Damped Vibrations and Characteristic Equations

The video tutorial explains free damped vibration by modeling a mass-spring system with a second-order differential equation. It covers calculating the mass, damping coefficient, and spring constant, and solving the equation using characteristic equations. Initial conditions are applied to find the particular solution for the displacement function.

What is the weight of the mass used in the free damped vibration example?

• 6 pounds

• 8 pounds

• 10 pounds

• 12 pounds

In the context of the video, what is the primary difference between damped and undamped vibrations?

• Damped vibrations have no spring constant

• Undamped vibrations have a gamma term

• Undamped vibrations have no mass

• Damped vibrations include a gamma term

How is the mass calculated in the example?

• By multiplying weight and gravity

• By subtracting gravity from weight

• By adding weight and gravity

• By dividing weight by gravity

What is the damping coefficient gamma in the example?

• 4 pound seconds per foot

• 5 pound seconds per foot

• 2 pound seconds per foot

• 3 pound seconds per foot

What is the spring constant k in the example?

• 14 pounds per foot

• 12 pounds per foot

• 16 pounds per foot

• 18 pounds per foot

What is the form of the characteristic equation used to find the general solution?

• r^2 + 16r + 64 = 0

• r^2 + 14r + 64 = 0

• r^2 + 12r + 64 = 0

• r^2 + 10r + 64 = 0

What are the roots of the characteristic equation in the example?

• r = -7 ± 2√7i

• r = -6 ± 2√7i

• r = -8 ± 2√7i

• r = -5 ± 2√7i

What is the value of c1 in the particular solution?

• 1

• 1/4

• 1/3

• 1/2

What is the value of c2 in the particular solution?

• 3/(4√7)

• 3/7

• 3/(2√7)

• 3/(√7)

What is the final form of the displacement function u(t)?

• u(t) = c1 e^(-6t) cos(2√7t) + c2 e^(-6t) sin(2√7t)

• u(t) = c1 e^(-7t) cos(2√7t) + c2 e^(-7t) sin(2√7t)

• u(t) = c1 e^(-6t) cos(3√7t) + c2 e^(-6t) sin(3√7t)

• u(t) = c1 e^(-5t) cos(2√7t) + c2 e^(-5t) sin(2√7t)