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A partical executes SHM with an amplltud...

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.

Text Solution

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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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