Effect of massive spring on static vs dynamic experiment

[Tulzz]

I've been reading about this but I couldn't find a conclusive answer.
I have to measure the spring constant (k) using a vertical spring-mass (M) configuration and using two different methods:

- static method: I calculate elongation vs weight and find a linear fit of the data (the slope is K)
- dynamic method: I measure the period and calculate K using K=w²*m

I don't know what mass i need to use to calculate K.

In the static measure, the weight i consider is: w = (springmass/2 + M)*g
In the dynamic method I am supposed to use: m = M + (springmass)/3

Am I missing something? I don't fully understand the static measure correction of the mass.
sorry for my english and for the format (first post)
Thanks!

 

[J Hann]

A small portion of the spring has mass = m / L dx
This portion of the spring has velocity = x * v / L
Now write dE and integrate x from 0 to L giving 1/2 (m/3) v^2

 

[Tulzz]

I guess i didn't express myself well. I know I have to use an effective mass when measuring K using a dynamic measurement technique. I don't fully understand the mass correction in the static case and why is it springmass/2 instead of springmass/3 as in the dynamic case.
Another question i have is if the effective mass correction still stands when considering viscous effects. I guess it does, because it just adds an external force that doesn't affect the kinetic energy in the lagrangian formulation. Am i right?
Thanks!

 

[J Hann]

I guess i didn't express myself well. I know I have to use an effective mass when measuring K using a dynamic measurement technique. I don't fully understand the mass correction in the static case and why is it springmass/2 instead of springmass/3 as in the dynamic case.
Another question i have is if the effective mass correction still stands when considering viscous effects. I guess it does, because it just adds an external force that doesn't affect the kinetic energy in the lagrangian formulation. Am i right?
Thanks!

I've never used this but perhaps you could write
dF = k x dm = k m x dx / L since dm = m dx / L
Integrating that will give F = k m L / 2 for the restoring force due to the mass of the spring.

 

[haruspex]

I've never used this but perhaps you could write
dF = k x dm = k m x dx / L since dm = m dx / L
Integrating that will give F = k m L / 2 for the restoring force due to the mass of the spring.

That works, but to clarify, x here is the distance that it would be from the top of the spring to a point in the spring, were the spring in its relaxed state (so not the actual distance).
To see why the dynamic effective mass is different, consider how fast the different parts of the spring move. Which part moves fastest? Where in the (compressed) spring is the linear density greatest?

 

[Tulzz]

Thanks a lot! I have one final questión: if I am studying the damped oscilator dynamics, do i still have to use the effective mass correction??

 

[haruspex]

Thanks a lot! I have one final questión: if I am studying the damped oscilator dynamics, do i still have to use the effective mass correction??

I'm fairly certain you should.