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A thrill-seeking cat with mass 4.00 \(\mathrm{kg}\) is attached by a harness to an ideal spring of negligible mass and oscillates verticallyin SHM. The amplitude is \(0.050 \mathrm{m},\) and at the highest point of the motion the spring has its natural unstretched length. Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring), the kinetic energy of the cat, the gravitational potential energy of the system relative to the lowest point of the motion, and the sum of these three energies when the cat is (a) at its highest point; ( \((b)\) at its lowest point; (c) at its equilibrium position.

Short Answer

Expert verified
(a) Highest: Total energy = 1.96 J. (b) Lowest: Total energy = 1.96 J. (c) Equilibrium: Total energy = 1.96 J.

Step by step solution

01

Understand the System

The mass of the cat is 4.00 kg, and it is oscillating vertically attached to a spring. The spring's natural length means no extension/compression at the highest oscillation point. The amplitude is the maximum displacement from the equilibrium position, which is 0.050 m.
02

Step 2a: Calculate Energies at the Highest Point

At the highest point, the spring is not stretched, so the elastic potential energy \( U_e = 0 \). The displacement is +0.050 m, which is the amplitude. At this point, kinetic energy \( KE = 0 \) as the cat isn't moving. The gravitational potential energy \( U_g \) is at its maximum: \( U_g = mgh = 4 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 0.050 \, \text{m} = 1.96 \, \text{J} \). Total energy \( E = U_e + KE + U_g = 1.96 \, \text{J} \).
03

Step 2b: Calculate Energies at the Lowest Point

At the lowest point (displacement -0.050 m), the elastic potential energy is maximum. \( U_e = \frac{1}{2} k x^2 \), where \( x = 0.050 \, \text{m} \). The gravitational potential energy \( U_g = 0 \), and kinetic energy \( KE = 0 \). Total energy \( E = \frac{1}{2} k (0.050)^2 \). Use conservation of energy to find total energy, which should remain 1.96 J.
04

Determine Spring Constant

Using energy conservation, at the lowest point: \[1.96 \, \text{J} = \frac{1}{2} k (0.050 \, \text{m})^2 \]. Solving for \( k \), \[ k = \frac{2 \times 1.96}{(0.050)^2} = 1568 \, \text{N/m}. \]
05

Step 4a: Verify Energies at the Lowest Point

Using \( k \approx 1568 \, \text{N/m}\), \( U_e = \frac{1}{2} \times 1568 \times (0.050)^2 = 1.96 \, \text{J} \). With \( KE = 0 \) and \( U_g = 0 \), total energy is \( 1.96 \, \text{J} \).
06

Step 4b: Calculate Energies at the Equilibrium Position

At equilibrium, potential energy is zero as it is the average point. All energy is kinetic: \( KE = 1.96 \, \text{J} \), meaning \( U_e = 0 \) and \( U_g = 0 \).
07

Verify Energy Conservation

Check that the sum of energies is constant: At all positions (highest, lowest, equilibrium), total energy \( E = 1.96 \, \text{J} \). This confirms energy conservation in the system.

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Key Concepts

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

Elastic Potential Energy
Elastic potential energy is the energy stored within a spring when it is compressed or stretched. In this particular exercise, when the cat attached to the spring reaches the highest point of its motion, the spring is at its natural length, which means there is no stretch. Thus, the elastic potential energy is zero at this point. However, when the cat is at the lowest point of its oscillation, the spring is fully stretched to the extent of the amplitude, creating the maximum elastic potential energy. The formula for elastic potential energy is given by \[ U_e = \frac{1}{2} k x^2 \] where \( k \) is the spring constant, and \( x \) is the displacement (0.050 m in this case). At this lowest point, we calculate it to be 1.96 J when \( k \) is found to be 1568 N/m.
Kinetic Energy
Kinetic energy refers to the energy an object possesses due to its motion. In the scenario of the oscillating cat, the kinetic energy is zero at both the highest and the lowest points of the motion. That is because at these points, the cat momentarily comes to rest before reversing its direction. At the equilibrium position, however, all the system's energy is in the form of kinetic energy because the net forces are balanced, and the cat is moving at its maximum speed. In this scenario, \[ KE = 1.96 \, \text{J} \] This ensures energy is conserved as the potential energies at equilibrium are zero.
Gravitational Potential Energy
Gravitational potential energy is the energy an object possesses because of its position in a gravitational field, usually Earth's. For the cat-spring system, this energy is maximum when the cat is at its highest point above the lowest position. The formula used to determine gravitational potential energy is: \[ U_g = mgh \] where \( m = 4.00 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), and \( h = 0.050 \, \text{m} \). This calculation results in a gravitational potential energy of 1.96 J at the highest point. At the lowest point, the gravitational potential energy is zero because this position is considered the reference level.
Energy Conservation
Energy conservation is a crucial principle in physics, asserting that the total energy in an isolated system remains constant. In the context of the thrill-seeking cat, the sum of elastic potential energy, kinetic energy, and gravitational potential energy equals 1.96 J at any point in the motion. Hence, even as energy shifts between different types, the total remains the same: - At the highest point: all energy as gravitational - At the equilibrium position: all energy as kinetic - At the lowest point: all energy as elastic potential This consistency in total energy highlights energy conservation in Simple Harmonic Motion.
Spring Constant
The spring constant \( k \) is a measure of the stiffness of the spring. A higher spring constant indicates a stiffer spring that requires more force to stretch or compress it by a certain distance. In the case at hand, we use the condition of energy conservation to determine the spring constant of 1568 N/m, ensuring that the total energy remains 1.96 J. By using the formula: \[ k = \frac{2 \times 1.96}{(0.050)^2} \] This calculated value aids in predicting how the spring will store and release energy throughout the motion of the cat, ultimately allowing us to better understand the dynamics of the SHM system.

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Most popular questions from this chapter

Two identical atoms in a diatomic molecule vibrate as harmonic oscillators. However, their center of mass, midway between them, remains at rest. (a) Show that at any instant, the momenta of the atoms relative to the center of mass are \(\vec{p}\) and \(-\vec{p} .\) (b) Show that the total kinetic energy \(K\) of the two atoms at any instant is the same as that of a single object with mass \(m / 2\) with a momentum of magnitude \(p .\) (Use \(K=p^{2} / 2 m . )\) This result shows why \(m / 2\) should be used in the expression for \(f\) in have masses \(m_{1}\) and \(m_{2},\) show that the result of part (a) still holds and the single object's mass in part (b) is \(m_{1} m_{2} /\left(m_{1}+m_{2}\right) .\) The quantity \(m_{1} m_{2} J\left(m_{1}+m_{2}\right)\) is called the reducedmass of the system.

A \(1.50-\mathrm{kg}\) ball and a \(2.00-\mathrm{kg}\) ball are glued together with the lighter one below the heavier one. The upper ball is attached to a vertical ideal spring of force constant \(165 \mathrm{N} / \mathrm{m},\) and the system is vibrating vertically with amplitude 15.0 \(\mathrm{cm} .\) The gine connecting the balls is old and weak, and it suddenly comes loose when the balls are at the lowest position in their motion. (a) Why is the glue more likely to fail at the lowest point than at any other point in the motion? (b) Find the amplitude and frequency of the vibrations after the lower ball has come loose.

Jerk. A guitar string vibrates at a frequency of 440 \(\mathrm{Hz}\) . A point at its center moves in SHM with an amplitude of 3.0 \(\mathrm{mm}\) and a phase angle of zero. (a) Write an equation for the position of the center of the string as a function of time. (b) What are the maximum values of the magnitudes of the velocity and acceleration of the center of the string? (c) The derivative of the acceleration with respect to time is a quantity called the jerk. Write an equation for the jerk of the center of the string as a function of time, and find the maximum value of the magnitude of the jerk.

A holiday ornament in the shape of a hollow sphere with mass \(M=0.015 \mathrm{kg}\) and radius \(R=0.050 \mathrm{m}\) is hung from a tree limb by a small loop of wire attached to the surface of the sphere. If the ornament is displaced a small distance and reased, it swings back and forth as a physical pendulum with negligible friction. Calculate its period. (Hint: Use the parallel-axis theorem to find the moment of inertia of the sphere about the pivot at the tree limb.)

A \(40.0 .\) N force strethes a vertical spring 0.250 \(\mathrm{m}\) . (a) What mass must be suspended from the spring so that the system will oscillate with a period of 1.00 \(\mathrm{s} ?(6)\) If the amplitude of the motion is 0.050 \(\mathrm{m}\) and the period is that specified in part (a), where is the object and in what direction is it moving 0.35 \(\mathrm{s}\) after it has passed the equilibrium position, moving downward? (c) What force (magnitude and direction) does the spring exert on the object when it is 0.030 \(\mathrm{m}\) below the equilibrium position, moving upward?

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