Simple Harmonic Motion: vertical springs - graphing period^2 & mass

crankine

1. The problem statement, all variables and given/known data

Using experimental values for mass and period of oscillations, I'm trying to graph period[tex]^{2}[/tex] (y-axis) against mass (x-axis) to get a gradient = 4(pi[tex]^{2}[/tex])/k so I can find k (spring constant).

The problem is that when I use each pair of results in the equation below, I get the right answer for k, but when I graph it and try to use the gradient its wrong!!!! I'm pretty confident I've substitued correctly etc.

If you want the exact data then I would put it up but I don't know how to put tables.

2. Relevant equations
T = 2pi√(mass/k):

T[tex]^{2}[/tex] = 4(pi[tex]^{2}[/tex])(mass/k) -> compared to y=mx, m=4(pi[tex]^{2}[/tex])/k
k = 4mass(pi[tex]^{2}[/tex])/(T[tex]^{2}[/tex]) -> substituting each pair of T and mass here I get the right answer


3. The attempt at a solution

I've tried swapping the axes and graphing T against m[tex]^{1/2}[/tex] and so on but nothing has got me any closer to the right answer.

Redbelly98

Welcome to Physics Forums.

That's weird, your equations and method all look correct.

Here's what I suggest: pick any 2 data points, and post them here. Also post the slope you calculate based on just those 2 points.

Then we can both look at your work and try to figure out what's going on.

crankine

(the right value for k is about 4 or 5)

mass 0.06
period 0.7
period^2 0.49
calculated value for k 4.834091952

mass 0.1
period 0.8
period^2 0.64
calculated value for k 6.168502751

graph of period^2 (y) and mass (x) gives gradient 1.2558, value for k from this is 31.4...

halp! :) thankyou

Redbelly98

Thanks for posting the data. Here are questions and comments:

• What are the units on the masses?
  
  
• It's difficult to get an accurate calculation of the slope from the data you give here, because the two periods are so close together (0.7 s and 0.8 s). Do you have 2 data points where the periods are not so close to each other?
  
  
• If we do use those 2 data points to calculate a slope, we get
  

  (0.64 - 0.49) / (0.1 - 0.06) = ____ ?

  (Hint: it's not 1.2558)

crankine

Thanks :)
All SI units so kg, seconds etc.
I didn't calculate the slope myself, I put it into excel - I think it made the points (0.64,0.49) and (0.1,0.06) i've sorted that so thats okay now.

BUT...... now the gradient = 3.75 now... so k = 10.5. This is still double what it should be, and double what I calculated!! argh.

Redbelly98

This is looking better, at least the math is correct for those two points. You could easily be a factor of two high or low when the two periods, 0.7s and 0.8s, are so close to one another.

How about 2 data points that are not so close together in period? That would be a lot closer to the actual slope.

crankine

No I don't I'm afraid as I only used 3 masses, so the period was 0.7, 0.75 and 0.8.
I think you're right though because when I added a theoretical higher point that fitted with the equation the gradient improved, so I'll go back and repeat the experiment with smaller and greater masses.
Thanks for your help :)