Recent content by 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...

Well: frequency=[ (1/2pi)sqrt(k/m) ] So: k=M(2pifrequency)^2 right?

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

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

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)

I had the same question. Can you please go step by step on how you acquired the answer, because I don't understand how to "final" the different liquids and getting a gauge pressure.

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...