simple harmonic motion in the simulation, the claw’s displacement, d, measures its position, x, from the equilibrium position, 0.000 m. thus, d = x, and the two variables can be used interchangeably. recall that the graph of applied force and displacement was a linear relationship, where the slope equaled the spring constant, k. for a linear relationship that crosses the origin, $y = mx$, where $y$ represents the variable plotted on the vertical axis, $m$ represents the slope, and $x$ represents the variable plotted on the horizontal axis. rewrite this equation to express the linear graph you plotted before, replacing displacement with position and applied force with the spring force. move the correct answer to each box. not all

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To solve this, we start by recalling the context of simple harmonic motion (SHM) for a spring. In SHM, the spring force $ F_s $ (which is the restoring force) is related to the displacement $ x $ (here, displacement $ d = x $) by Hooke's Law.

### Step 1: Recall Hooke's Law and the linear relationship
The linear relationship given is $ y = mx $, where $ y $ is the vertical - axis variable, $ m $ is the slope, and $ x $ is the horizontal - axis variable. In the context of the spring - mass system:
 - The spring force ($ F_s $) is the variable on the vertical axis (analogous to $ y $).
 - The position (displacement, $ x $) is the variable on the horizontal axis (analogous to $ x $ in $ y = mx $).
 - The slope $ m $ is equal to the spring constant $ k $, but we also know from Hooke's Law that the spring force $ F_s=-kx $ (the negative sign indicates that the spring force is a restoring force, opposite to the direction of displacement). However, if we are just considering the magnitude of the relationship for the linear graph (or if we are plotting the applied force which is equal in magnitude to the spring force for the purpose of this linear relationship analysis, ignoring the sign for the form of the equation $ y = mx $):

We replace $ y $ with $ F_s $ (spring force) and $ m $ with $ k $ (spring constant) and $ x $ with $ x $ (position, since $ d=x $). But from Hooke's Law, the correct relationship between spring force and displacement (position) is $ F_s=-kx $, but if we consider the applied force (which is equal in magnitude to the spring force for the linear graph, and the problem says "rewrite this equation to express the linear graph you plotted before, replacing displacement with position and applied force with the spring force"), and the original linear relationship is $ y = mx $ (passing through the origin). In the case of the spring, the spring force $ F_s $ and displacement $ x $ have a linear relationship. From Hooke's Law, $ F_s=-kx $, but if we are talking about the magnitude or the relationship as a linear graph (where the slope is related to the spring constant), and since the applied force and displacement have a linear relationship with slope equal to the spring constant $ k $, and we are to replace applied force with spring force and displacement with position ($ x $):

The linear equation $ y = mx $ becomes $ F_s=kx $ (if we consider the magnitude of the force - displacement relationship for the graph, or if the applied force is equal to the spring force in magnitude for the purpose of this linear fit, ignoring the direction - related negative sign for the equation form $ y = mx $). But more accurately, from Hooke's Law, the spring force $ F_s=-kx $, but if we are following the problem's instruction to use the linear relationship $ y = mx $ and replace $ y $ with spring force ($ F_s $) and $ x $ with position ($ x $) and the slope $ m $ with the spring constant $ k $ (with the understanding of the direction of the force accounted for in the sign, but the problem might be looking for the form based on the given linear relationship $ y = mx $):

So we substitute $ y=F_s $ (spring force), $ m = k $ (spring constant), and $ x=x $ (position) into $ y = mx $.

### Step 2: Write the final equation
Substituting the values into $ y = mx $, we get $ F_s=kx $ (or if we consider the restoring nature, $ F_s=-kx $, but based on the problem's description of the linear relationship $ y = mx $ (passing through the origin) and replacing applied force (which is equal to the spring force in magnitude for the graph) with spring force and displacement with position, the equation is $ F_s = kx $ (where $ F_s $ is the spring force, $ k $ is the spring constant, and $ x $ is the position (displacement)).

# Answer:
$ F_s=kx $ (or $ F_s = -kx $ depending on the sign convention for the direction of the spring force, but based on the given linear relationship $ y = mx $, the answer is $ \boldsymbol{F_s = kx} $)

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QUESTION IMAGE

Question

Explanation:

Step 1: Recall Hooke's Law and the linear relationship

Step 2: Write the final equation

Answer: