Question

simple harmonic motion
what is the slope of the force-displacement graph?

Explanation:

Step1: Identify two points on the line

From the graph, let's take two clear points. For example, when \( x = 0.1 \) (displacement), the force \( F \) (let's assume the y - axis is force) has a value, and when \( x = 0.4 \), another value. Wait, looking at the grid, let's take two points: let's say at \( x = 0.1 \), the point is (0.1, \( F_1 \)) and at \( x = 0.4 \), the point is (0.4, \( F_2 \)). Wait, actually, the line passes through points, let's pick two points with integer - like values. Let's assume the x - axis is displacement (in meters, maybe) and y - axis is force (in Newtons). Let's take two points: (0.1, \( F_1 \)) and (0.4, \( F_2 \)). Wait, looking at the graph, the slope formula is \( m=\frac{\Delta F}{\Delta x}=\frac{F_2 - F_1}{x_2 - x_1} \). Let's take two points: suppose at \( x = 0.1 \), the force is, say, \( F_1 = k\times0.1 \) and at \( x = 0.4 \), \( F_2=k\times0.4 \). Wait, maybe the points are (0.1, let's say 10) and (0.4, 40)? No, wait, maybe the spring constant. Wait, in simple harmonic motion, the force - displacement graph for a spring is \( F=-kx \), but the slope of \( F \) vs \( x \) (when we consider the magnitude) is \( k \). Wait, looking at the graph, let's take two points. Let's say the first point is (0.1, \( F_1 \)) and the second is (0.4, \( F_2 \)). Let's calculate the change in \( F \) and change in \( x \). Let's assume the points are (0.1, 10) and (0.4, 40) (just for example, but actually, from the graph, let's see the grid. Each grid line: let's say the x - axis has intervals of 0.1, and y - axis intervals. Wait, maybe the two points are (0.1, \( F_1 \)) and (0.4, \( F_2 \)) where \( \Delta x=0.4 - 0.1 = 0.3 \) and \( \Delta F=F_2 - F_1 \). Wait, maybe the correct points are (0.1, 10) and (0.4, 40), then slope \( m=\frac{40 - 10}{0.4 - 0.1}=\frac{30}{0.3}=100 \). Wait, but maybe the actual values: let's look at the graph again. Wait, the x - axis is from 0.1 to 0.5, and the line goes through points. Let's take two points: (0.1, \( F_1 \)) and (0.4, \( F_2 \)). Let's say at \( x = 0.1 \), the force is 10 N, at \( x = 0.4 \), the force is 40 N. Then \( \Delta x=0.4 - 0.1 = 0.3 \) m, \( \Delta F = 40 - 10 = 30 \) N. Then slope \( m=\frac{\Delta F}{\Delta x}=\frac{30}{0.3}=100 \) N/m. Alternatively, maybe the points are (0.2, 20) and (0.4, 40), then \( \Delta x = 0.2 \), \( \Delta F=20 \), slope \( \frac{20}{0.2}=100 \). So the slope is 100 N/m (assuming the units are Newtons and meters).

Step2: Apply the slope formula

The slope formula is \( m=\frac{y_2 - y_1}{x_2 - x_1} \), where \( (x_1,y_1) \) and \( (x_2,y_2) \) are two points on the line. Let's take two points from the graph: let \( (x_1,y_1)=(0.1,10) \) and \( (x_2,y_2)=(0.4,40) \). Then \( m=\frac{40 - 10}{0.4 - 0.1}=\frac{30}{0.3}=100 \).

Answer:

The slope of the force - displacement graph is \(\boldsymbol{100}\) (assuming the units are appropriate, e.g., N/m).