Help with formula

[imastudent615]

Hey-
For this year's science fair, my partner and I did a variation of the Simple Harmonic Motion in a Spring-Mass System project. We have conducted the experiment, but the formula is confusing us. It appears that the formula we should use is m = k( T^2/4pi^2)- me (equation 3 in the directions). Is that correct? Assuming that it is, a couple of the variables are giving us trouble. First of all, what is the equivalent mass, me, of the spring (what does that mean and how do we find it?)?
I was reading the instructions on the site, and it made sense until i got to the graph part. Do you think you could explain that (basically finding k and me)?
Also the mass (added to the spring (m)) is only the weights right? not the weights and the spring?
Thank you so much in advance for clearing up my confusion

[kgudger]

Hello and welcome to the forums.

First, the effective mass is:

real springs contribute some of their own weight to the load ... the total mass pulling down on the spring is actually comprised of two masses, the added weight, m, plus a fraction of the mass of the spring, which we will call the mass equivalent of the spring, me

The equation 3 you mentioned is a way for us to find the effective mass of the spring. Since we can't measure this directly, we use equation 3 and a graph to find it. We'll use the simple equation for a line, y = mx + b. This holds for any straight line, and hopefully your results will give us a straight line.

In the case of equation 3, Y is the added mass, X is T^2/4pi^2, k, the spring constant, will be the slope of the line on this graph, and 'b' (tada!) is the effective mass of the spring (and the y intercept). (Note: you might want to use a spreadsheet or a graphing calculator to generate your graph.)

Let us know how your results look!
Keith

[imastudent615]

I had another question.
In the example charts, it states the Load (mass added to spring) in g is 24 and 30, but then when it converts these numbers into kilograms, it becomes 0.007 and .028 respectively. Wouldn't you just divide the 24 and 30 by 1000? Where do the .007 and .028 come from?

[Craig_Bridge]

Are the "example charts" available online somewhere that we can take a look at? There is either a misprint on the numbers or units or you are misinterpreting the charts. A "Load" can be expressed in force units such as Newtons in which case there is a linear unit conversion involved; however, there is a 20% difference between the mass in grams and a 400% difference in the numbers (.007 and 0.028). 20 squared is 400, so there is some square law factor involved. My best guess is the chart is for some physics property that involves a square law so the most likely explaination is that you are misinterpreting the chart.

[ChrisG]

Here is the web page that describes this project and has the chart in question:
https://www.sciencebuddies.org/science- ... p064.shtml

[Craig_Bridge]

After reviewing the chart under step 5 Column A, I conclude that there is either a misprint or a horrible choice of arbitrary data that is confusing because the data shown in step 4 is inconsistent with data shown in step 5. If column A is supposed to correspond to the mass added to the spring, "m" from equation 2 and 3, and the data is supposed to be consistent with step 4, the entries should be 0.024 and 0.030. I will call this to the attention of the staff and they will get back with the orignal author.