1.

The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where, E is the total energy)A. `(U_(max))/(2)`B. `(U_(max))/(3)`C. `(U_(max))/(4)`D. `2 U_(max)`

Answer» Correct Answer - C
PE at `x = (A)/(2)` is `(1)/(2) m omega^(2) xx (A^(2))/(4)`
`therefore" "PE = (1)/(4)[(1)/(2)m omega^(2)A^(2)]=(U_(max))/(4)`


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