Transcribed Image Text: Suppose a mass spring system with damping is modeled by the IVP y+8y’+by =0; ylos =2; y’CO) =-1 a) If you were to graph the equation of motion using a graphing that you will see on your screen. Calculator, which is the graph And why? ylt) (I) (1x) 21 2 (I1) (Iv) 2- t 2 2 b) which best deseribes this system? Explain why. mass spring (I) underdamped (II) overdanmped (m) critically damped (m) undam ped E) NONE
A mass-spring system with damping can be mathematically modeled by the following differential equation:
y” + 8y’ + λy = 0
Where y represents the displacement of the mass from its equilibrium position, y’ represents the velocity of the mass, y” represents the acceleration of the mass, and λ is a constant associated with the damping.
To solve this initial value problem, we need to find the solution to the differential equation, given the initial conditions y(0) = 2 (y-initial) and y'(0) = -1 (y-derivative initial).
a) To graph the equation of motion using a graphing calculator, we need to solve the differential equation. However, since the question does not provide a specific value for λ, we cannot determine the exact solution. Nevertheless, we can analyze the general behavior of the system based on the value of λ.
The roots of the characteristic equation associated with the differential equation are given by:
r^2 + 8r + λ = 0
To classify the behavior of the mass-spring system, we need to consider the discriminant D = 8^2 – 4λ.
If D < 0, the roots are complex, and the system is underdamped. If D = 0, the roots are real and identical, and the system is critically damped. If D > 0, the roots are real and different, and the system is either overdamped or undamped (if λ = 0).
Unfortunately, without the specific value of λ, we cannot determine the exact behavior of the system. However, we can still analyze the possible cases.
b) If we consider the three possible cases mentioned above, we can discuss the behavior of the system.
(i) Underdamped: In this case, the roots are complex and the system exhibits oscillatory behavior with a decreasing amplitude over time. The mass-spring system will go back and forth, gradually decreasing in magnitude until it reaches equilibrium.
(ii) Overdamped: In this case, the roots are real and different, and the system exhibits damped behavior without oscillation. The mass-spring system will return to equilibrium without any oscillatory motion.
(iii) Critically damped: In this case, the roots are real and identical, and the system exhibits the most rapid return to equilibrium without oscillation. The mass-spring system will reach equilibrium in the shortest amount of time.
(iv) Undamped: If λ = 0, the system becomes undamped. In this case, the roots are pure imaginary and the system exhibits periodic, undamped oscillation. The mass-spring system will oscillate indefinitely without returning to equilibrium.
Based on the information given, it is not possible to definitively determine the specific behavior of the system. However, we can make an educated guess based on the given initial conditions.
If the initial displacement (y-initial) is 2 and the initial velocity (y-derivative initial) is -1, it suggests that the system is likely underdamped. The negative velocity at t = 0 indicates that the mass is moving away from equilibrium, and the positive displacement suggests that the restoring force will pull the mass back towards equilibrium. Therefore, an underdamped system is the most reasonable assumption.
To summarize, based on the given information and assumptions, the graph of the equation of motion is most likely to represent an underdamped mass-spring system.