Mass vs. Average Period Squared

AI Thread Summary
The discussion centers on plotting average period squared against mass, determining which variable belongs on the x-axis and y-axis. It is clarified that mass should be the independent variable (x-axis) since the period depends on mass. The user encounters discrepancies in calculating the spring constant (k) from two different graphs, with one yielding a negative slope, which is incorrect. The importance of using the correct values for period and frequency is emphasized, as well as ensuring proper units for calculations. The conversation concludes with a confirmation that energy stored in a spring is measured in Joules, provided the spring constant and displacement are correctly defined.
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I have to plot the average period squared vs the mass, now my question is which goes to the x-axis and which goes to the y?
 
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X: Independent variable
Y: Dependent variable.

So, did you change mass to find different periods or did you change the periods of find different masses?
 
i added mass to see the variation in oscillation periods
 
That means that the period DEPENDS on the mass, making mass the independent variable and period the dependent variable.

Remember:
X is what you change
Y is what you measure
 
for example to calculate the k constant i believe the formula is

k = (4pi²)/slope

but the slope of the graph has to be t² and kilograms correct? i have the values in t² and the mass in grams
 
If you want the spring constant to come out as Newtons per meter then yes the mass has to be in kilograms.
 
Is it supposed to give a huge difference from the elongation method?

My assignment says i have to do 2 tables where one measures force against displacement and than this one above

for force vs displacement i get k as 34.6 while when i do it with the formula i got above i get k as-8.77, is this normal?
 
You can't calculate the slope the same way for force vs. displacement. Remember, F = kx, so when you graph that equation you are going to end up with a line anyways. What do you notice about the slope?
 
The instructions say that the slope of the graph force vs displacement which i got a slope of 34.6 N/m and when i did the slope calculation of the oscillation graph i got -4.5 for the slope where i imputed in the formula

k = (4pi²)/slope

which gave me -8.77 N/m
 
  • #10
If you plot displacement against weight, yes. If you plot it against mass, the k = m * 9.8
 
  • #11
The first graph i have
y axis - Force (N)
x axis - displacement (m)

Where i got the slope of 34.6 which the instructions say its the k constant

the second graph
y axis - t²
x axis - mass (kg)

where i got slope of -4.5

which i imputed here

k = (4pi²)/slope

which gave me -8.77 N/m

Now i was supposed to calculate the difference between the two

with these values i get a difference of -125% which i think is a huge difference
 
  • #12
You swapped X and Y on the first graph. You change force and measure displacement.

I'm not quite sure how you ended up with a negative slope, there should be a strong positive correlation between mass and period squared. Besides, the spring constant can't be negative. Can you list some data points?
 
  • #13
First graph

x f(N)
0.98
1.96
2.94
3.92

y displacement (m)
0.03
0.06
0.088
0.115Second graph

x mass(kg)
.3
.4
.5
.6

y t²(s)
2.72
2.13
1.66
1.41

i used this website to help input data to get info
http://www.shodor.com/unchem/math/lls/leastsq.html
 
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  • #14
Ok the first graph seems fine but I can't understand why there's a negative slope on the second one. Can you make sure you're using periods and not frequencies? That's the only thing I can think of that would decrease like that.
 
  • #15
we had to count how many secs it took to do 25 oscillations

.3kg - took 15.19 sec
.4kg - 17.16 sec
.5kg - 19.38 sec
.6kg- 21.09 sec

than i took the average period T (sec)

divided 25 (oscillations) by the secs it took and squared them

25/15.19 = 1.65 = t² = 2.72
25/17.16 = 1.46 = t² = 2.13
25/19.38 = 1.29 = t² = 1.66
25/21.09 = 1.19 = t² = 1.41and that's how i got my values.

in a way the graph makes sense, the higher the mass the less time it takes to do an oscillation but i know o have something wrong on this graph
 
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  • #16
Ah, that's it. An oscillation divided by time is a frequency. You want to use the reciprocal of that (period). I did that for the excel chart I made of your data and got both measurements of k almost identical.
 
  • #17
So i am supposed to divide these values by one and than square them, is that what i did wrong?

25/15.19 = 1.65 = t² = 2.72
25/17.16 = 1.46 = t² = 2.13
25/19.38 = 1.29 = t² = 1.66
25/21.09 = 1.19 = t² = 1.41
 
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  • #18
Or just divide the values you already have by one since you already squared them. If you don't understand why, frequency is how many oscillations, revolutions, etc. the object makes per second, thus having the unit of hertz. Period is how many seconds it takes the object to make one revolution.

The frequency of the Earth is 1/24 hours = 0.04 per hour (meaning that it makes 1/24 of a rotation per hour) and its period is 24 hours (meaning that it takes 24 hours to make one revolution.
 
  • #19
Your a life saver, i have been around for so long that i was getting mad lol I just have 1 more question i had to calculate the energy stored in one of the spring, the units of the energy stored is in Joules correct?
 
  • #20
Yes, so long as the spring constant is N/m and the displacement is m.
 
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