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Explain by plats the position of particle executing simple harmonic motion at different time. |
Answer» Solution :The figure shows the positions of a particle executing S.H.M. are at discrete value to time, each interval of time being `(T)/(4)" where "phi = 0` and T is the period of motion.  For a given SHM, A is the amplitude, then the velocity and position of a particle at time t is determine by the phase `(omega t+ phi)` of cosine function. In the equation of SHM `x(t) = A cos (omega t+phi), phi =0" and " omega = (2pi)/(T)`, `x(t)= A cos ((2pi)/(T)t)` Now `t=0` `x(t)= A cos (0) = +A` Taking `t=(T)/(4)` `x(t)= A cos"" ((2pi)/(T)xx(T)/(4)) = Acos ""(pi/2)= 0` Taking `t=(T)/(2)` `x(t)= A cos ""((2pi)/(T)xx(T)/(2)) = Acos ""pi = -A` Taking `t=(3T)/(4)` `x(t)= A cos ""((2pi)/(T)xx(3T)/(4)) = Acos ""(3PI)/(2)= 0` Taking `t=T` `x(t)= A cos ""((2pi)/(T)xxT) = Acos ""2pi = +A` Taking `t=(5T)/(4)` `x(t)= A cos ""((2pi)/(T)xx(5T)/(4)) = Acos ""(5pi)/(2)= 0` From figure got following observations : (i) After period T, periodic (REPEATS) motion takes place. (ii) Period T remains fixed no MATTER what location you choose as the INITIAL (t=0) location. (III) The speed is maximum for zero displacement (at x=0) and zero at the extremes of motion (x= A). 
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