Why Plot T² Against 1/K for a Straight Line Graph in Oscillation Studies?

[Peter G.]

So, I have to assess the relationship between the Constant of a Spring and the Time Period of Oscillations.

I have researched the equation that links the two and I know that to have a straight line graph I need to plot:

T² against 1/K

However, I do not know why; how I get to that conclusion.

I have read about log-log graphs and played around with the concept to see if it would yield the plot I mentioned above. I tried plotting for the log of T against the log of K and I did get a straight line plot. I checked if my gradient was equal to -0.5 and my y-intercept was equal to (log 2∏-0.5 log m) and they were, but I still do not understand why I have to plot T² against 1/K.

Can anyone please shed some light onto this please?

Thanks in advance!

 

[rock.freak667]

You are asked to plot that because of the derivation of the periodic time formula for a spring oscillation.

The final equation is

$$T = 2 \pi \sqrt{\frac{m}{k}}$$

If you square both sides you will get


$$T^2 = \frac{4 \pi ^2 m}{k}$$

Which is the form Y=MX where Y = T² and X = 1/k.

Thus plotting Y vs X i.e. T² vs. 1/k will give you a straight line. So you can get the mass oscillating if you desire from the gradient of the graph you plotted.

 

[Chestermiller]

I like your idea of using a log-log plot much better. I have over 40 years of experience as a practicing engineer, and that is the way I would do it. Congratulations on thinking out of the box.

Chet