Problem 29 - College Physics

Question

\(\bullet\) A 0.150 \(\mathrm{kg}\) toy is undergoing SHM on the end of a horizontal spring with force constant 300.0 \(\mathrm{N} / \mathrm{m} .\) When the object is 0.0120 \(\mathrm{m}\) from its equilibrium position, it is observed to have a speed of 0.300 \(\mathrm{m} / \mathrm{s}\) . Find (a) the total energy of the object at any point in its motion, (b) the amplitude of the motion, and (c) the maximum speed attained by the object during its motion.

Step-by-Step Solution

Answer

(a) 0.02835 J; (b) 0.0137 m; (c) 0.615 m/s

Step 1: Find the Total Mechanical Energy

The total mechanical energy \( E \) of an object in simple harmonic motion (SHM) is conserved and can be expressed as the sum of its kinetic energy \( K \) and potential energy \( U \). The kinetic energy \( K \) is given by \( K = \frac{1}{2}mv^2 \), and the potential energy \( U \) due to a spring is \( U = \frac{1}{2}kx^2 \). Substitute the given values: \( m = 0.150 \, \text{kg} \), \( v = 0.300 \, \text{m/s} \), \( k = 300.0 \, \text{N/m} \), and \( x = 0.0120 \, \text{m} \).Calculate:\[ K = \frac{1}{2} (0.150) (0.300)^2 = 0.00675 \, \text{J} \]\[ U = \frac{1}{2} (300.0) (0.0120)^2 = 0.0216 \, \text{J} \]The total energy \( E \) is:\[ E = K + U = 0.00675 + 0.0216 = 0.02835 \, \text{J} \]

Step 2: Calculate the Amplitude of Motion

The total mechanical energy \( E \) is entirely potential energy at the turning points of the motion, where the displacement is equal to the amplitude \( A \). Thus, \( E = \frac{1}{2} k A^2 \). Substitute the known values to solve for \( A \):\[ 0.02835 = \frac{1}{2}(300.0)A^2 \]Solve for \( A \):\[ A^2 = \frac{0.02835}{150.0} \]\[ A = \sqrt{\frac{0.02835}{150.0}} = 0.0137 \, \text{m} \]

Step 3: Determine the Maximum Speed

The maximum speed \( v_{\text{max}} \) occurs when all the total energy \( E \) is kinetic energy \( K \). Thus, \( E = \frac{1}{2} mv_{\text{max}}^2 \). We already know \( E = 0.02835 \, \text{J} \). Use this to solve for \( v_{\text{max}} \):\[ 0.02835 = \frac{1}{2} (0.150) v_{\text{max}}^2 \]Solve for \( v_{\text{max}} \):\[ v_{\text{max}}^2 = \frac{0.02835}{0.075} \]\[ v_{\text{max}} = \sqrt{\frac{0.02835}{0.075}} = 0.615 \, \text{m/s} \]

Key Concepts

Total Mechanical Energy

In simple harmonic motion (SHM), total mechanical energy represents the sum of kinetic and potential energy, remaining constant throughout the motion. This energy is only exchanged between kinetic and potential forms as the object oscillates. The kinetic energy of an object is given by the formula: \( K = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity.The potential energy stored in the spring can be calculated using: \( U = \frac{1}{2}kx^2 \), where \( k \) is the spring constant and \( x \) is the displacement from equilibrium.By integrating these values:- Kinetic energy \( K = 0.00675 \, \text{J} \)- Potential energy \( U = 0.0216 \, \text{J} \)The total mechanical energy \( E \) becomes:\[ E = K + U = 0.00675 + 0.0216 = 0.02835 \, \text{J} \].Understanding that energy conservation applies here helps us predict various points of movement, from maximum stretch (potential energy) to maximum speed (kinetic energy). It underlies essential concepts in oscillations and is pivotal when analyzing any oscillatory motion. Keep in mind that although the form of energy might change, the total mechanical energy remains constant in an ideal no-friction system.

Amplitude of Motion

Amplitude plays a crucial role in understanding SHM, representing the greatest displacement from the equilibrium position. It indicates how far an object travels during each oscillation. At the peak of the motion, all energy is stored as potential energy, expressed by:\[ E = \frac{1}{2} k A^2 \], where \( A \) is the amplitude.By inserting known values, we targeted amplitude:\[ 0.02835 = \frac{1}{2}(300.0)A^2 \]Solving for \( A \), we find:\[ A^2 = \frac{0.02835}{150.0} \]\[ A = \sqrt{\frac{0.02835}{150.0}} = 0.0137 \, \text{m} \]This indicates that the toy stretches 0.0137 meters away from the equilibrium point. This measure is vital because it sets the boundary of oscillatory motion, defining the extent within which all kinetic and potential energy transformations will occur. Amplitude impacts oscillation duration, influencing periods and frequencies noticeable in seasonal weather shifts and pendulum swings.

Maximum Speed

In SHM, an object experiences maximum speed when passing through its equilibrium point. Here, potential energy nears zero, and kinetic energy maximizes. The relationship of speed with total energy is given by:\[ E = \frac{1}{2} mv_{\text{max}}^2 \].Utilizing this, we draw:\[ 0.02835 = \frac{1}{2} (0.150) v_{\text{max}}^2 \]And solve:\[ v_{\text{max}}^2 = \frac{0.02835}{0.075} \]\[ v_{\text{max}} = \sqrt{\frac{0.02835}{0.075}} = 0.615 \, \text{m/s} \]This maximum speed of 0.615 meters per second offers insights into dynamics at the equilibrium point. When the speed peaks, kinetic energy also peaks, demonstrating how the energy shifts and the pivotal role speed plays during oscillation. Recognizing maximum speed aids in predicting object behavior, gauging passage timing, and hinting at possible external influences on harmonic systems. Understanding these speed dynamics bridges perception with physical realities in daily life experiences, such as playground swings and seismic activities.