That is asimple Question
The first one in hich the shock absorbers are not working
It is asituation similar to that of without any damping
Now omega = sqrt(k/m)
therefore the motion of the spring represents the motion of the car
therfore the the height h is given by the following equation:
x =Acos(omega*t)
this a cosine function
u can sketch it easily
Now what does this mean. It means that the spring will oscillate for ever without any stop in its oscillation about its mean position
Now the problem is that the cosine function has some negative values. The cosine function has it most negative value at x = - A. This means that the car is returning to its original postion. Below I am attaching the both pieces of the graphs. Theya are not neatly drawn
h = 2A (maximum height reached by the car)
It will continue oscillating forever because there is no damping force
No for the second case where the shock absorber is functioning normally. For passenger comfort the oscillation is critically damped or slightly underdamped
again the position of the spring determines the position of the car
when the situation is critcally damped the system it comes to its equlibrium psoition without any oscillation
since this is not the case asked
we will assume the oscilation to ve underdamped
x =Ae^(-b/2m)t *(omega_dt)
This will represent an oscillation that is underdamped
This piece of function can be drawm easily using advanced calculus concepts
But I recommend u to go through your book.
the above equation gives the motion about the mean point. Now take a look of the graph from your book and there will be certain negative value of x
Choose the most negative value of x and dram a line perpendicualr to the y axis. Name the line as your reference line
now u have all positive values of y coordinates . this is the height reached by the car as a function of time
Please click on the picture to view it
The second diagram represents the fist case and the first represents the second case
-----Hope that Helps