To tackle the problem of a particle moving in a fixed plane along the trajectory defined by the equation \(3\omega t\), we need to break it down into several parts. Let's go through each part step by step.

Visualizing the Trajectory

First, we need to sketch the trajectory of the particle. The equation \(3\omega t\) suggests a linear relationship with time, where \(3\omega\) is a constant that scales the time variable \(t\). This indicates that the particle moves in a straight line rather than a curved path.

• The trajectory can be represented as a straight line on a graph where the x-axis is time \(t\) and the y-axis is the position of the particle.
• At \(t = 0\), the position is \(0\). As time increases, the position increases linearly, indicating uniform motion.

Determining the Force Acting on the Particle

Next, let's find the force acting on the particle. According to Newton's second law, the force \(F\) acting on an object is given by the equation:

F = m \cdot a

Here, \(a\) is the acceleration of the particle. Since the trajectory is linear, we can find the velocity \(v\) by differentiating the position with respect to time:

v = \frac{d(3\omega t)}{dt} = 3\omega

The acceleration \(a\) is the derivative of velocity:

a = \frac{dv}{dt} = 0

Since the acceleration is zero, the force acting on the particle is:

F = m \cdot 0 = 0

Calculating Potential Energy

Potential energy (PE) is often associated with conservative forces, such as gravitational or elastic forces. In this case, since there is no force acting on the particle, we can assume that the potential energy is constant. If we denote the potential energy at a reference point as \(U_0\), then:

PE(t) = U_0

Finding Total Energy as a Function of Time

The total energy \(E\) of the particle is the sum of its kinetic energy (KE) and potential energy (PE). The kinetic energy is given by:

KE = \frac{1}{2} m v^2

Substituting the velocity we found earlier:

KE = \frac{1}{2} m (3\omega)^2 = \frac{9}{2} m \omega^2

Thus, the total energy is:

E(t) = KE + PE = \frac{9}{2} m \omega^2 + U_0

Assessing the Nature of Motion

To determine if the motion is periodic, we need to consider the nature of the trajectory. Since the particle moves in a straight line with constant velocity, it does not return to a starting point after a fixed interval. Therefore, the motion is not periodic.

Summarizing Key Points

• The trajectory is a straight line represented by \(3\omega t\).
• The force acting on the particle is zero.
• The potential energy is constant, \(PE(t) = U_0\).
• The total energy is \(E(t) = \frac{9}{2} m \omega^2 + U_0\).
• The motion is not periodic.

This breakdown provides a comprehensive understanding of the particle's motion, its energy dynamics, and the nature of its trajectory. If you have any further questions or need clarification on any part, feel free to ask!