Car's vibration after being compressed, and released on springs

AI Thread Summary
The discussion focuses on calculating the frequency of a car's vibration after it is compressed by the weight of seven people. The car drops 0.87 cm when the average mass of the passengers is considered, leading to the calculation of the spring constant (k) using the formula k = mg/x. The initial frequency (f1) is calculated as 5.34 Hz, but an error occurs when trying to find the frequency (f2) for the car's empty mass of 1100 kg. The user realizes they neglected to account for the total mass of the passengers in their calculations. The thread concludes with a request for assistance in correcting the frequency calculation.
rinarez7
Messages
27
Reaction score
0
1. When seven people whose average mass is 83 kg sit down in a car they find that the car drops down 0.87cm lower on its springs. Then they get out of the car and bounce it up and down. The acceleration of gravity os 9.8m/s^2.
What is the frequency of the car's vibration if its mass (empty) is 1100kg? Answer in units of Hz.



2. f= 1/2pi (sqrt k/m)

k=F/x= mg/x

m1f1^2 = m2f2^2




3. First I solved for k: k= mg/ x = 1183 kg (9.8) / .0087m= 1332574.713 N/m

Then I solved for f1= 1/2pi (sqrt k/m)= 1/2pi (sqrt (1332574.713/ 1183kg)= 5.34 Hz

Now, when I tried to use these numbers in m1f1^2= m2f2^2 and solve for f2
1183kg(5.34)^2 = 1100kg (f2)^2
f2= 5.539 Hz
But this is wrong. Help!
 
Physics news on Phys.org
Hi,

F=kx > G=kx > (83*7)g = k *.0087

f1= 1/2pi (sqrt k/m)= 1/2pi(sqrt(k/1100))
 
Thank you! I missed the 83 kg * 7 people.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top