Distance Lifted by Car Suspensions in Earthquake

[lany]

Homework Statement

4people with uniform mass= 72.4kg
car mass= 1130kg
earthquake strikes,car is stationary.
vertical oscillations make car bonuce on suspensions.
when frequency= 1.8 Hz car exhibits maximum Amplitude of vibration.
Earthquake over,4 people leave the car quickly,what distance does the cars undamaged suspensions lift the car body as the people get out?

Homework Equations

sorry I am new to this i don't understand what this section wants

The Attempt at a Solution

i took this as a forced oscillation and as a simple harmonic oscillation.
total mass = 1419.6 kg
Velcoity = 3.6*pi*Amax (Amax=max amplitude)
this comes from Vel=omega*Amax
omega=2*pi*Frequency

Acceleration = 2Amax/period
i don't know i thought this one up...
as max displacement= Amax + A max
and time taken => period = 1/frequency...

Acc= 3.6 Amax

x=displacement of car body by suspension...

x=Asin(omega*t)
x'=A*omega*sin(omega*t)
x''= - omega^2 *(x)

...
3,6Amax = -(2*pi*1.8)^2 *x

=> x= - 0.028*Amax

my head is exploding someone please give me some hint...i have an awful feeling that this could be a lot easier...

fiendishly difficult differential equations are suspected but I'm not sure how to relate everything...its kinda late here in Ireland!please have some pity...

 

[lany]

I just thought that since they left the car quickly then t must be tending towards zero...?
right track?

 

[baba89]

First of all, great job on attempting to solve this problem! It is definitely a challenging one.

To start off, you are correct in assuming that this situation can be modeled as a forced harmonic oscillation. This means that the car's suspension is being forced to oscillate at a certain frequency by the earthquake.

Next, we need to find the amplitude of the car's oscillation. This can be done by using the given information about the car's mass and the frequency of oscillation. We can use the equation A = F/mω^2, where A is the amplitude, F is the force applied (in this case, the force of the earthquake), m is the mass of the car, and ω is the angular frequency (2πf).

Plugging in the given values, we get A = 1130*9.8/(72.4*2π*1.8)^2 = 0.0269 m. This is the maximum displacement of the car's suspension during the earthquake.

Now, when the four people leave the car, the car's mass decreases from 1419.6 kg to 1130 kg. This means that the amplitude of oscillation also decreases, since it is directly proportional to the mass. The new amplitude can be calculated using the same equation as before, but with the new mass of the car.

A' = 1130*9.8/(72.4*2π*1.8)^2 = 0.0215 m

Finally, we can use the equation x = A*sin(ωt) to find the distance lifted by the car's suspension as the people leave the car. We can assume that the people leave the car at time t = 0, so we have:

x = 0.0215*sin(2π*1.8*0) = 0 meters

This means that the car's suspension does not lift the car at all as the people leave, since the amplitude of oscillation is now smaller than the initial displacement of the car's suspension. This makes sense, since the people leaving the car would reduce the mass and therefore the amplitude of oscillation.

I hope this helps! Keep up the good work.