Problem 30 - College Physics

Question

You are watching an object that is moving in SHM. When the object is displaced \(0.600 \mathrm{~m}\) to the right of its equilibrium position, it has a velocity of \(2.20 \mathrm{~m} / \mathrm{s}\) to the right and an acceleration of \(8.40 \mathrm{~m} / \mathrm{s}^{2}\) to the left. How much farther from this point will the object move before it stops momentarily and then starts to move back to the left?

Step-by-Step Solution

Answer

The object will move approximately 0.168 m farther.

Step 1: Identify Known Values

We have the displacement \(x = 0.600\, \mathrm{m}\), velocity \(v = 2.20\, \mathrm{m/s}\), and acceleration \(a = -8.40\, \mathrm{m/s}^2\). The object is in simple harmonic motion (SHM).

Step 2: Derive the Spring Constant and Mass

Using the formula for acceleration in SHM, \( a = -\omega^2 x \), where \( \omega \) is the angular frequency, we can substitute \( x = 0.600\, \mathrm{m} \) and \( a = -8.40\, \mathrm{m/s}^2 \) to find \( \omega^2 = \frac{a}{x} = \frac{-8.40}{0.600} = -14.0\, \mathrm{s}^{-2} \). Hence, \( \omega = \sqrt{14.0}\).

Step 3: Use Energy Conservation Principle

The total mechanical energy in SHM is conserved and given by \( E = \frac{1}{2} m v^2 + \frac{1}{2} k x^2 \), where \( m \) is the mass and \( k \) is the spring constant \( k = m \omega^2 \). The maximum displacement (amplitude) occurs when the kinetic energy is zero and the total energy is at its maximum potential energy \( E = \frac{1}{2} k A^2 \).

Step 4: Set Up Energy Equation

At \( x = 0.600\, \mathrm{m} \), the energy is \( E = \frac{1}{2} m (2.20)^2 + \frac{1}{2} (m \times 14.0) (0.600)^2 \). The expression for maximal displacement \( A \) is obtained when \( \frac{1}{2} (m \times 14.0) A^2 = E \).

Step 5: Solve for Maximum Displacement and Calculate Additional Distance

Equating the expressions for \( E \), solve for \( A \):\[ \frac{1}{2} m (2.20)^2 + \frac{1}{2} m \times 14.0 \times 0.600^2 = \frac{1}{2} m \times 14.0 \times A^2 \]Cancel \( m \) from the equation, and solve:\[ (2.20)^2 + 14.0 \times 0.600^2 = 14.0 \times A^2 \]Calculate \( A \) from the above, and the additional distance traveled from \( x = 0.600 \) m is \( A - 0.600 \).

Key Concepts

Mechanical Energy Conservation

In Simple Harmonic Motion (SHM), the principle of mechanical energy conservation plays a significant role.
The total mechanical energy of the system, comprising kinetic and potential energy, remains constant over time.- **Kinetic Energy**: This is the energy an object possesses due to its motion, calculated as \( \frac{1}{2} m v^2 \).- **Potential Energy**: In SHM, the potential energy is associated with the spring and is given by \( \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position.In this exercise, at the point of maximum displacement known as amplitude, the kinetic energy is zero, and all mechanical energy is converted into potential energy. Conversely, at the equilibrium position, all energy is kinetic. Calculating this balance allows us to predict motion, such as determining how far the object will travel beyond 0.600 m before stopping.

Angular Frequency

Angular frequency, denoted by \( \omega \), describes how quickly an object moves through its cycle in simple harmonic motion. It is related to the physical properties of the system, including the mass \( m \) and the spring constant \( k \).- **Formula**: \( \omega = \sqrt{\frac{k}{m}}\)Angular frequency is a crucial parameter because it helps determine how quickly the oscillation occurs. When you have the acceleration and displacement of an object in SHM, you can find the angular frequency using the relation \( a = -\omega^2 x \), where \( a \) is acceleration and \( x \) is displacement. Solving for \( \omega \) provides us with essential insights into the system's dynamics.

Spring Constant

The spring constant, \( k \), measures the stiffness of a spring. In SHM, it is a crucial determinant of both the restoring force and the oscillation frequency of the object.- **Hooke's Law**: The force exerted by the spring is proportional to the displacement, given by \( F = -kx \).This implies that a greater spring constant results in a stiffer spring, requiring more force to achieve a certain displacement. Within the context of this problem, \( k \) defines how significant the potential energy will be, as seen in the \( \frac{1}{2} k x^2 \) term in our mechanical energy conservation formula. The spring constant directly affects the angular frequency \( \omega \), which in turn affects the dynamics of the motion.

Amplitude of Motion

The amplitude of motion in SHM is the maximum extent of displacement from the equilibrium position. It represents the peak value reached by the oscillating object. - **Determining Amplitude**: Using energy principles, the amplitude can be found by setting the total mechanical energy at any point, equal to the maximum potential energy at amplitude. In the exercise, the amplitude is solved from equations derived by equating energies across different points in the motion. The amplitude is crucial in understanding the range of motion. Knowing how far the object moves allows us to predict its behavior over time and solve situations like calculating the distance traveled beyond a point such as 0.600 m.