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Understanding potential energy in the context of simple harmonic motion is key to solving problems involving oscillating particles. In this scenario, potential energy represents the stored energy due to an object's position in a force field, such as the spring force in this motion. For a particle moving in simple harmonic motion with a displacement from the equilibrium position given by the equation \( x(t) = (5.0 \, \text{m}) \cos \left[\frac{\pi}{3} \text{rad/s} \cdot t - \frac{\pi}{4} \text{rad}\right] \), the potential energy \( U \) at any point \( x \) is given by:

• \( U = \frac{1}{2} m \omega^2 x^2 \)

Here, \( m \) is the particle's mass, \( \omega \) is the angular frequency, and \( x \) is the displacement.To find the position where the potential energy equals half of the total energy, the equation \( \frac{1}{2} m \omega^2 x^2 = \frac{1}{4} m \omega^2 A^2 \) is used, where \( A \) is the amplitude of motion. This is derived by setting the potential energy to half of the particle's total energy and solving for \( x \).

Total energy in simple harmonic motion is the sum of kinetic and potential energy, and it remains constant in an ideal system (without damping). This total energy can be expressed as:

• \( E = \frac{1}{2} m \omega^2 A^2 \)

In this equation, \( m \) is the mass of the particle, \( \omega \) is the angular frequency, and \( A \) is the amplitude. Essentially, this formula represents the maximum energy when the particle is at its most extreme position.In our problem, the total energy is constant, but the distribution between potential and kinetic energy changes with position in the cycle. When the potential energy is half of the total energy, it typically implies that the other half is kinetic, indicating the particle's speed is neither at maximum nor zero, but somewhere in between.

Angular frequency is a crucial concept in understanding simple harmonic motion. It refers to how rapidly the oscillating motion completes its cycles and is related to the period of motion.The angular frequency \( \omega \) is given in terms of radians per second, and it is calculated from the rate of change of the phase angle in the motion equation. In our problem, the expression \( x(t) = (5.0 \, \text{m}) \cos \left[\frac{\pi}{3} \text{rad/s} \cdot t - \frac{\pi}{4} \text{rad}\right] \) reveals that the angular frequency is \( \omega = \frac{\pi}{3} \text{ rad/s} \).The concept can also be connected to the physical aspects of the problem:

• The higher the angular frequency, the faster the particle oscillates.
• A change in angular frequency would affect both the period of oscillation and the energy distribution at any point in the cycle.

Understanding angular frequency helps students grasp how quickly or slowly different energy states are reached during the oscillation.