Why Does the Slope of Log T vs. Log m_eff Equal 0.5 in Oscillation Experiments?

[mattpd1]

Homework Statement

In the lab, we had a hanging spring. We proceeded to add mass to the spring (starting at 100g, and increasing by 50g until 500g) and timed the period of oscillation, T, for each mass added.

We also found the spring constant by finding the slope of Displacement vs. Mass loaded and setting it equal to g/k. Then solve for k.

Our ultimate goal was to solve for effective mass, m_eff.

My problem is one of the questions at the end of the lab... It says

"Plot T vs. m_eff on log-log paper. Why should the slope of log T vs. log m_eff equal 0.5? Consider how you derived m_eff... Why does the slope HAVE to equal 0.5?"

Homework Equations

We are given the equation:
$$T=2\pi \sqrt{\frac{m_{eff}}{k}}$$

this can be rearranged:
$$\frac{T^2k}{2\pi ^2}=m_{eff}$$

$$m_{eff}=m_{loaded}+m_{excess}$$

Theoretically:
$$m_{excess}=\frac{1}{3}m_{spring}$$

The Attempt at a Solution

I have made the graph, and it does look like a slope of 0.5, but I don't know why. Can you help? If you need any more info about the lab, let me know.

 

[quantumdude]

We are given the equation:
$$T=2\pi \sqrt{\frac{m_{eff}}{k}}$$

Take the log of both sides of that equation, and keep in mind that:

$$\sqrt{a}=a^{1/2}$$

 

[mattpd1]

like this?
$$log(T)=log(2\pi )+\frac{1}{2}log(\frac{m_{eff}}{k})$$

So, I end up with en equation for a line with a slope of 1/2? Which question does this answer though? I think this answers the question why the slope SHOULD equal 1/2, but what about the other part that asks why it MUST equal 1/2 (because of how m_eff was derived)?

By the way, thank you.