A particle executes simple harmonic motion of ampliltude A. At what distance from the mean position is its kinetic energy equal to its potential energ
A particle executes simple harmonic motion of ampliltude A. At what distance from the mean position is its kinetic energy equal to its potential energy?
A. 0.51 A
B. 0.61 A
C. 0.71 A
D. 0.81 A
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Correct Answer - C
`KE=PE`
`(1)/(2)mw^(2)(A^(2)-x^(2))=(1)/(2)mw^(2)x^(2)`
`A^(2)-x^(2)=x^(2)`
`A=2x^(2)`
`A=sqrt(2)x`
`x=(A)/(sqrt(2))=0.707 A`.
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