Hooke's Law and Centripetal Motion

[CAG0625]

Homework Statement

A 0.20 kg mass hangs vertically from a spring and an elongation below the support point of the spring of 9.50 cm is recorded. With 1.00 kg hanging on the spring, a second elongation of 12.00 cm is recorded. Calculate the spring constant k in Newtons per meter. (Note the equilibrium position is not 0.) Then at the elongation of 9.50 cm, a frequency of 300 rev/min is recorded. Then at elongation of 12.00 cm, frequency of 400 rev/min is recorded. Calculate the effective mass of the system in kilograms. (convert to SI. w=f*2pi/60 switches from rev/min to rad/sec.)

Homework Equations

W=mg
W=-kd

The Attempt at a Solution

W=(0.20kg)(980.35 cm/s^2) = 196.07N
W=(1.00kg)(980.35cm/s^2)=980.35N
196.07N=-k(0.095m)=-2063.89N/m
980.35N=-k(0.12m) = -8169N/m
Is that all correct and I do not know where to go from here.

 

[haruspex]

W=(0.20kg)(980.35 cm/s^2) = 196.07N

kg cm/s^2 is not going to give Newtons. Better to convert the cm to m, not the other way around.

196.07N=-k(0.095m)=-2063.89N/m

You are misreading the question. I'm not sure if it is misleading because it refers to 'elongation' when it means length, or maybe you are supposed to allow for the spring's own mass:

(Note the equilibrium position is not 0.)

 

[andrewkirk]

Your first two lines in 3 give you the force F for two different extensions x below the unweighted length.
You don't know x, but if you assume the unweighted length of the hanging spring is h then you have two values of x+h.
Using Hooke's Law you can get two independent equations in h and k. Then solve for h and k.