Question 3.12: The following data are given for a damped single degree of f...
The following data are given for a damped single degree of freedom mass– spring system, mass m = 5 kg, damping coefficient c = 20 N · s/m, and stiffness coefficient k = 5 × 10³ N/m. The mass is subjected to a force which depends on the velocity and can be written as F = b \dot{x} , where b is a constant. Determine the system response in the following two cases of the forcing functions; (1) F = 50 \dot{x} N, and (2) F = 400 \dot{x} N.
The equation of motion of the system is
m\ddot{x} + (c − b) \dot{x}+ kx = 0
Case 1. F = 50\dot{x}
In this case, the differential equation of motion is
5\ddot{x} + (20 − 50)\dot{x} + (5)(10³)x = 0
This is the case of negative damping in which the characteristic equation is given by
5p² − 30p + 5 × 10³ = 0The roots of the characteristic equation are
p_{1} = 3.0 + 31.48i, \quad p_{2} = 3.0 − 31.48i
The roots are complex conjugates and the solution can then be written as
x(t) = Xe^{3.0t} \sin(31.48t + \phi)
where X and \phi are constants that depend on the initial conditions. Because of the exponential growth in the solution, the system is unstable and the motion is oscillatory with amplitude that increases with time.
Case 2. F = 400\dot{x}
In this case, the differential equation of motion is given by
or
5\ddot{x} − 380\dot{x} + 5 × 10³x = 0
The coefficient of \dot{x} in this equation is also negative. The characteristic equation is
5p² − 380p + 5 × 10³ = 0which has the roots
p_{1} = 59.0713, \quad p_{2} = 16.9287
Since the roots are real and distinct, the solution can be written as
x(t) = A_{1}e^{p_{1}t} + A_{2}e^{p_{2}t} = A_{1}e^{59.0713t} + A_{2}e^{16.9287t}
where A_{1} and A_{2} are constants that depend on the initial conditions. The solution in this case is nonoscillatory exponential growth.