How do i get the frequency of undamped motion?

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SUMMARY

The discussion centers on calculating the frequency of undamped motion for an aircraft's landing gear system. The maximum force exerted during landing is 7500N, with a vertical spring stiffness of 600N/mm and a damping coefficient of 38 x 10^3 Ns/m. The correct formula for frequency (fn) is derived as fn = 1/2π * √(k/m), leading to a final calculation of 4.45 Hz after correcting the spring constant to 600,000 N/m. The mass of the system is calculated to be approximately 764.53 kg based on the force of gravity.

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  • Understanding of Simple Harmonic Motion (SHM)
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  • Knowledge of differential equations
  • Ability to manipulate and solve equations involving spring constants and damping coefficients
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ThePeculiarEngineer

Homework Statement


The single wheel of an aircraft can undergo a max of 7500N at a vertical velocity of 8 m/s on landing. The vertical spring moves in SHM and has a stiffness of 600N/mm. The systems vertical damper has a damping coefficient of 38 x 10^3 Ns.m-1

Homework Equations


F=Kx
F=Kx+cv=ma
ω = √ k/m

The Attempt at a Solution


I've tried "F=Kx+cv=ma" but when i solve for "Kx+cv" I am not quite sure how to then solve for the mass. the question doesn't even give the mass
 
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Hello peculiar, :welcome:
Seems to me you have a differential equation with two initial conditions: ##x (=0)## and ##\dot x = 8 ## m/s

Show your work in detail and check if there is a given you haven't used (he said mysterically
 
BvU said:
Hello peculiar, :welcome:
Seems to me you have a differential equation with two initial conditions: ##x (=0)## and ##\dot x = 8 ## m/s

Show your work in detail and check if there is a given you haven't used (he said mysterically
Thank you for replying BvU

Ah I see, is x=0 because it is in the equilibrium position?
 
I expect you are allowed to call it that. I don't know how realistic this exercise is. 750 kg isn't much.
 
Hey, BvU how does this look?

F=7500
k=600Nmm=0.6Nm

m =F/g=7500/9.81=764.53kg

Frequency(fn) of undamped motion

fn=1/2pi *√ k/m
so...
fn=1/2pi *√ 0.6Nm/764.53Kg=4.45*10^-3 Hz
 
ThePeculiarEngineer said:
k=600Nmm=0.6Nm
Excuse me ? Could you check this thoroughly :rolleyes: ?
 
Ah sorry, here are the changes

F=7500
k=600Nmm=6*10^5Nm

m =F/g=7500/9.81=764.53kg

Frequency(fn) of undamped motion

fn=1/2pi *√ k/m
so...
fn=1/2pi *√ 6*10^5Nm/764.53Kg=4.45 Hz
 

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