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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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