Dampened harmonic motion question

AI Thread Summary
The discussion focuses on solving a damped harmonic motion problem involving an automobile's suspension system. The mass of the automobile is 2000 kg, and the suspension sags 10 cm under its weight. To find the spring constant (k), the initial force equation mg = -ky is suggested, with the weight of each wheel considered at 500 kg. For the damping constant (b), the relationship between the period of damped and undamped oscillations is emphasized, noting that they are nearly identical. The conversation highlights the importance of correctly applying the equations of motion for damped oscillation.
sigmavirus
Messages
24
Reaction score
0

Homework Statement


Automobile of mass m=2000kg has a suspension that sags 10 cm when the entire auto is placed on it. The amp. of oscillation decreases 50% with every complete oscillation. Solve for (a) spring constant k and (b)the damping constant b for one wheel (assuming each wheel supports 500 kg).


Homework Equations


Fspring=-ky
Fdrag=-bv
y(t)=ymaxe-bt/2m


The Attempt at a Solution


I think i should use \SigmaF=ma for the initial force, which would be gravity ant the spring. so mg=-ky then solve for k. i am not sure how correct this is, though. then for part b, i don't know what to do about t in the equation and i was considering using a variable (so t/2 for the 2nd oscillation)
 
Physics news on Phys.org
You need to develop the damped harmonic oscillator equation of motion. Otherwise you don't know how long one complete oscillation lasts.

Though if you haven't studied this, I can tell you that the period of a damped oscillation, is very VERY close to that of the undamped oscillation. (I'm talking differences in the third or fourth digit here)

\omega_{damped}\approx \omega_0=\sqrt\frac{k}{m}
 
i solved it using a period of 2\pi. thanks for the help!
 
sigmavirus said:
i solved it using a period of 2\pi. thanks for the help!

Are you sure? Since that doesn't have the right units.

Did you mean \frac{2\pi}{\omega} ? If so, then you're correct. :)
 
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