Hooke' Law questions.

  • Thread starter Splint
  • Start date
  • #1
10
0
Hi,

I've been doing an experiment and I'm now putting the data into a spreadsheet.

I started by hanging a spring vertically and adding weights noting the displacement from the unweighted equilibrium point.

I put into my spreadsheet a column for finding K using -k=F/delta x. the thing is all of the values for K are different. I was under the impression that the K constant was meant to remain the same unless the spring stretched beyond its elastic limit or was cut short..

Some numbers:

400g added in total, delta x = 3.75cm so .4Kg*9.8=3.92N and 3.75cm*.01=.0375m so F/n = 104.5Nm
650g added in total, delta x = 11.75cm so .65Kg*9.8=6.37N and 11.75cm*.01=.1175m so F/n = 54.2Nm
1000g added in total, delta x = 23.0cm so 1Kg*9.8=9.8N and 23.0cm*.01=.23m so F/n = 42.6Nm

Is this correct?

Next related question:

K = Gd^4 / 8nD^3 is the spring design equation.

K=?Nm
G= the shear modulus of the spring material = 79.3Gpa
d=wire diameter = 0.96mm = 0.00096m
n = number of active coils in the spring = 204
D = mean coil diameter = 10.16mm = 0.01016m

K= (79.3(0.00096)^4) / (8(204)(0.01016)^3) = 3.935 E-8 Nm which is nothing like the figures above and it is one single value.

BTW, the weight of the spring is insignificant compared to the weights added.

Just to add to the confusion I used the same spring and timed oscillations with various weights attached and used K=4Pi^2mass/time^2

E.g.
mass = 650g*.001*9.8 = 6.37N
t = 30 oscillations in 27.9 seconds = .93s
K = 4Pi^2(6.37) / 0.93^2 = 290Nm which is nowhere near either of the other values.

Any idea where I'm going wrong?

Thanks
Splint
 

Answers and Replies

  • #2
327
2
Isn't the equation [tex]F = - k \Delta x[/tex] valid only for a massless spring?

I suggest you add a constant force to account for the mass of the spring. That is, the total force [tex]F[/tex] is equal to [tex]F = F_{ext} + F_{spring}[/tex] where [tex]F_{ext}[/tex] is the force you apply externally and [tex]F_{spring}[/tex] is the force that gets applied to the spring because of it's own mass. You can find the unknown [tex]F_{spring}[/tex] using two of the three equations. See if you get a more consistent value for [tex]k[/tex].

And by the way, isn't the units of the spring constant N/m ?

In the spring design equation you forgot to convert the value for the shear modulus from GPa to Pa.

In the third equation, the mass should be in kg and not N. You shouldn't have multiplied 0.65 kg by the gravitational constant.
 
  • #3
10
0
Thanks, that was helpful.
 

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