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\(\bullet\) 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.

Short Answer

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

Step by step solution

01

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:\[ E = K + U = 0.00675 + 0.0216 = 0.02835 \, \text{J} \]
02

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 \( A \):\[ A^2 = \frac{0.02835}{150.0} \]\[ A = \sqrt{\frac{0.02835}{150.0}} = 0.0137 \, \text{m} \]
03

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 \( v_{\text{max}} \):\[ v_{\text{max}}^2 = \frac{0.02835}{0.075} \]\[ v_{\text{max}} = \sqrt{\frac{0.02835}{0.075}} = 0.615 \, \text{m/s} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

\(\bullet\) A 0.500 kg glider on an air track is attached to the end of an ideal spring with force constant \(450 \mathrm{N} / \mathrm{m} ;\) it undergoes simple harmonic motion with an amplitude of 0.040 \(\mathrm{m} .\) Compute (a) the maximum speed of the glider, (b) the speed of the glider when it is at \(x=-0.015 \mathrm{m},(\mathrm{c})\) the magnitude of the maximum acceleration of the glider, (d) the acceleration of the glider at \(x=-0.015 \mathrm{m},\) and \((\mathrm{e})\) the total mechanical energy of the glider at any point in its motion.

\(\bullet\) \(\bullet\) A harmonic oscillator is made by using a 0.600 kg frictionless block and an ideal spring of unknown force constant. The oscillator is found to have a period of 0.150 s. Find the force constant of the spring.

\(\bullet\) A 1.35 \(\mathrm{kg}\) object is attached to a horizontal spring of force constant 2.5 \(\mathrm{N} / \mathrm{cm}\) and is started oscillating by pulling it 6.0 \(\mathrm{cm}\) from its equilibrium position and releasing it so that it is free to oscillate on a frictionless horizontal air track. You observe that after eight cycles its maximum displacement from equilibrium is only 3.5 \(\mathrm{cm}\) . (a) How much energy has this system lost to damping during these eight cycles? (b) Where did the "lost" energy go? Explain physically how the system could have lost energy.

\(\bullet\) \(\bullet\) An apple weighs 1.00 \(\mathrm{N}\) . When you hang it from the end of a long spring of force constant 1.50 \(\mathrm{N} / \mathrm{m}\) and negligible mass, it bounces up and down in SHM. If you stop the bouncing and let the apple swing from side to side through a small angle, the frequency of this simple pendulum is half the bounce frequency. (Because the angle is small, the back-and-forth swings do not cause any appreciable change in the length of the spring.) What is the unstretched length of the spring (with the apple removed)?

\(\bullet\) \(\bullet\) Inside a NASA test vehicle, a 3.50 -kg ball is pulled along by a horizontal ideal spring fixed to a friction-free table. The force constant of the spring is 225 \(\mathrm{N} / \mathrm{m} .\) The vehicle has a steady acceleration of \(5.00 \mathrm{m} / \mathrm{s}^{2},\) and the ball is not oscillating. Suddenly, when the vehicle's speed has reached \(45.0 \mathrm{m} / \mathrm{s},\) its engines turn off, thus eliminating its acceleration but not its velocity. Find (a) the amplitude and (b) the frequency of the resulting oscillations of the ball. (c) What will be the ball's maximum speed relative to the vehicle?

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