What is the potential energy at 1/2A in simple harmonic motion?

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The discussion centers on calculating the potential energy (U) of a particle in simple harmonic motion when its position is at (1/2A). The correct formula for potential energy is identified as U = 1/2kx^2, and the substitution of x = 1/2A is correctly noted. However, there is confusion regarding the relationship between U and the total energy (E), with a miscalculation suggested in the interpretation of the results. Participants emphasize the importance of clarity in the calculations and understanding the distinction between potential energy and total energy. The conversation highlights the need for careful application of formulas in physics problems.
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A particle of mass is connected to a spring with a force constant K. The particle undergoes simple harmonic with an amp A. What is the potential energy of the partic when the position is (1/2A)?



Homework Equations


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


The Attempt at a Solution



U=1/2kx^2=1/2k(1/2A)^2=1/4(1/2KA^2)=1/4E= K=E-U= 3/4? Not really sure at all can somebody help i don't think i did it correctly
 
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Welcome to PF;
You started strong - the formula for PE is correct.
You substituted x=A/2 fine.

You have been asked to find U - so stop when you have.
 
Besides, E-U = 3/4 makes no sense.
 
1/4(1/2KA^2) doesn't equal 1/4E, it equals 1/4 Umax.
gotta be careful.
 

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