Determining the mass of a holder through a Period Squared vs. Mass Added graph?

  • #1
Determining the mass of a holder through a "Period Squared vs. Mass Added" graph?

This question is for Physics Lab.

The second part of the lab dealt with simple harmonic motion. We weighed a mass holder with spring attached using a pan balance. We then added .2 kg and increased the weight to this spring and used a smart timer to time the period.

Apparently, I'm supposed to be able to confirm the mass of the holder and spring by using the graph of "Period Squared vs. Mass Added."

Why would the y-intercept give the mass of the holder? That doesn't make sense to me.

I'm supposed to perform a sample calculation using the equation [tex]T^2 = (4\pi^2/k)(m_{added} + m_{holder})[/tex]. However, if I calculate the period using the equation [tex]T = 2\pi\sqrt(M/k)[/tex], the answer is not at all similar to what I measured. If I use the value of period squared based on the experiment, then I get a negative answer.
 

Answers and Replies

  • #2
hotvette
Homework Helper
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erik-the-red said:
Why would the y-intercept give the mass of the holder? That doesn't make sense to me.
Hint: [itex]T^2 = (4\pi^2/k)(m_{added} + m_{holder})[/itex] is a linear equation of the form y = mx + b. Multiply it out.
 
  • #3
Thanks. The equation fits in with the graph in terms of axes.
 
  • #4
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Then, what does the negative value of x-intercept means?
 
  • #5
Integral
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Science Advisor
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Hummm..... It means the x intercept is negative!

Perhaps you could provide a bit more information.

Does your question have anything to do with the thread you posted in?
 
  • #6
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How could one get the error in mass here?

I have my T[tex]^{2}[/tex] formula matched up to y=ax+b.
A program Curve Expert will give me values for the coefficients a and b.
I am finding the standard deviation using Curve Expert and thus will get the error in spring constant k. (as a=4pi^2/k) This should carry through when calculating the mass of the spring in b?
 

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