Solving Spring Energy: Half Amplitude & Kinetic Energy %

[go2cnavy]

For an Ideal spring...At half Amplitude, what % of the energy is kinetic?

I know at A the kinetic energy it instantly zero which makes max potential energy (.5kx^2)

I also know that after released, at the instant passising through the equilibrium point kinetic energy is max and potential energy is zero.

From the energy conservation approach Et = Ei so, .5mv^2 = .5kA^2


I have no clue where to start this problem. Maybe I am making it too difficult

 

[OlderDan]

For an Ideal spring...At half Amplitude, what % of the energy is kinetic?

I know at A the kinetic energy it instantly zero which makes max potential energy (.5kx^2)

I also know that after released, at the instant passising through the equilibrium point kinetic energy is max and potential energy is zero.

From the energy conservation approach Et = Ei so, .5mv^2 = .5kA^2


I have no clue where to start this problem. Maybe I am making it too difficult

You know how to calculate the potential energy at full amplitude when it is the total energy. You should also be able to calculate it at half amplitude when it is not the total energy. Any energy that is not potentail energy at half amplitude (or anywhere else) is kinetic energy.

 

[kyle8921]

.Dear colleague,

Thank you for reaching out with your question about solving spring energy. I am happy to provide some insight and guidance on this topic.

Firstly, it is important to note that for an ideal spring, the energy is conserved throughout its motion. This means that the total energy at any point in time is equal to the initial energy of the system. In other words, the sum of the potential energy and kinetic energy at any point in time is always equal to the initial energy of the system.

Now, let's consider the question at hand - at half amplitude, what percentage of the energy is kinetic? To answer this, we need to understand the relationship between potential energy and kinetic energy in a spring.

At the maximum amplitude, the spring is fully compressed or stretched, and all of its energy is in the form of potential energy. As the spring begins to oscillate, the potential energy is converted into kinetic energy as the spring moves back and forth. At the halfway point, the spring is at its equilibrium position and has no potential energy. Therefore, at this point, all of the energy is in the form of kinetic energy.

To determine the percentage of energy that is kinetic at half amplitude, we can use the equation you mentioned - .5mv^2 = .5kA^2. This equation relates the kinetic energy (represented by .5mv^2) to the potential energy (represented by .5kA^2) at any point in time.

At half amplitude, the displacement of the spring (represented by A) is equal to half of the maximum displacement. Therefore, we can rewrite the equation as .5mv^2 = .5k(0.5A)^2. Simplifying this, we get .5mv^2 = .125kA^2.

Now, we can see that at half amplitude, the kinetic energy is equal to one-eighth of the initial potential energy (.125 = 1/8). This means that at half amplitude, the kinetic energy represents 12.5% of the total energy.

I hope this explanation helps you understand the concept of energy conservation in a spring and how to determine the percentage of kinetic energy at any point in its motion. If you have any further questions, please do not hesitate to reach out.

Best regards,