Springs and potential/kinetic energy

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Homework Help Overview

The discussion revolves around a mass-spring system undergoing simple harmonic motion. The original poster presents a problem involving the calculation of amplitude and the locations where kinetic and potential energy are equal.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between kinetic and potential energy in the context of oscillation, questioning the use of speed at different positions and the implications for energy conservation.

Discussion Status

Participants are actively engaging with the problem, with some offering insights into the nature of total energy in the system and its constancy throughout the motion. There is a recognition of the need to clarify the relationship between energy types at various positions.

Contextual Notes

There is a noted confusion regarding the use of the speed at the equilibrium position for calculations related to energy at different points in the oscillation. Participants are also considering the implications of energy conservation principles in their reasoning.

Carrie
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Homework Statement


A 0.50-kg mass is attached to a spring of spring constant 20 N/m along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position.

(a) What is the amplitude of vibration?
I already solved for the amplitude and got 0.24 meters.

(b) At what location are the kinetic energy and the potential energy the same?
?

Homework Equations


E = 1/2mv^2 + 1/2kx^2

The Attempt at a Solution


E = 1/2mv^2 + 1/2kx^2

I thought I would set the kinetic energy and potential energy equal to each other, so I got:

1/2(.5)(1.5)^2 = 1/2(20)x^2
From that, I got x = .24 meters.

But that's wrong and the answer is 0.17 meters. I'm confused as to where I went wrong. :(

Thank you!
 
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Carrie said:
I thought I would set the kinetic energy and potential energy equal to each other, so I got:

1/2(.5)(1.5)^2 = 1/2(20)x^2

Why did you use 1.5 m/s for the speed here?
 
Oh, the 1.5 m/s is for the equilibrium position, which I guess isn't necessarily the same position?
 
Right. The location where the KE equals the PE is not the equilibrium position.
 
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I'm sorry, but now I'm really lost. How do I find v then, or am I using the wrong the wrong equation or something?
 
What can you say about the total energy E at different points of the motion?
 
The total energy stays the same.
 
Carrie said:
The total energy stays the same.
Yes. If you pick any two points of the motion, then E is the same for those two points. You probably used that idea when answering part (a).

See if you can use the same idea to help solve (b).
 
Ohhh okay, so if at equilibrium, the total energy (when PE is 0) is 1/2 mv^2, which is 0.56 J, then when they're the same:

0.56 (total energy in the system) = 2* 1/2kx^2 - I'm not sure if the theory is right, but since KE = PE, then KE+PE becomes PE + PE and so you can multiply by two. Maybe. Because I did get the answer this way, but I also want to make sure I didn't just get lucky.
x = 0.17 m.
 
  • #10
You are thinking correctly. Good work.
 
  • #11
Awesome! Thank you so much for your help!
 

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