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7 min readβ’june 18, 2024
Kanya Shah
Daniella Garcia-Loos
Kanya Shah
Daniella Garcia-Loos
The energy of a system is conserved.
A system with an internal structure can have internal energy, and changes in a systemβs internal structure can result in changes in internal energy
Here are some key things to know about the internal energy of a simple harmonic oscillator:
A system with internal structure can have potential energy. Potential energy exists within a system if the objects within that system interact with conservative forces.
Here are some key things to know about the potential energy of a simple harmonic oscillator:
The internal energy of a system includes the kinetic energy of the objects that make up the system and the potential energy of the configuration of objects that make up the system.
Here are some key things to know about the kinetic energy of a simple harmonic oscillator:
This topic is pretty much just an application of the energy types and conversions we covered in Unit 4: Energy. The main idea is that through SHM, the energy is converted from potential to kinetic and back again throughout the motion. The maximum potential energy occurs when the spring is stretched (or compressed) the most, and the maximum kinetic energy occurs at the equilibrium point.Β
If we were to make a graph of energy vs time, it would look like this:
Example Problem 1:
A mass of 1 kg is attached to a spring with a spring constant of 50 N/m and is allowed to oscillate vertically in a frictionless environment. The mass is initially displaced 0.2 meters from its equilibrium position and released from rest. What is the total energy of the oscillator at the maximum displacement from the equilibrium position?
Solution:
The total energy of a simple harmonic oscillator is the sum of its potential energy and kinetic energy.
The potential energy of a simple harmonic oscillator is given by the equation: U = 1/2*kx^2, where U is the potential energy, k is the spring constant, and x is the displacement of the oscillator from its equilibrium position.
The kinetic energy of a simple harmonic oscillator is given by the equation: K = 1/2*mv^2, where K is the kinetic energy, m is the mass of the oscillator, and v is the velocity of the oscillator.
In this problem, the mass of the oscillator is 1 kg, the spring constant is 50 N/m, and the displacement from the equilibrium position is 0.2 meters.
At the maximum displacement from the equilibrium position, the velocity of the oscillator is zero and the potential energy is at a maximum.
Therefore, the total energy of the oscillator at the maximum displacement from the equilibrium position is: U + K = (1/2)(50 N/m)(0.2 m)^2 + 0 = 1 J
This means that the total energy of the oscillator at the maximum displacement from the equilibrium position is 1 J.
Example Problem 2:
A mass of 2 kg is attached to a spring with a spring constant of 100 N/m and is allowed to oscillate vertically in a frictionless environment. The mass is initially displaced 0.5 meters from its equilibrium position and released from rest. What is the total energy of the oscillator at the equilibrium position?
Solution:
The total energy of a simple harmonic oscillator is the sum of its potential energy and kinetic energy.
The potential energy of a simple harmonic oscillator is given by the equation: U = 1/2*kx^2, where U is the potential energy, k is the spring constant, and x is the displacement of the oscillator from its equilibrium position.
The kinetic energy of a simple harmonic oscillator is given by the equation: K = 1/2*mv^2, where K is the kinetic energy, m is the mass of the oscillator, and v is the velocity of the oscillator.
In this problem, the mass of the oscillator is 2 kg, the spring constant is 100 N/m, and the displacement from the equilibrium position is 0.5 meters.
At the equilibrium position, the velocity of the oscillator is zero and the potential energy is at a minimum.
Therefore, the total energy of the oscillator at the equilibrium position is: U + K = 1/2(100 N/m)(0.5 m)^2 + 0 = 12.5 J
This means that the total energy of the oscillator at the equilibrium position is 12.5 J.
π₯Watch: AP Physics 1 -Β Problem Solving q+a Simple Harmonic Oscillators
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7 min readβ’june 18, 2024
Kanya Shah
Daniella Garcia-Loos
Kanya Shah
Daniella Garcia-Loos
The energy of a system is conserved.
A system with an internal structure can have internal energy, and changes in a systemβs internal structure can result in changes in internal energy
Here are some key things to know about the internal energy of a simple harmonic oscillator:
A system with internal structure can have potential energy. Potential energy exists within a system if the objects within that system interact with conservative forces.
Here are some key things to know about the potential energy of a simple harmonic oscillator:
The internal energy of a system includes the kinetic energy of the objects that make up the system and the potential energy of the configuration of objects that make up the system.
Here are some key things to know about the kinetic energy of a simple harmonic oscillator:
This topic is pretty much just an application of the energy types and conversions we covered in Unit 4: Energy. The main idea is that through SHM, the energy is converted from potential to kinetic and back again throughout the motion. The maximum potential energy occurs when the spring is stretched (or compressed) the most, and the maximum kinetic energy occurs at the equilibrium point.Β
If we were to make a graph of energy vs time, it would look like this:
Example Problem 1:
A mass of 1 kg is attached to a spring with a spring constant of 50 N/m and is allowed to oscillate vertically in a frictionless environment. The mass is initially displaced 0.2 meters from its equilibrium position and released from rest. What is the total energy of the oscillator at the maximum displacement from the equilibrium position?
Solution:
The total energy of a simple harmonic oscillator is the sum of its potential energy and kinetic energy.
The potential energy of a simple harmonic oscillator is given by the equation: U = 1/2*kx^2, where U is the potential energy, k is the spring constant, and x is the displacement of the oscillator from its equilibrium position.
The kinetic energy of a simple harmonic oscillator is given by the equation: K = 1/2*mv^2, where K is the kinetic energy, m is the mass of the oscillator, and v is the velocity of the oscillator.
In this problem, the mass of the oscillator is 1 kg, the spring constant is 50 N/m, and the displacement from the equilibrium position is 0.2 meters.
At the maximum displacement from the equilibrium position, the velocity of the oscillator is zero and the potential energy is at a maximum.
Therefore, the total energy of the oscillator at the maximum displacement from the equilibrium position is: U + K = (1/2)(50 N/m)(0.2 m)^2 + 0 = 1 J
This means that the total energy of the oscillator at the maximum displacement from the equilibrium position is 1 J.
Example Problem 2:
A mass of 2 kg is attached to a spring with a spring constant of 100 N/m and is allowed to oscillate vertically in a frictionless environment. The mass is initially displaced 0.5 meters from its equilibrium position and released from rest. What is the total energy of the oscillator at the equilibrium position?
Solution:
The total energy of a simple harmonic oscillator is the sum of its potential energy and kinetic energy.
The potential energy of a simple harmonic oscillator is given by the equation: U = 1/2*kx^2, where U is the potential energy, k is the spring constant, and x is the displacement of the oscillator from its equilibrium position.
The kinetic energy of a simple harmonic oscillator is given by the equation: K = 1/2*mv^2, where K is the kinetic energy, m is the mass of the oscillator, and v is the velocity of the oscillator.
In this problem, the mass of the oscillator is 2 kg, the spring constant is 100 N/m, and the displacement from the equilibrium position is 0.5 meters.
At the equilibrium position, the velocity of the oscillator is zero and the potential energy is at a minimum.
Therefore, the total energy of the oscillator at the equilibrium position is: U + K = 1/2(100 N/m)(0.5 m)^2 + 0 = 12.5 J
This means that the total energy of the oscillator at the equilibrium position is 12.5 J.
π₯Watch: AP Physics 1 -Β Problem Solving q+a Simple Harmonic Oscillators
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