InterviewSolution
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InterviewSolution
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A partical executes SHM with an amplltude of `10 cm` and friquency `2 Hz, at t = 0`, the partical is at point where potential energy and kinetic energy are same. Find the equation of displacement of partical. |
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Answer» Let `x = A sin (omega t + phi)` : then `v = (dx)/(dt) = A omega cos (omega t + phi)` `KE = (1)/(2) mv^(2) = (1)/(2) mA^(2) omega^(2) cos ^(2) (omega t + phi)` `(KE)_(max) = (1)/(2) mA^(2) omega^(2) [for cos ^(2) omega t + phi)]` `PE = (1)/(2) mA^(2) - KE = (1)/(2) mA^(2) - mA^(2) omega^(2) cos ^(2) (omega t + phi)` `= (1)/(2) mA^(2) omega^(2) sin ^(2) (omega t + phi)` According to the problem `KE = PE`: therefore `= (1)/(2) mA^(2) omega^(2) sin ^(2) (omega t + phi) = (1)/(2) mA^(2) omega^(2) cos ^(2) (omega t + phi)` `tan^(2) (omega t + phi) = 1 implies tan^(2) (omega t + phi) = tan^(2) (pi)/(4)` `omega t + phi = pi//4 implies phi = pi//4 (t = 0)` `x = A sin (omega t + phi)` Here , `A = 10 cm = 0.1 m` `omega = 2 pi f = 2pi xx 2 = 4 pi rad//s` `phi = pi//4` hence, the equation of SHM is `x = 0.1 sin (4 pi + (pi)/(4))` |
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