Question

When a 82.0 kg. person climbs into an 1800 kg. car, the car's springs compress vertically 1.7 cm. What will be the frequency of vibration when the car hits a bump

Answer 1

The total mass is 1800+85= 1885 kg.
If there are 4 springs, the mass per each spring is
m = 1885/4 =4705 kg.
Hook’s law
F= - kx,
k•x =m•g,
k=m•g/x = 4705•98/0.017 =2.7•10^6 N/m
Angular frequency
ω = sqrt(k/m) = sqrt(2.7•10^6/1885) = 37.9 rad/s.
f = ω/2• π =6 Hz

Answer 2

The total mass is 1800+85= 1885 kg.
If there are 4 springs, the mass per each spring is
m = 1885/4 =471.25kg.
Hook’s law
F= - kx,
k•x =m•g,
k=m•g/x = 471,25•9.8/0.017 =2.7•10^5 N/m
Angular frequency
ω = sqrt(k/m) = sqrt(2.7•10^5/1885) = 12 rad/s.
f = ω/2• π =1.9Hz

Answer 3

that is what I did and I am still getting my problem wrong

Answer 4

Now it has to be correct...
m1 = 82 kg, m2 =1800 kg,
x =1.7 cm = 0.017 m.

It’s important that only the added weight that causes
the spring to compress 1.7 cm.
k = m1•g/x = 82•9.8/0.017=4.73•10^4 N/m,
m = m1 +m2 =1800+85= 1885 kg.
The total weight will oscillate.
ω = sqrt(k/m) =
= sqrt(4.73•10^4 /1885) = 5 rad/s.
f = ω/2• π = 5/2• π = 0.8 Hz