Calculating Spring and Damping Constants for a Car Suspension System

[ChazyChazLive]

Homework Statement

The suspension system of a 2200 kg automobile "sags" 14 cm when the chassis is placed on it. Also, the oscillation amplitude decreases by 55% each cycle. Estimate the values of (a) the spring constant k and (b) the damping constant b for the spring and shock absorber system of one wheel, assuming each wheel supports 550 kg.

Homework Equations

The Attempt at a Solution

I found k by making mg=-kx
k=38500 N/m
part B i found it to be 876 kg/s.
However, this answer is wrong.
I'm not sure what's wrong here but the formula I used works with everyone elses example.
e ^ (-bt / 2m) = 55/100
I found T to be 0.751 using T=2pi radical (m/k)

 

[NBAJam100]

What did you use for t when solving for b?

 

[Thaakisfox]

The oscillation damping rate is BY 55% every cycle, hence we have:

x(t+T)=x(t)-0.55\cdot x(t)=0.45\cdot x(t)

 

[ChazyChazLive]

To NBAJam100: I used the T I found
To Thaakisfox: I don't really understand what you wrote. It looks confusing. Not sure how to apply it.

 

[Thaakisfox]

Well you want to find the ratio of the amplitudes between cycles.
At time t let the position be x(t). Let the period of oscillation be T. Then the position after one cycle will be: x(t+T).

But we know the x(t) function, so:

\frac{x(t+T)}{x(t)}=e^{-bT/2m}

But it is also given that the amplitude decreases BY 55% every cycle, so:

x(t+T)=x(t)-0.55x(t)=0.45x(t) \Longrightarrow \frac{x(t+T)}{x(t)}=0.45

Combining these you will get the result...

 

[ChazyChazLive]

Oh, I see. That makes a lot of sense. Then I just plug it in and I got 1170kg/s. My confusion with the capital and lower t's got me mixed up. Thankyou very much. =]