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Eagles342
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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)?
E=1/2kA^2
1/2kdelta^2=1/2mv^2+1/2kx^2
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
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