If the KE of a particle performing a SHM of amplitude A is `(3)/(4)` of its total energy, then the value of its displacement is
A. `x = +- (A)/(2)`
B. `x = +- (A)/(4)`
C. `x = +- (sqrt(3))/(2)A`
D. `x = +- (A)/(sqrt(2))`

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1 Answers

Correct Answer - A
`(1)/(2)m omega^(2)(A^(2)-x^(2))=(3)/(4)((1)/(2)m omega^(2)A^(2))`
`(A^(2)-x^(2))=(3)/(4)A^(2)`
`therefore" "x^(2) = (A^(2))/(4)`
`therefore" "x = +- (A)/(2)`

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