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Spring Constant from Best Fit Line - sqrt(m) versus T

  • Thread starter icefall5
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  • #1
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Homework Statement


Construct a plot of T versus [itex]\sqrt{m}[/itex]. Fit a straight line through your data and use Eq. (3.2.3) to obtain the spring constant k from the slope, as well as an uncertainty estimate for k.

T is the period of an inertial balance with various masses of mass m. Periods are measured in seconds -- we measured 10 oscillations and divided by 10. Masses are measured in grams; mass was given for each mass.

Homework Equations


Eq. (3.2.3): [itex]T = 2 \pi \sqrt{\frac{m}{k}}[/itex]


The Attempt at a Solution


Solving this equation for k gives [itex]\sqrt{k} = \frac{2\pi \sqrt{m}}{T}[/itex]. I graphed [itex]\sqrt{m}[/itex] on the y-axis, because with rise over run for the slope this makes sense. Then doing 2*pi times that number and then squaring that number gave me a spring constant of 167207, which is incredibly off (the equation for my trendline, done using Excel, is [itex]\sqrt{m} = 65.08T - 11.779[/itex]).

What am I missing? Thank you!
 

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Answers and Replies

  • #2
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[strike]After tons upon tons upon tons of Googling, I determined that the spring constant is correct if the input mass is given in kg, not g. Dividing by 1000 gives roughly 167, which seems much more correct. Woohoo![/strike]

Not woohoo. 167 is still incredibly high for the spring constant. The ultimate goal of the lab is to determine the mass of an unknown mass, and this spring constant would make the mass 88 kg. I know that the mass was between 300 and 500 g.
 
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  • #3
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What are your values? I mean, which value is mass and which is T? For example, what is 10 in the Excel table? A mass in grams, a period in seconds or the time for 10 periods?
 
  • #4
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Sorry for not specifying this initially. The top row is the values on the x-axis: the period T in seconds. The bottom row is the values on the y-axis: [itex]\sqrt{m}[/itex] in g. The masses were 100 g, 200 g, 300 g, 400 g, and 500g.
 
  • #5
SammyS
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[strike]After tons upon tons upon tons of Googling, I determined that the spring constant is correct if the input mass is given in kg, not g. Dividing by 1000 gives roughly 167, which seems much more correct. Woohoo![/strike]

Not woohoo. 167 is still incredibly high for the spring constant. The ultimate goal of the lab is to determine the mass of an unknown mass, and this spring constant would make the mass 88 kg. I know that the mass was between 300 and 500 g.
167 Newtons/meter ≈ 0.95 lb/in .
 
  • #6
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I looked at your data and is not really following the pendulum formula with a constant k.
k seem to double when going from the small mass to the large one.
No surprise that what you get from the linear fit does not make sense.

If you plot 1/(4pi^2T^2) versus 1/m (the slope should be k) you can see that the points are not on a straight line.
 
  • #7
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the original statement asks you to plot T against √m.
This should give a straight line through the origin with gradient = 2π/√k.
Finding k should then be straight forward
 
  • #8
SammyS
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Did you measure the spring constant directly?
 

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