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Figures depicts two circular motions. The radius of the circle, the period of revolution, the initial position and the sense of revolution are indicated on the figures. Obtain the simple harmonic motions of the x-projection of the radius vector of the rotating particle P in each case. |
Answer» Solution :(a) As shown in figure, SUPPOSE a particle moving with uniform circular motion is at `P_1` at time t=0 and makes and angle `PHI = 45^(@)((pi)/(4)RAD)` with POSITIVE x-axis. After time `t_1` it covers an angle `omega t` and reaches at `P_2` and makes an angle `theta = omega t + phi` with positive x-axis.  Makes an angle `theta = (2pi)/(T)t+ (pi)/(4)` The projection of OP on the x-axis at time t is given by `x (t) = A COS ((2pi)/(T)t+(pi)/(4))" have "OP_(2) = A cos (omega t+phi)` It `t= 4s` then, `x(t) = A cos ((2pi)/(4)t+(pi)/(4)) = A cos ((pi)/(4)+(pi)/(4)) = A sin (pi)/(4)`. Which is a SHM of amplitude A, period 4s and an initial phase `= (pi)/(4) rad`. (b) As shown in figure, suppose a particle moving with uniform circular motion is on P at time t=0 and makes an angle `phi = 90^(@) =(pi/2)` with positive x-axis after a time t, it covers an angle of `omega t` and reaches at `P_1` with clockwise direction and makes an angle of `theta = (pi)/(2) - omega t`. The projection of `P._1` on the x-axis `OP._1`.  `x(t) = B cos theta = B cos((pi)/(2)- omega t)= B sin omega t` `= B cos (omega t-(pi)/(2))""[therefore cos(-theta)= cos theta]` `therefore x(t) = B cos ((pi)/(15)t- (pi)/(2))` This represents a SHM of amplitude B, period T= 30s and an initial phase of `-(pi)/(2)`. |
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