Question

simple harmonic motion this is a multi - part item. use the slider to determine the applied force needed for displacements of 0.100 m, 0.200 m, 0.300 m, 0.400 m, and 0.500 m. plot the points on the coordinate grid. then, graph a line by selecting two points.

Explanation:

To solve this problem related to Simple Harmonic Motion (specifically Hooke's Law, \( F = kx \), where \( F \) is the applied force, \( k \) is the spring constant, and \( x \) is the displacement), we follow these steps:

Step 1: Identify the Spring Constant (\( k \))

From the simulation, the spring constant \( k \) is given as \( 200 \, \text{N/m} \) (this is typically set in the "Spring Constant" slider or display).

Step 2: Calculate Force for Each Displacement

Using Hooke’s Law \( F = kx \), we calculate the force for each displacement (\( x = 0.100 \, \text{m}, 0.200 \, \text{m}, 0.300 \, \text{m}, 0.400 \, \text{m}, 0.500 \, \text{m} \)):

• For \( x = 0.100 \, \text{m} \):

\( F = (200 \, \text{N/m})(0.100 \, \text{m}) = 20.0 \, \text{N} \)

• For \( x = 0.200 \, \text{m} \):

\( F = (200 \, \text{N/m})(0.200 \, \text{m}) = 40.0 \, \text{N} \)

• For \( x = 0.300 \, \text{m} \):

\( F = (200 \, \text{N/m})(0.300 \, \text{m}) = 60.0 \, \text{N} \)

• For \( x = 0.400 \, \text{m} \):

\( F = (200 \, \text{N/m})(0.400 \, \text{m}) = 80.0 \, \text{N} \)

• For \( x = 0.500 \, \text{m} \):

\( F = (200 \, \text{N/m})(0.500 \, \text{m}) = 100.0 \, \text{N} \)

Step 3: Plot the Points

On the coordinate grid, plot the points:

• \( (0.100 \, \text{m}, 20.0 \, \text{N}) \)
• \( (0.200 \, \text{m}, 40.0 \, \text{N}) \)
• \( (0.300 \, \text{m}, 60.0 \, \text{N}) \)
• \( (0.400 \, \text{m}, 80.0 \, \text{N}) \)
• \( (0.500 \, \text{m}, 100.0 \, \text{N}) \)

Step 4: Draw the Line

Select two points (e.g., \( (0.100, 20) \) and \( (0.500, 100) \)) and draw a straight line through them. The line should pass through all plotted points (since \( F \) is directly proportional to \( x \) for a spring, the graph is linear).

Final Answer (Force Values for Each Displacement)

• Displacement \( 0.100 \, \text{m} \): \( \boldsymbol{20.0 \, \text{N}} \)
• Displacement \( 0.200 \, \text{m} \): \( \boldsymbol{40.0 \, \text{N}} \)
• Displacement \( 0.300 \, \text{m} \): \( \boldsymbol{60.0 \, \text{N}} \)
• Displacement \( 0.400 \, \text{m} \): \( \boldsymbol{80.0 \, \text{N}} \)
• Displacement \( 0.500 \, \text{m} \): \( \boldsymbol{100.0 \, \text{N}} \)

(For plotting, use these \( (x, F) \) pairs and draw a linear graph.)

Answer:

To solve this problem related to Simple Harmonic Motion (specifically Hooke's Law, \( F = kx \), where \( F \) is the applied force, \( k \) is the spring constant, and \( x \) is the displacement), we follow these steps:

Step 1: Identify the Spring Constant (\( k \))

From the simulation, the spring constant \( k \) is given as \( 200 \, \text{N/m} \) (this is typically set in the "Spring Constant" slider or display).

Step 2: Calculate Force for Each Displacement

Using Hooke’s Law \( F = kx \), we calculate the force for each displacement (\( x = 0.100 \, \text{m}, 0.200 \, \text{m}, 0.300 \, \text{m}, 0.400 \, \text{m}, 0.500 \, \text{m} \)):

• For \( x = 0.100 \, \text{m} \):

\( F = (200 \, \text{N/m})(0.100 \, \text{m}) = 20.0 \, \text{N} \)

• For \( x = 0.200 \, \text{m} \):

\( F = (200 \, \text{N/m})(0.200 \, \text{m}) = 40.0 \, \text{N} \)

• For \( x = 0.300 \, \text{m} \):

\( F = (200 \, \text{N/m})(0.300 \, \text{m}) = 60.0 \, \text{N} \)

• For \( x = 0.400 \, \text{m} \):

\( F = (200 \, \text{N/m})(0.400 \, \text{m}) = 80.0 \, \text{N} \)

• For \( x = 0.500 \, \text{m} \):

\( F = (200 \, \text{N/m})(0.500 \, \text{m}) = 100.0 \, \text{N} \)

Step 3: Plot the Points

On the coordinate grid, plot the points:

• \( (0.100 \, \text{m}, 20.0 \, \text{N}) \)
• \( (0.200 \, \text{m}, 40.0 \, \text{N}) \)
• \( (0.300 \, \text{m}, 60.0 \, \text{N}) \)
• \( (0.400 \, \text{m}, 80.0 \, \text{N}) \)
• \( (0.500 \, \text{m}, 100.0 \, \text{N}) \)

Step 4: Draw the Line

Select two points (e.g., \( (0.100, 20) \) and \( (0.500, 100) \)) and draw a straight line through them. The line should pass through all plotted points (since \( F \) is directly proportional to \( x \) for a spring, the graph is linear).

Final Answer (Force Values for Each Displacement)

• Displacement \( 0.100 \, \text{m} \): \( \boldsymbol{20.0 \, \text{N}} \)
• Displacement \( 0.200 \, \text{m} \): \( \boldsymbol{40.0 \, \text{N}} \)
• Displacement \( 0.300 \, \text{m} \): \( \boldsymbol{60.0 \, \text{N}} \)
• Displacement \( 0.400 \, \text{m} \): \( \boldsymbol{80.0 \, \text{N}} \)
• Displacement \( 0.500 \, \text{m} \): \( \boldsymbol{100.0 \, \text{N}} \)

(For plotting, use these \( (x, F) \) pairs and draw a linear graph.)