Suspension strut design (diff eq and damping systems)

[uiucbro]

Homework Statement

The mass of a car that acts on one wheel is 100 kg. The
elasticity (spring) constant in the suspension system of that wheel is k= 10^4 N/m:
Design the strut (find the friction/resistance constant c)
such that any vertical motion of the wheel (set up for example by going
over a bump or pothole on the road) will die out in the shortest amount
of time.

Homework Equations

mx''+cx'+kx=F(t) where F is some force

The Attempt at a Solution

Of the three different damping systems (under damped, critically damped, over damped), I believe critically damped allows for the shortest amount of time until the system is back to normal. Critically damped is when c^2-4*m*k=0, so therefore c=2000. I fear that there's more to this as this question is worth a lot of points. Is there anything I missed?

 

[mighty2000]

Thank you for your question. Your understanding of critically damped systems is correct. However, there are a few more considerations to keep in mind when designing the strut for the given car.

Firstly, the mass of the car (100 kg) is acting on one wheel, but we must also consider the weight distribution of the car and the position of this particular wheel. This will affect the overall force acting on the wheel and, consequently, the required value for c.

Secondly, the road conditions and the type of bumps or potholes that the car may encounter should also be taken into account. A higher c value may be needed for rougher roads, while a lower c value may suffice for smoother roads.

Lastly, it is important to consider the material and design of the strut itself. A well-designed and high-quality strut can have a significant impact on the overall damping of the system, and thus the value of c.

In summary, while your calculation for the value of c is correct, it is important to consider other factors such as weight distribution, road conditions, and strut design in order to accurately determine the optimal value for c in this scenario.