k F X m HE 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: d'x(t) c dx(t) k + +x(t) == F(t) dt² m dt m m

Given 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/m

Q2 The angular position c of a mass is controlled by a servo system through a referen signal °r. The moment of inertia of moving parts referred to the load shaft, J, is 150 kebo and damping torque coefficient referred to the load shafi(B, is 4.5 x 103 Nw-m fraden The torque developed by the motor at the load s 7.2 x 10* Nw-m per radian of error (a)Obtain the response of the system to a step input of 1 rad and determine the peak time, peak overshoot and frequency of transient oscillations. Also find the steady state error for à constant angular velocity of 1 revolution / minute. (b)If a steady torque of 1000 Nwm is applied at the load shaft, determine the steady Kan state error. 5(J5+ B) 03 For the system shown, apply a proportional controller to move the time constant to a sixth of its open loop value. R(s) + E(s) U(s) Kp 'Y(s) s +0.333 10

Consider the system presented in the following figure K, = 500N / m K, = 2000N / m M = 1Kg %3D X, = 20mm %3D Vo = Omm/s %3D The stiffness of the equivalent spring (in N/m) is equal to Choose... The natural frequency in rad/s is equal to The amplitude of vibration (in m) is Choose... Choose... The phase shift of the vibration is Choose...

For the following water liquid level system, 9mi(t) is the mass flow rate input deviation to the system and the restrictions can be considered linear resistances about the equilibrium: 9mi h2 Ro R1 a) Using basic fluids principles, develop the Dynamic Equations for the deviation of the two liquid level heights from their equilibrium height as a function of the input mass flow rate, Imi-

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 - Explain Van-Neumann stability by finite difference schemes for this equations a- Laplace equation. J²u dyz (x, y) = 0 J²u dx2(x, y) + b- poison equation. J²u J²u { (x, y) + əy² (x, y) = f (x, y) əx²

4. Consider a mass of 500 g placed at the end of a spring with stiffness constant 100 N/m and hanging at rest. The mass is displaced downward by 1.0 cm and released from rest. When the motion sets in, a time-varying force F(t) = cos³ 2t is applied to the spring-mass system. Solve the nonhomogeneous 2nd order differential equation representing the motion. That is, solve for x(t). [Note: Take the downward direction as the negative x-axis.]

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…

Equation of motion of a suspension system is given as: Mä(t) + Cx(t) + ax² (t) + bx(t) = F(t), where the spring force is given with a non-linear function as K(x) = ax²(t) + bx(t). %3D a. Find the linearized equation of motion of the system for the motion that it makes around steady state equilibrium point x, under the effect of constant F, force. b. Find the natural frequency and damping ratio of the linearized system. - c. Find the step response of the system ( Numerical values: a=2, b=5, M=1kg, C=3Ns/m, Fo=1N, xo=0.05m