms is attached. This mass is displaced 0.45 meters below equilibrium and then launched with an initial velocity of 2 meters/second. Write the equation of motion in the form x(t) = c₁ cos(wt) + c₂ sin(wt). Do not leave unknown constants in your equation. x(t) = Rewrite the equation of motion in the form ä(t) = A sin(wt + ), where 0 ≤ ¢ < 2π. Do not leave unknown constants in your equation. x(t) =The stress profile shown below is applied to six different biological materials: Log Time (s] The mechanical behavior of each of the materials can be modeled as a Voigt body. In response to o,= 20 Pa applied to each of the six materials, the following responses are obtained: 2 of Maferial 6 Material 5 0.12 0.10 Material 4 0.08 Material 3 0.06 0.04 Material 2 0.02 Material 1 (a) Which of the materials has the highest Young's Modulus (E)? Why? Log Time (s) (b) Using strain value of 0.06, estimate the coefficient of viscosity (n) for Material 6. Stress (kPa) Strain
- Consider the following system mä(t) + cx(t) + kx(t) = F,8(t) + F28(t – t1) Assume zero initial conditions. The units are in Newtons. If m = 42.7334 kg, c = 44.7394 Ns/m, k = 6776.8559 N/m, F1 = 47.5525 N.s, F2 = -5.3708 N.s and t = 3.9737 s, the velocity of the system at t2 = 2.3387 s is1. Aspring mass system serving as a shock absorber under a car s suspension, supports the M = 1000 kg mass or the car. For this shock absorber, k = 1x 10°/m and c = 2x 10' N s/m. Tne car drives over a corrugated road with rorce F = 2x 10' sin(@t) N. Use your notes to model the second order difrerential equation suited to this application. Simplity the equation with the coerricient of X" as one. Soive X (tne general solution) in terms or @ using the complimentary and particular solution method. In determining the coernicients of your particular solution, it will be required that you assume o -1xo or 1- w x -m. Do not use Matiab as its solution will not be identiriable in the solution entry. Do not determine the value of W. You must Indicate In your solution! 1. Tne simpliried dirrerential equation in terms or the displacement X you will be solving 2. Ine m equation and complimentary solution Xe 3. Tne choice ror the particular solution and the actual particular solution Xp 4. Express…1. A spring mass system serving as a shock absorber under a car's suspension, supports the M=1000kgmass of the car. For this shock absorber,k=1000N/m and c=2000N s/m. The car drives over a corrugated road with force F=2000sin(wt)N. Use your notes to model the second order differential equation suited to thisapplication. Simplify the equation with the coefficient of x'' as one. Solve x (the general solution) interms of using the complimentary and particular solution method. In determining the coefficients ofyour particular solution, it will be required that you assume w2 -1=w or . Do not 1-w2=-wuse Matlab as its solution will not be identifiable in the solution entry. Do not determine the value of w.You must indicate in your solution:1. The simplified differential equation in terms of the displacement x you will be solving2. The m equation and complimentary solution3. The choice for the particular solution and the actual particular solution xp4. Express the solution x as a piecewise…
- 1:19 P N{ l 100% I A docs.google.com A block of mass m = 2 kg is attached to a spring with spring constant k = 200 N/m, and is set to oscillate on a friction-less horizontal surface. At time t = 0 its position is xO = 0 and its velocity is vo = +5 m/s. If the displacement is a cosine function, x(t)=A cos(wt+p), then which answer is correct about the oscillation amplitude A and the phase constant,p,: A = 0.25 m, and p = - T/2 A = 0.5 m, p = 0 A = 0.5 m, and p = r/2 A = 0.5 m, p = -T/2 A = 0.25 m, and p = t/2 Other: A body oscillates with simple harmonic frictionless motion along the x-axis. Its displacement varies ith time according to x(t)=0.5 cos(Tt). The ues of the velocity (in m/s) and acceleration (in m/s^2) of the 8 II3. The relationship between arterial blood flow and blood pressure in a single artery satisfies the following first-order differential equation: dP(t) + dt RC mmHg (cm³/s) P(t) = where Qin is the volumetric blood flow, R is the peripheral resistance, and C is arterial compliance (all constant). Qin-60 cm³/s and the initial arterial pressure is 6 mmHg. Also, assume R = 4 and C= 0.4- Oin cm³ mmHg (a) Find the transient solution Ptran(t) for the arterial pressure. The unit for P(t) is mmHg. (b) Determine the steady-state solution Pss(t) for the arterial pressure. (c) Determine the total solution P(t) assuming that the initial arterial pressure is 0.62. •A 5-kg object is constrained to move along a straight line. Its initial speed is 12 m/s in one direction, and its final speed is 8 m/s in the opposite Complete the graph of force versus time with direction. F (N) (s) appropriate values for both variables (Figure 7-26). Several answers are correct, just be sure that your answer is internally consistent. Figure 7-26 Problem 62
- 1. Verify Eqs. 1 through 5. Figure 1: mass spring damper In class, we have studied mechanical systems of this type. Here, the main results of our in-class analysis are reviewed. The dynamic behavior of this system is deter- mined from the linear second-order ordinary differential equation: where (1) where r(t) is the displacement of the mass, m is the mass, b is the damping coefficient, and k is the spring stiffness. Equations like Eq. 1 are often written in the "standard form" ď²x dt2 r(t) = = tan-1 d²r dt2 m. M +25wn +wn²x = 0 (2) The variable wn is the natural frequency of the system and is the damping ratio. If the system is underdamped, i.e. < < 1, and it has initial conditions (0) = zot-o = 0, then the solution to Eq. 2 is given by: IO √1 x(1) T₁ = +b+kr = 0 dt 2π dr. dt ل لها -(wat sin (wat +) and is the damped natural frequency. In Figure 2, the normalized plot of the response of this system reveals some useful information. Note that the amount of time Ta between peaks is…A:YV 36 ll 46.ll Test 3.pdf F1 F2 a F3 A d. C a F4 E Consider the following values: - F5 a = 12 m; b = 10 m; c = 10 m; d = 8 m; F1 = 6 kN; F2 = 7 kN; F3 = 8 kN; F4 = 9 kN; F5 = 10 kN; a = 30° 0 = 45° B = 60° 1] What is the resultant moment of the five forces acting on the rod about point A? a) - 111.6 kN.m b) 112.6 kN.m c) - 184.6 kN.m d) - 109.6 kN.m e) - 104.1 kN.m f) None of them 2] What is the resultant moment of the five forces acting on the rod about point B? a) -95.8 kN.m b) 42.7 kN.m c) 80.9 kN.m d) 10.9 kN.m e) 51.8 kN.m f) None of them 3] What is the resultant moment of the five forces acting on the rod about point C? a) 32.6 kN.m b) - 21.1 kN.m c) 17.4 kN.m d) - 16.5 kN.m e) 15.6 kN.m f) None of them 4] What is the resultant moment of the five forces acting on the rod about point D? a) 37.6 kN.m b) 73.8 kN.m c) 71.6 kN.m d) 52.1 kN.m e) 21.1 kN.m f) None of them 5] What is the moment of the force F2 about point E? a) 90.7 kN.m b) - 88.1 kN.m c) 54.6 kN.m d) 103.1 kN.m e)…In the Fig. 2 below, let Ki = K2 = K and ti = t=t. %3D T -T X Fig. 2 (a) Let T= 0 °C and T= 200 °C. Solve for T: and unknown rates of heat flow in term of k and t. MEC_AMO_TEM_035_02 Page 2 of 11 Finite Element Analysis (MECH 0016.1) – Spring - 2021 -Assignment 2-QP (b) Let T- 400 °C and let fs have the prescribed value f. What are the unknowns? Solve for them in term of K, t, and f.