Problem 32 - College Physics

Question

\(\bullet\) \(\bullet\) 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 an additional 0.240 m before momentarily stopping.

Step 1: Understanding SHM

Simple Harmonic Motion (SHM) implies that the object is oscillating back and forth about an equilibrium position. It can be characterized by its amplitude, maximum speed, and acceleration. The object's acceleration a when displaced is given by: \[ a = -\omega^2 x \]where \( \omega \) is the angular frequency, and \( x \) is the displacement from the equilibrium.

Step 2: Calculate Angular Frequency

The given acceleration \( a = -8.40 \ \mathrm{m/s}^2 \) when \( x = 0.600 \ \mathrm{m} \). Plugging these values into the equation:\[ -8.40 = -\omega^2 \times 0.600 \]Solving for \( \omega \), we get:\[ \omega^2 = \frac{8.40}{0.600} = 14 \rightarrow \omega = \sqrt{14} \approx 3.74 \, \mathrm{rad/s} \]

Step 3: Energy Conservation Principle

In SHM, mechanical energy is conserved. At \( x = 0.600 \ \mathrm{m} \), the object has kinetic energy \( KE \) and potential energy \( PE \). However, at maximum displacement (amplitude), the object has only potential energy and zero kinetic energy. Thus:\[ \frac{1}{2}m\omega^2A^2 = \frac{1}{2}mv^2 + \frac{1}{2}m\omega^2x^2 \]where \( A \) is the amplitude and \( v = 2.20 \ \mathrm{m/s} \). Mass \( m \) cancels out.

Step 4: Solve for Amplitude

Substitute values into the energy conservation equation:\[ \omega^2A^2 = v^2 + \omega^2x^2 \]\[ 14A^2 = (2.20)^2 + 14(0.600)^2 \]\[ 14A^2 = 4.84 + 14 \times 0.36 \]\[ 14A^2 = 4.84 + 5.04 = 9.88 \]\[ A^2 = \frac{9.88}{14} \approx 0.706 \]\[ A = \sqrt{0.706} \approx 0.840 \, \mathrm{m} \]

Step 5: Calculate Additional Displacement

The maximum displacement from the equilibrium position is 0.840 m. The object is currently at 0.600 m from equilibrium. Therefore, the additional distance it will move before momentarily stopping is:\[ \text{Additional distance} = A - x = 0.840 \ \mathrm{m} - 0.600 \ \mathrm{m} = 0.240 \ \mathrm{m} \]

Key Concepts

Angular Frequency

In Simple Harmonic Motion (SHM), angular frequency \( \omega \) is an essential concept as it describes how quickly an object oscillates around its equilibrium position. Angular frequency is linked to the number of oscillations or rotations per unit time. The unit for angular frequency is radians per second \( \text{rad/s} \).

To find the angular frequency in our scenario, we use the formula for acceleration in SHM:

• \( a = -\omega^2 x \)
• where \( a \) is the acceleration, \( x \) is the displacement from equilibrium, and \( \omega \) is the angular frequency.

Given that \( a = -8.40 \ \mathrm{m/s^2} \) and \( x = 0.600 \ \mathrm{m} \), we substitute these values into the formula to calculate \( \omega \):

• \( -8.40 = -\omega^2 \times 0.600 \)
• Solving for \( \omega \), we find \( \omega = \sqrt{14} \approx 3.74 \ \mathrm{rad/s} \).

Understanding \( \omega \) helps us realize how quickly the object would complete one oscillation cycle. This knowledge is foundational as we move into problems that involve predicting movement and calculating further positions.

Energy Conservation

In the context of SHM, energy conservation is a powerful tool that helps us understand how energy transforms during an object's motion. The total mechanical energy remains constant as the object oscillates, expressed as the sum of its kinetic energy (KE) and potential energy (PE).

When the object is at a displacement \( x \) from its equilibrium,

• It has both kinetic energy \( KE = \frac{1}{2} mv^2 \)
• and potential energy \( PE = \frac{1}{2} m\omega^2x^2 \)

where \( v \) is the velocity at that displacement.
At maximum displacement, or amplitude \( A \), all the energy is potential, and kinetic energy is zero. Thus, total energy at any point is:

• \( \frac{1}{2} m \omega^2 A^2 = \frac{1}{2} m v^2 + \frac{1}{2} m \omega^2 x^2 \).

Mass \( m \) cancels out, allowing us to focus on just velocities and displacements when solving for amplitude or other quantities. This principle shows that: as the object moves, energy switches between kinetic and potential, maintaining a balance, critical for predicting the motion over time.

Amplitude Calculation

Amplitude \( A \) in SHM is the maximum distance the object moves from its equilibrium position. Understanding this concept aids in predicting how far the object will travel before it reverses direction.

The amplitude is found using the energy conservation equation, integrating all the known energies at a particular point:

• The formula to find amplitude becomes: \( \omega^2 A^2 = v^2 + \omega^2 x^2 \).
• With \( v = 2.20 \ \mathrm{m/s} \), \( x = 0.600 \ \mathrm{m} \), and \( \omega = 3.74 \ \mathrm{rad/s} \).

Plugging in these values:

• \( 14A^2 = (2.20)^2 + 14(0.600)^2 \)
• Solving gives: \( A^2 = \frac{9.88}{14} \approx 0.706 \) so \( A \approx 0.840 \ \mathrm{m} \).

This calculation shows how far the object travels from the equilibrium before reversing. The extra distance it moves, beyond 0.600 m, can then be calculated. Understanding amplitude is key to solving problems involving motion limits in SHM.