By Jaidan Bhakta

spring_data <- data.frame(
  Displacement = seq(0, 1, by = 0.1),
  Force = seq(0, 50, by = 5),
  EPE = c(0.0, 0.25, 0.50, 1.125, 0.80, 
        1.25, 1.80, 2.45, 3.20, 4.05, 5.0)
)
library(ggplot2)
library(plotly)

ggplot(spring_data, aes(x = Displacement, y = Force)) + 
  geom_point() +
  geom_smooth(method = "lm", se = FALSE) + 
  labs(title = "Applied Force v. Displacement", 
       x = "Displacement (m)", y = "Force (N)") +
  theme_minimal()

Demonstrates Hooke’s Law:

\[ F = -k * x \]

• In this case, k is the slope of this graph, which is 50 Newtons per meter
• In physics, we call k or the slope the spring constant
• In physics, this applied force (F) is negative because it is always in the opposite direction of displacement

ggplot(spring_data, aes(x = Displacement, y = EPE)) +
  geom_point() +
  stat_smooth(method = "lm", se = FALSE,
              formula = y ~ poly(x,2), color = "red") +
  labs(title = "Elastic Potential Energy vs. Displacement", 
       x = "Displacement (m)", y = "Elastic Potential Energy (J)") +
  theme_minimal()

Demonstrates the quadratic relationship between EPE and Displacement:

\[ EPE = \frac{1}{2} * k * x^2 \]

• Again, k is the spring constant in this equation as well
• The elastic potential energy (EPE) is directly proportional to the square of the displacement (x) of the compressed spring
• The coefficient in this case: k/2

Another quadratic relationship:

Insert Hooke’s law equation (x = F/k) into EPE equation: \[ EPE = \frac{1}{2} * k * (\frac{F}{k})^2 \] Final simplified equation: \[ EPE = \frac{F^2}{2*k} \] - Here, the elastic potential energy (EPE) is directly proportional to the square of the force applied (F) to the spring. The coefficient is smaller in this case: 1/(2k)