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A particle starts oscillating simple harmonically from its equilibrium position then, the ratio of kinetic energy and potential energy of the particle at the time `T//12` is: `(T="time period")`A. `1 : 4`B. `2 : 1`C. `3 : 1`D. `4 : 1` |
Answer» Correct Answer - C When `t = (T)/(12)`, then `x = A "sin"(2pi)/(T)xx(T)/(12)=(A)/(2)` `KE = (1)/(2)mv^(2)=(1)/(2)m omega^(2)(r^(2)-x^(2))` `= (1)/(2)m omega^(2)(A^(2)-(A^(2))/(4))` `=(3)/(4)((1)/(2)momega^(2)A^(2))` `PE = (1)/(2)m omega^(2)x^(2)=(1)/(4)((1)/(2)m omega^(2)A^(2))` `(KE)/(PE)=(3)/(1)` |
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