Problem 77 - University Physics with Modern Physics

Question

A particle with mass \(m\) is acted on by a conservative force and moves along a path given by \(x=x_{0} \cos \omega_{0} t\) and \(y=y_{0} \sin \omega_{0} t\) where \(x_{0}, y_{0},\) and \(\omega_{0}\) are constants. (a) Find the components of the force that acts on the particle. (b) Find the potential energy of the particle as a function of \(x\) and \(y .\) Take \(U=0\) when \(x=0\) and \(y=0 .\left(\text { c) Find the total energy of the particle when (i) } x=x_{0 .}\right.\) \(y=0\) and \(\left(\text { ii) } x=0, y=y_{0}\right.\)

Step-by-Step Solution

Answer

Force components: \(F_x = -m \omega_{0}^2 x\), \(F_y = -m \omega_{0}^2 y\). Potential energy: \(U(x, y) = -\frac{1}{2} m \omega_{0}^2 (x^2 + y^2)\). Total energy at given points: zero.

Step 1: Express Velocity Components

The velocity components are the derivatives of position with respect to time. Therefore, we have:- \(v_x = \frac{dx}{dt} = -x_{0} \omega_{0} \sin \omega_{0} t\)- \(v_y = \frac{dy}{dt} = y_{0} \omega_{0} \cos \omega_{0} t\)

Step 2: Calculate Force Components

Given that the force acting on a particle in simple harmonic motion is derived from the position to velocity and acceleration relationship, we use: - Force in x-direction: \(F_x = m a_x = -m \omega_{0}^2 x_{0} \cos \omega_{0} t\)- Force in y-direction: \(F_y = m a_y = -m \omega_{0}^2 y_{0} \sin \omega_{0} t\)

Step 3: Relate Force to Potential Energy

The relationship between force and potential energy in a conservative force field is given by:- \(F_x = -\frac{\partial U}{\partial x}\)- \(F_y = -\frac{\partial U}{\partial y}\)

Step 4: Integrate Force to Find Potential Energy

Integrating each force component will give us the potential energy function. For the x-component:\[ U(x) = \int F_x \, dx = \int -m \omega_{0}^2 x \, dx = -\frac{1}{2} m \omega_{0}^2 x^2 + C_x \]For the y-component:\[ U(y) = \int F_y \, dy = \int -m \omega_{0}^2 y \, dy = -\frac{1}{2} m \omega_{0}^2 y^2 + C_y \]Thus, the total potential energy is:\[ U(x, y) = -\frac{1}{2} m \omega_{0}^2 x^2 - \frac{1}{2} m \omega_{0}^2 y^2 + C\]

Step 5: Determine Constant of Integration

The condition given is that \(U = 0\) at \(x = 0\) and \(y = 0\). Using this condition to solve for \(C\):\[-\frac{1}{2} m \omega_{0}^2 (0) -\frac{1}{2} m \omega_{0}^2 (0) + C = 0 \Rightarrow C = 0\]Thus, the potential energy is: \[ U(x, y) = -\frac{1}{2} m \omega_{0}^2 x^2 - \frac{1}{2} m \omega_{0}^2 y^2 \]

Step 6: Write Total Energy Equation

The total mechanical energy \(E\) is the sum of kinetic \(T\) and potential \(U\) energies:\[ T = \frac{1}{2} m (v_x^2 + v_y^2) \]\[ T = \frac{1}{2} m ((-x_{0} \omega_{0} \sin \omega_{0} t)^2 + (y_{0} \omega_{0} \cos \omega_{0} t)^2) \]\[ = \frac{1}{2} m \omega_{0}^2 (x_{0}^2 \sin^2 \omega_{0} t + y_{0}^2 \cos^2 \omega_{0} t) \]\[ E = T + U = \frac{1}{2} m \omega_{0}^2 (x_{0}^2 \sin^2 \omega_{0} t + y_{0}^2 \cos^2 \omega_{0} t) - \frac{1}{2} m \omega_{0}^2 (x^2 + y^2) \]

Step 7: Solve for Total Energy at Given Points

Substituting the specified values:(i) \(x = x_0\), \(y = 0\):\[E = \frac{1}{2} m \omega_{0}^2 (0 + y_{0}^2) - \frac{1}{2} m \omega_{0}^2 (x_{0}^2 + 0) = 0\](ii) \(x = 0\), \(y = y_0\):\[E = \frac{1}{2} m \omega_{0}^2 (x_{0}^2 + 0) - \frac{1}{2} m \omega_{0}^2 (0 + y_{0}^2) = 0\]

Step 8: Final Answer Verification

Since both energy components are zero, we confirm that the point energies at (i) \(x_0\), \(0\) and (ii) \(0\), \(y_0\) are zero.

Key Concepts

Potential Energy

Potential energy is a fascinating concept that relates to the position of an object within a force field. In the case of conservative forces, potential energy can be understood as stored energy that depends on the position of an object.

When dealing with conservative forces, such as gravity or spring forces, potential energy functions play a crucial role. They allow us to determine how much energy an object has based on its position within the field. A key characteristic of potential energy is that it is path-independent, meaning it only depends on the initial and final positions, not on the path taken between them.

In the context of our exercise:

• The particle's potential energy depends on its positions along the x and y axes.
• The formula for potential energy derived from the exercise is given by
  \[ U(x, y) = -\frac{1}{2} m \omega_{0}^2 x^2 - \frac{1}{2} m \omega_{0}^2 y^2 \]
• The potential energy is zero at the origin, as specified by the given condition \(U = 0\) at \(x = 0\) and \(y = 0\).

The negative sign in the potential energy formula indicates the type of force field interacting with the particle, typically resulting in restoring forces like those seen in harmonic motion. It is always integral to remember this relationship when dealing with potential energy and conservative forces.

Simple Harmonic Motion

Simple Harmonic Motion (SHM) is a term used to describe the oscillatory motion of particles or objects that follow a periodic path. It is characterized by the fact that the restoring force acting to return the object to its equilibrium position is directly proportional to the displacement.

In our scenario, the particle undergoes SHM as it moves along a path defined by oscillating positions in both the x and y directions given by \(x=x_{0} \cos \omega_{0} t\) and \(y=y_{0} \sin \omega_{0} t\).

• This motion results in the particle moving through an elliptical path in the xy-plane.
• Simple Harmonic Motion arises due to conservative forces, ensuring that energy is conserved as potential energy and kinetic energy interact.
• The parameters involved like the amplitude (\(x_{0}, y_{0}\)) and angular frequency (\(\omega_{0}\)) dictate the specifics of the particle's path and velocity.

Understanding SHM helps us predict the position and velocity of objects at any point in time, which is crucial in many physical and engineering applications. The idea of a restoring force ensures that objects return to their equilibrium position, making SHM an elegant model for analyzing periodic motions.

Kinetic Energy

Kinetic energy refers to the energy of motion. For any object in motion, we can determine its kinetic energy based on its mass and velocity. It is mathematically expressed by the formula: \(T = \frac{1}{2} m v^2\).

In our problem, the kinetic energy of the particle is calculated by considering its velocity components as it moves along the SHM path:

• The velocity components are derived as \(v_x = -x_{0} \omega_{0} \sin \omega_{0} t\) and \(v_y = y_{0} \omega_{0} \cos \omega_{0} t\).
• The total kinetic energy becomes \(T = \frac{1}{2} m \omega_{0}^2 (x_{0}^2 \sin^2 \omega_{0} t + y_{0}^2 \cos^2 \omega_{0} t)\).

Kinetic energy and potential energy are intimately linked, especially in systems involving SHM.

As the particle moves back and forth in its path, the energy transitions from kinetic to potential and back. This process can be observed through:

At maximum displacement, the kinetic energy is minimum, and potential energy is at its peak.

When passing through the equilibrium position, the kinetic energy reaches its maximum, and potential energy falls to zero.

This beautiful dance between kinetic and potential energy ensures the total mechanical energy remains constant, a hallmark of conservative systems. Recognizing this energy interplay aids in understanding how systems conserve energy and enables predicting future states in dynamic systems.