Kinetic Energy from a Simple Harmonic motion and spring

In summary: Glad I could help.In summary, a 0.26-kg block on a horizontal frictionless surface attached to an ideal massless spring with a spring constant of 190 N/m executes simple harmonic motion when pulled from its equilibrium position at x=0.00m to a displacement of x=+0.080m and released from rest. Using the equations KE=1/2mw^2A^2sin^2(wt+phi), a=(k/m)x, and w=(k/m)^1/2, we can find the kinetic energy of the block at a displacement of x=1.4x10^-2m. To do so, we must first find the amplitude, A, and the phase
  • #1
Brittany King
12
0

Homework Statement


A 0.26-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 190 N/m. The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is x = 1.4×10−2 m, what is the kinetic energy of the block?

Homework Equations



KE=1/2mw^2A^2sin^2(wt+phi)
a=(k/m)x
w=(k/m)^1/2

The Attempt at a Solution


I have everything for the equation except for phi and and the amplitude A. I have found a=10.23 m/s2 and w=27.0327rad/s but I am stuck on how to find phi and A. Is there a relationship between the initial equilibrium position for finding A or phi?

Thanks for the help!
 
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  • #2
What's wrong with ##PE = 1/2 k x^2##?
You have the amplitude, so you know the total energy. (Yes, you do have the amplitude)
 
  • #3
Brittany King said:
Is there a relationship between the initial equilibrium position for finding A or phi?
You do have A.
Brittany King said:
The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest
Brittany King said:
KE=1/2mw^2A^2sin^2(wt+phi
What will be the KE of the block at t=0? Put it in this equation to get phi.
 
  • #4
BiGyElLoWhAt said:
What's wrong with ##PE = 1/2 k x^2##?
You have the amplitude, so you know the total energy. (Yes, you do have the amplitude)

To use PE=1/2kx^2 wouldn't I need the x at
cnh1995 said:
You do have A.What will be the KE of the block at t=0? Put it in this equation to get phi.

Is A= 0.014m?
 
  • #5
Brittany King said:
To use PE=1/2kx^2 wouldn't I need the x at 0.014m? If that is my A?Is A= 0.014m?
 
  • #6
No. When it's stretched initially, prior to letting go, what is the KE? At which point in the cycle of a general harmonic oscillator is the KE equal to this value?

Then you're looking for the KE at the point x = .014m. So you know x for the PE, as you know at which point you want to look at.
 
  • #7
BiGyElLoWhAt said:
No. When it's stretched initially, prior to letting go, what is the KE? At which point in the cycle of a general harmonic oscillator is the KE equal to this value?

Then you're looking for the KE at the point x = .014m. So you know x for the PE, as you know at which point you want to look at.

Oh I got it now! Thanks so much ! :)
 
  • #8
No problem.
 

1. What is kinetic energy in the context of simple harmonic motion and a spring?

Kinetic energy is the energy that an object possesses due to its motion. In the context of simple harmonic motion and a spring, it refers to the energy that is generated as a result of the back and forth movement of the object attached to the spring.

2. How is kinetic energy related to the amplitude of the simple harmonic motion?

The amplitude of the simple harmonic motion directly affects the kinetic energy of the object. As the amplitude increases, the object moves with a greater velocity, resulting in a higher kinetic energy. Similarly, a smaller amplitude will result in a lower kinetic energy.

3. Can the kinetic energy of an object in simple harmonic motion be calculated?

Yes, the kinetic energy of an object in simple harmonic motion can be calculated using the formula KE = 1/2 mv², where m is the mass of the object and v is its velocity.

4. How does the spring constant affect the kinetic energy of an object in simple harmonic motion?

The spring constant, denoted by k, is directly proportional to the kinetic energy of an object in simple harmonic motion. A higher spring constant will result in a higher kinetic energy, while a lower spring constant will result in a lower kinetic energy.

5. Is there a relationship between the frequency of oscillation and the kinetic energy of an object in simple harmonic motion?

Yes, there is a direct relationship between the frequency of oscillation and the kinetic energy of an object in simple harmonic motion. As the frequency increases, the object moves back and forth at a faster rate, resulting in a higher kinetic energy. Similarly, a lower frequency will result in a lower kinetic energy.

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