What is the Amplitude of a Mass-Spring System?

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A mass of 0.25 kg is attached to a spring with a spring constant of 35 N/m and oscillates on a frictionless surface with a speed of 1.04 m/s at equilibrium. The total energy of the system can be calculated using kinetic energy since there is no potential energy at the equilibrium point. The user struggled to find the amplitude using various equations but ultimately realized that all energy at equilibrium is kinetic. The clarification provided helped the user understand the solution better. The discussion emphasizes the importance of recognizing energy types in oscillating systems.
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A block of mass 0.25 kg is connected to a spring with a spring constant (k) of 35 N/m. It is oscillating on a frictionless horizontal surface. It's speed as it passes through equilibrium is 1.04 m/s. What is the total energy of the system?

I know that E=0.5(k)(A)^2. I just can't figure out the Amplitude. I know I can do it. I just don't know what kind of equation to use to get the amplitude. I have tried using x=A cos (wt) but that wasn't right. I have also tried vmax=A(k/m), that wasn't right either. I'm confused.
 
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Since the body is connected to only one spring (if it was oscillating while attached to two springs which were both under compression at all times, then that would be different) then there is no compression or extension in the spring at the equilibrium point, hence there is no potential energy in the spring at that point.
All the energy is therefore just kinetic energy. You have the mass and the velocity at the equilibrium point, so you can find the KE.
 
Thanks so much for your help. I miss the most simple solutions sometimes. It helped a lot for you to explain it. Thanks again. I got it right.:smile:
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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