How do i get the frequency of undamped motion?

[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

 

[BvU]

Hello peculiar,
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

 

[ThePeculiarEngineer]

Hello peculiar,
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?

 

[BvU]

I expect you are allowed to call it that. I don't know how realistic this exercise is. 750 kg isn't much.

 

[ThePeculiarEngineer]

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

 

[BvU]

k=600Nmm=0.6Nm

Excuse me ? Could you check this thoroughly ?

 

[ThePeculiarEngineer]

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