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Elastic potential energy is the energy stored in an elastic object when it experiences deformation, such as through stretching or compressing a spring. The formula for calculating elastic potential energy is given by \( U = \frac{1}{2} k x^2 \). Here, \( k \) is the spring constant, which measures the stiffness of the spring, and \( x \) is the displacement from the spring’s equilibrium position.
This potential energy is dependent on the square of the displacement, meaning even small stretches can significantly impact energy. You can visualize this concept with a graph where the displacement is on the x-axis and the elastic potential energy on the y-axis. It forms a parabolic curve, indicative of how energy changes quadratically with displacement.
To calculate the energy at a specific displacement, substitute the displacement value into the equation. For example, with a spring constant of 10.0 N/m, if the spring is displaced by 0.3 meters, the elastic potential energy is \( U = \frac{1}{2} \times 10.0 \times (0.3)^2 = 0.45 \) joules.

The spring constant, denoted by \( k \), is a crucial part of describing a spring’s behavior. It tells us how strongly a spring will resist deformation. The larger the spring constant, the stiffer the spring.
In our exercise, the spring constant is 10.0 N/m, implying the spring is relatively firm. This figure is used in calculating various spring-related concepts, including elastic potential energy, as it determines the energy stored for a given displacement.
When dealing with problems involving the harmonic motion of a spring, the spring constant helps in determining how much the spring elongated or compressed will affect the stored energy and the force exerted by the spring.
A practical view of the spring constant can be thought of in terms of hooks like characteristics, where the restoring force by the spring is linearly proportional to how far it has been displaced from its equilibrium position, determined by \( F = -kx \).

The amplitude of oscillation is the maximum extent of displacement from the equilibrium point during harmonic oscillation. In this exercise, it's crucial for determining the energy states of the oscillating system.
The amplitude can be calculated using the principle that the total energy of a harmonic oscillator is conserved. That means, at the amplitude, all the energy is stored as potential energy, and the kinetic energy is momentarily zero.
Given the total energy is 0.200 J, and knowing the expression for potential energy, you can find amplitude \( A \) from \( \frac{1}{2} k A^2 = 0.200 \). Solving for \( A \) provides \( A = \sqrt{\frac{2 \times 0.200}{10.0}} \approx 0.2 \) m, showing how much the system can stretch from its center equilibrium position without further energy input.

The phase angle, denoted as \( \phi \), in the context of harmonic oscillation, indicates the specific starting conditions of the system. It helps in specifying the initial position and velocity of an oscillator at \( t = 0 \).
The displacement of the oscillator as a function of time can be expressed by \( x = A\cos(\omega t + \phi) \). Here, \( \omega \) is the angular frequency. Observations of initial conditions, such as a negative initial displacement paired with a positive initial velocity, can help determine the phase angle.
For this scenario, since the initial velocity is positive and the displacement is negative, this implies \( \cos(\phi) < 0 \) and \( \sin(\phi) > 0 \), leaning into the third quadrant of the unit circle, giving a phase angle of \( \phi = \frac{3\pi}{2} \) radians. Understanding this helps in aligning theoretical models of oscillation with real-world behavior of springs in harmonic motion.