Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Recently in my AP Physics class I did a lab in which I measured k for a spring by setting up an oscillating system with it, and timing the period, repeating for different masses. Since $T=2\pi\sqrt{\frac mk}$, we were supposed to relate $T^2=\frac{4\pi^2}km=Cm$ for some $C$. This is a proportionality equation in $T^2$ and $m$, which we were to plot on a graph and find the slope $C=\frac{4\pi^2}k$, from which we can find $k=\frac C{4\pi^2}$.

Now my lab partner and I found the $k$ in two different ways; while I simply averaged $k_i={C_i\over4\pi^2}$ for each trial, he used statistics software to find the equation $\hat{T}^2\approx C_am+\beta$, where $\beta\approx0$ and $\bar{k}={C_a\over4\pi^2}$. He claimed that the $\beta$ term helped to eliminate error, leading to a better $k$. We recalculated $k$ by direct use of Hooke's Law and found that the percent difference was lower in my partner's case, leading me to believe that he may be correct in his method.

So, my question is, physics.stackexchange: How does using the linreg analysis on the data produce better results? Or, if this is a fluke of data and my results should have been better, why shouldn't it?

Thank you in advance.

share|improve this question
1  
Hi actorclavilis, and welcome to Physics Stack Exchange! Just wanted to say that I really like this question :-) –  David Z Dec 10 '11 at 1:46
add comment

1 Answer

up vote 6 down vote accepted

The main problem is to determine what corresponds to zero mass of the harmonic oscillator. Remember that a fraction of the spring mass also participates in the motion. By introducing an intercept $\beta$, your friend takes into account that the true zero of the mass parameter $m$ may be shifted from what you think it is. So an affine model $T^2=Cm+\beta$ is more precise than a linear model $T^2=Cm$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.