Engineering Vibrations (W21)
Due date: 03/04/2021
Problem 1 (10 points) Consider the two degrees of freedom mechanical system shown
in figure 1 such that k1 = k2 = k and m2 = 10m1 = 10m.
1. Derive the normal modes.
2. Find the complete solution of the system. Why is it difficult to calculate the con-
stants in the solution using the initial conditions? Explain.
3. Discuss the relationship between the frequencies and the normal modes ampli-
tudes. Provide a physical interpretation.
Figure 1: Two degrees of freedom System
Problem 2 (10 points) Consider the weakly coupled mechanical system shown in figure
2. Let k be the stiffness of the spring and m1 = m2 = m. Given that the initial conditions
θ1(0) = 0
θ˙1(0) = A
θ2(0) = 0
θ˙2(0) = 0
1. Compute the complete solution of the system linearized around θ1 = θ2 = 0.
2. Given the numerical values in the following table, plot θ1(t) and θ2(t) on the same
figure for 0 < t < 100s. Give a physical interpretation of what is happening.
Parameter Numerical Value Unit
l 1 [m]
d 0.5 [m]
m 3 [Kg]
g 9.8 [m/s2]
A 5 [rad/s]
k 10 [N/m]
Figure 2: Weakly Coupled System
Problem 3 (10 Points)
1. Simulate the Nonlinear model of problem 2 with the same initial conditions. Plot
θ1(t) and θ2(t) on the same figure and compare with the results of problem 2.
2. Simulate the Nonlinear model of problem 2 with the following initial conditions.
θ1(0) = −pi/6
θ˙1(0) = pi
θ2(0) = pi/6
θ˙2(0) = pi
Plot θ1(t) and θ2(t) on the same figure and compare with part 1.
Problem 4 (Extra credit-10 points) Consider the two degrees of freedom mechanical
system shown in figure 3. A rod of mass m2, uniformly distributed along its length
2l, is hinged at its center of mass. One of its two ends is attached to spring 1 whose
stiffness constant is k1. Also, spring 1 is attached to a body with mass m1 which in turn
is attached, via spring 2 whose stiffness constant is k2, to a wall. The unstretched length
of the two springs is l/2. Clearly, this is a two degrees of freedom system having x (the
stretched length of spring 2) and θ (the angle between the rod and the vertical line) as
1. Show that the equations of motion are given by
m1x¨ = f1(x, θ), f1(x, θ) = −(k1 + k2)x− k1l(1−
3− 2(cosθ − sinθ))
θ¨ = f2(x, θ), f2(x, θ) = −k1l(l − x+ l√
3− 2(cosθ − sinθ) )(sinθ + cosθ)
Hint: Use the law of cosines and law of sines.
2. Linearize the two functions f1(x, θ) and f2(x, θ) around (x¯ = 0, θ¯ = 0).
3. Put the linearized equations of motion in matrix form and give the mass matrix
M and the stiffness matrix K. Now, let k1 = k2 = k and m1 = m2 = m. Find the
solution for the normal modes using the matrix method.
Note: This device is in the horizontal plane, so the force of gravity is not a factor in this
Figure 3: Two degrees of freedom System