Springs and potential/kinetic energy

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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?
 
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?
 
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.
 
Awesome! Thank you so much for your help!