Springs and potential/kinetic energy

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A 0.50-kg mass attached to a spring with a spring constant of 20 N/m oscillates on a frictionless surface, achieving a speed of 1.5 m/s at the equilibrium position. The amplitude of vibration was calculated to be 0.24 meters. To find the location where kinetic energy (KE) equals potential energy (PE), the total energy of the system was determined to be 0.56 J at equilibrium. By setting KE equal to PE, the correct position was found to be 0.17 meters. The discussion clarified the relationship between total energy, KE, and PE in simple harmonic motion.
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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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