Question

simple harmonic motion; labels: applied force, spring force, displacement, equilibrium position, velocity; sliders for spring constant (k) and applied force (f); variables: x, -, fₛ, k, +; spring diagram

Explanation:

To solve for the spring force (\( F_s \)) in simple harmonic motion, we use Hooke's Law, which states that the force exerted by a spring is given by:

\[ F_s = -kx \]

where:

• \( k \) is the spring constant (stiffness of the spring),
• \( x \) is the displacement from the equilibrium position,
• The negative sign indicates that the force is opposite to the direction of displacement (restoring force).

Step 1: Identify the values of \( k \) and \( x \)

From the simulation:

• Spring constant, \( k = 200 \, \text{N/m} \) (as seen in the "Spring Constant" slider),
• Displacement, \( x = 0.2 \, \text{m} \) (as seen in the displacement arrow).

Step 2: Apply Hooke's Law

Substitute \( k = 200 \, \text{N/m} \) and \( x = 0.2 \, \text{m} \) into the formula:

\[ F_s = - (200 \, \text{N/m}) \times (0.2 \, \text{m}) \]

\[ F_s = -40 \, \text{N} \]

The negative sign indicates the force is a restoring force (opposite to the displacement direction). If we only consider the magnitude, the spring force is \( 40 \, \text{N} \).

Answer:

The spring force \( F_s \) is \(\boldsymbol{-40 \, \text{N}}\) (or \( 40 \, \text{N} \) in magnitude, directed opposite to the displacement).