The natural frequency of a mass-spring system.

[Blackplague]

Homework Statement

A certain spring elongates 9mm when it is suspended vertically and a block of mass M is hung on it. The natural frequency of this mass-spring system is:
a)0.014 b) 5.3Hz c) 31.8Hz d) 181.7 e) need to know M

x=9mm
mass=M

Homework Equations

I don't know any relevant equations for this. It only gives me "x" as an actually constant and not a variable like the mass has.

The Attempt at a Solution

 

[kuruman]

Draw a free body diagram of the hanging mass when it is at equilibrium and use it to get the spring constant. Once you have the spring constant, you can find the natural frequency.

 

[Blackplague]

I only get Mg as the force exerting downward. so i get:
F=kx then Mg=kx. I realize that I get .009m as the x value. but what do I do from here:
Mg=k(.009m)

 

[kuruman]

So what is k in terms of the other quantities?

 

[Blackplague]

Well since the x is .009m then i would have to say the spring constant will have to be in N/m units. right?

 

[Blackplague]

Well since the x is .009m then i would have to say the spring constant will have to be in N/m units. right?

Actually it would have to be k=(mg)/x

 

[kuruman]

And how is the natural frequency f related to all this?

 

[Blackplague]

Well:
frequency=[ (1/2pi)sqrt(k/m) ]

So: k=M(2pifrequency)^2 right?

 

[kuruman]

Actually it would have to be k=(mg)/x

So if you put in this expression for k in your expression for f, what do you get?

 

[Blackplague]

Then I get:
f= 1/(2pi) multiplied by squrt[(Mg/x)/(M)]
and if i simplify that i get:
f= 1/(2pi) multiplied by squrt [Mg/Mx]
then i get:
f= 1/(2pi) multiplied by squrt [g/x]

So i put in constants and get:
f= 1/(2pi) multiplied by squrt [9.8/.009m]
f=5.25
OH MY! The Answer is b thank you so much! I have been stuck on this problem for 2 hours. THANK YOU!