6.5. A student hangs masses on a spring and measures the spring?s extension as a

Question

6.5. A student hangs masses on a spring and measures the spring?s extension as a function of the applied force in order to find the spring constant k. Her measurements are: There is an uncertainty of 0.2 in each measurement of the extension. The uncertainty in the masses is negligible. For a perfect spring, the extension delta L of the spring will be related to the applied force by the relation k1L = F, where F = mg, and delta L=L - L0, and L0 is the unstretched length of the spring. Use these data and the method of least squares to find the spring constant k, the unstretched length of the spring L0, and their uncertainties. Find x^2 for the fit and the associated probability.

Explanation / Answer

Using Excel, we do:

y values: the weights (N)

Thus, we need to multiply the masses by 9.8 m/s^2.

x values: the length of stretch (m)

Thus, we need to divide each measurement by 100.

As

F = k(delta L)

Using =SLOPE(y values, x values), we have k:

k = 1.578E5 N/m [SPRING CONSTANT, ANSWER]

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For the goodness of fit, we use the formula

X^2 = Sum(O - E)^2 / (sigma)^2

Using the regression line, we can get the expected values of the stretch for each data point. We actually got

F = 1.578E5L - 5696.5

---> L = [F + 5696.5]/1.578E5

Here, using technology to get sigma for the y values,

sigma = 0.01505 m

Thus, using the expected values of this equation,

X^2 = 10.608 [chi-squared, ANSWER]

THe probability of this, with 8 - 1 = 7 degrees of freedom, USING =CHIDIST(chi squared, degrees of freedom) is

p = 0.1567 [probability, ANSWER]

DONE!

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