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Chapter 15: Problem 19

A 50.0 -g object connected to a spring with a force constant of \(35.0 \mathrm{N} / \mathrm{m}\) oscillates on a horizontal, frictionless surface with an amplitude of \(4.00 \mathrm{cm} .\) Find (a) the total energy of the system and (b) the speed of the object when the position is \(1.00 \mathrm{cm} .\) Find \((\mathrm{c})\) the kinetic energy and \((\mathrm{d})\) the potential energy when the position is \(3.00 \mathrm{cm} .\)

Short Answer

Expert verified
The total energy of the system is 0.14 J. The speed of the object when the position is \(1.00 \mathrm{cm}\) is 0.60 m/s. The kinetic energy when the position is \(3.00 \mathrm{cm}\) is 0.035 J and the potential energy is 0.105 J.

Step by step solution

01

Calculate the Total Energy

Total energy in the system of an oscillator is the sum of kinetic and potential energy, at any point during the motion. And maximum potential energy is when the object is at the extreme ends. This Total Energy is given by the formula: \( E = \frac{1}{2} k A^{2} \), where 'E' is the total energy, 'k' is the spring constant and 'A' is the amplitude of oscillation.
02

Calculate the Speed of the Object

The speed of the object can be found using the law of conservation of energy. When the object is not at an extreme position, it has both kinetic and potential energy. Therefore the total energy equals combined kinetic and potential energies. Mathematically, \( v = \sqrt{(\frac{2}{m}) (E - \frac{1}{2}kx^{2})} \), where 'm' is the mass of the object, 'v' is the speed, 'x' is the position of the object.
03

Calculate the Kinetic Energy and Potential Energy

The kinetic energy can be calculated using the formula: \( KE = \frac{1}{2} m v^{2} \). The potential energy is given by: \( PE = \frac{1}{2} k x^{2} \), where 'k' is the spring constant and 'x' is the position of the object. We can use the speeds calculated in Step 2, and by substituting the respective positions we can find the kinetic and potential energies at those positions.

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Key Concepts

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

Spring Constant
The spring constant, often represented by the symbol 'k', is a measure of a spring's stiffness. A larger spring constant means a stiffer spring, which requires more force to stretch or compress it by a given amount. This force constant plays a crucial role in determining how a spring will behave in a system of oscillations.
  • The spring constant is measured in newtons per meter (N/m).
  • It is pivotal in calculating the total energy in a spring system, through the formula for potential energy.
In our exercise, the given spring constant is 35.0 N/m. This indicates a moderately stiff spring that requires a relatively substantial force to be displaced by a small distance. When you hear about oscillations involving springs, understanding the spring constant is essential to gauging the system's properties.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. In oscillating systems, like a mass attached to a spring, kinetic energy is highest when the object is moving fastest, typically as it passes through the equilibrium position.
  • The formula for kinetic energy is:

  • At any given point in the oscillation,
  • The kinetic energy will change as the object's position changes. When the position is given, we can determine the speed and use it to compute the kinetic energy. In our problem, when the object is at a specific position, like 1.00 cm or 3.00 cm, its kinetic energy will differ based on the balance of potential and mechanical energies at those points.
Potential Energy
Potential energy in spring-mass systems is primarily stored energy related to the position of the object. Unlike kinetic energy, potential energy is highest at the extreme ends of oscillation. It decreases as the object moves towards the equilibrium position.
  • Given by the formula: \( PE = \frac{1}{2} k x^{2} \), where 'x' is the object's displacement from the equilibrium.
  • This potential energy represents how much energy is stored due to the object's position.
For instance, when the object is at a distance of 3.00 cm away from the equilibrium, the potential energy is calculated by substituting the values into the potential energy equation. This helps us understand how the energy transitions between potential and kinetic as the object moves through positions in its path.
Conservation of Energy
The conservation of energy principle states that energy in a closed system remains constant, neither created nor destroyed, only changing forms. In the context of oscillating systems, the total mechanical energy, a combination of kinetic and potential energy, stays the same throughout the motion.
  • When the object is at an extreme point, the energy is purely potential.
  • Conversely, at the equilibrium point, the energy is entirely kinetic.
  • In between, the energy seamlessly transitions between the two forms.
To solve problems involving oscillations, this principle is incredibly useful. For example, when calculating the speed of the object at different positions, we use the conservation of energy to find how the total energy distribute between potential and kinetic, allowing for insightful calculations of the dynamics involved.

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Most popular questions from this chapter

After a thrilling plunge, bungee-jumpers bounce freely on the bungee cord through many cycles (Fig. P15.22). After the first few cycles, the cord does not go slack. Your little brother can make a pest of himself by figuring out the mass of each person, using a proportion which you set up by solving this problem: An object of mass \(m\) is oscillating freely on a vertical spring with a period \(T\). An object of unknown mass \(m^{\prime}\) on the same spring oscillates with a period.\(T^{\prime} .\) Determine (a) the spring constant and (b) the unknown mass.

A lobsterman's buoy is a solid wooden cylinder of radius \(r\) and mass \(M .\) It is weighted at one end so that it floats upright in calm sea water, having density \(\rho .\) A passing shark tugs on the slack rope mooring the buoy to a lobster trap, pulling the buoy down a distance \(x\) from its equilibrium position and releasing it. Show that the buoy will execute simple harmonic motion if the resistive effects of the water are neglected, and determine the period of the oscillations.

An automobile having a mass of \(1000 \mathrm{kg}\) is driven into a brick wall in a safety test. The bumper behaves like a spring of force constant \(5.00 \times 10^{6} \mathrm{N} / \mathrm{m}\) and compresses \(3.16 \mathrm{cm}\) as the car is brought to rest. What was the speed of the car before impact, assuming that no mechanical energy is lost during impact with the wall?

A \(0.500-\mathrm{kg}\) object attached to a spring with a force constant of \(8.00 \mathrm{N} / \mathrm{m}\) vibrates in simple harmonic motion with an amplitude of \(10.0 \mathrm{cm} .\) Calculate (a) the maximum value of its speed and acceleration, (b) the speed and acceleration when the object is \(6.00 \mathrm{cm}\) from the equilibrium position, and (c) the time interval required for the object to move from \(x=0\) to \(x=8.00 \mathrm{cm}\).

Damping is negligible for a \(0.150-\mathrm{kg}\) object hanging from a light \(6.30-\mathrm{N} / \mathrm{m}\) spring. A sinusoidal force with an amplitude of \(1.70 \mathrm{N}\) drives the system. At what frequency will the force make the object vibrate with an amplitude of \(0.440 \mathrm{m} ?\)
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