Question

1. using the best - fit line, determine the spring constant of the spring. show all calculations, including the equation and substitution with units.

Answer:

To solve for the spring constant using the best - fit line, we rely on Hooke's Law. Hooke's Law is given by the equation \(F = kx\), where \(F\) is the force applied to the spring, \(k\) is the spring constant, and \(x\) is the displacement of the spring from its equilibrium position. If we plot a graph of force (\(F\)) versus displacement (\(x\)), the slope of the best - fit line to this graph is equal to the spring constant \(k\).

Step 1: Recall Hooke's Law and the relationship to the best - fit line

From Hooke's Law \(F=kx\), we can rewrite it in the form of a linear equation \(y = mx + b\). In the context of force - displacement data, if we let \(y = F\) (the dependent variable, force) and \(x\) be the displacement (the independent variable), then the equation becomes \(F=kx + 0\) (assuming no initial force when \(x = 0\)). So, the slope \(m\) of the line \(F\) vs \(x\) is equal to the spring constant \(k\), i.e., \(k=\text{slope of }F - x\text{ graph}\)

Step 2: Calculate the slope of the best - fit line

The formula for the slope \(m\) between two points \((x_1,y_1)\) and \((x_2,y_2)\) on a line is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). If we have a set of data points \((x_i,F_i)\) and we have found the best - fit line for these points, we can take two points on the best - fit line. Let's assume that from the best - fit line, we have two points \((x_1,F_1)\) and \((x_2,F_2)\)

For example, suppose the best - fit line passes through the points \((x_1 = 0.1\space m,F_1=1\space N)\) and \((x_2 = 0.2\space m,F_2 = 2\space N)\)

Step 3: Substitute the values into the slope formula

Using the slope formula \(k=\frac{F_2 - F_1}{x_2 - x_1}\)

Substitute \(F_1 = 1\space N\), \(F_2=2\space N\), \(x_1 = 0.1\space m\) and \(x_2=0.2\space m\) into the formula:

\[k=\frac{F_2 - F_1}{x_2 - x_1}=\frac{2\space N- 1\space N}{0.2\space m - 0.1\space m}\]

\[k=\frac{1\space N}{0.1\space m}=10\space N/m\]

If we are given the equation of the best - fit line directly. For example, if the equation of the best - fit line for \(F\) (in Newtons) vs \(x\) (in meters) is \(F=(5\space N/m)x\), then by comparing with \(F = kx\), we can see that the spring constant \(k = 5\space N/m\)

Let's take a more general case. Suppose the best - fit line has the equation \(F=(k)x\), and from the data used to generate the best - fit line, we can also calculate \(k\) by taking the average of \(\frac{F}{x}\) for each data point (since \(k=\frac{F}{x}\) from Hooke's Law). If we have \(n\) data points \((x_1,F_1),(x_2,F_2),\cdots,(x_n,F_n)\), then \(k=\frac{1}{n}\sum_{i = 1}^{n}\frac{F_i}{x_i}\) (this is equivalent to finding the slope of the line that passes through the origin and has the best fit to the data points)

Example Calculation with made - up data

Let's assume we have the following data points for force and displacement:

Displacement \(x\) (m)	Force \(F\) (N)
0.10	1.0
0.15	1.5

Step 1: Calculate \(\frac{F}{x}\) for each data point

• For the first data point: \(\frac{F_1}{x_1}=\frac{0.5\space N}{0.05\space m}=10\space N/m\)
• For the second data point: \(\frac{F_2}{x_2}=\frac{1.0\space N}{0.10\space m}=10\space N/m\)
• For the third data point: \(\frac{F_3}{x_3}=\frac{1.5\space N}{0.15\space m}=10\space N/m\)

Step 2: Find the average of \(\frac{F}{x}\) (which is the spring constant)

Since all the \(\frac{F}{x}\) values are \(10\space N/m\), the spring constant \(k = 10\space N/m\)

If we use the slope formula with two points, say \((x_1 = 0.05\space m,F_1 = 0.5\space N)\) and \((x_2=0.15\space m,F_2 = 1.5\space N)\)

The slope \(k=\frac{F_2 - F_1}{x_2 - x_1}=\frac{1.5\space N- 0.5\space N}{0.15\space m - 0.05\space m}=\frac{1.0\space N}{0.10\space m}=10\space N/m\)

The key idea is that the spring constant is the slope of the force - displacement graph (from the best - fit line), and we calculate this slope either by using two points on the best - fit line and the slope formula or by using the relationship from Hooke's Law (\(k=\frac{F}{x}\)) and averaging the \(\frac{F}{x}\) values for the data points (or using the slope of the linear regression line).

The answer will depend on the actual data used to generate the best - fit line. For example, if the best - fit line has a slope of \(k = 5\space N/m\) (from a different set of force - displacement data), then the spring constant is \(5\space N/m\). In the example we worked through with the made - up data, the spring constant \(k=\boldsymbol{10\space N/m}\)