A point performs simple harmonic oscillation of period T and the equation of motion is given by x= a sin(omega t+(pi)/(6)). After the elapse of what fraction of the time period, the velocity of the point will be equal to half of its maximum velocity?

A point performs simple harmonic oscillation of period T and the equat

1.	A point performs simple harmonic oscillation of period T and the equation of motion is given by x= a sin(omega t+(pi)/(6)). After the elapse of what fraction of the time period, the velocity of the point will be equal to half of its maximum velocity?
Answer» `(T)/(3)` `(T)/(12)` `(T)/(8)` `(T)/(6)` Solution :Velocity at time t, `V=(v_("max"))/(2)` `root(OMEGA)(a^(2)-x^(2))= (a omega)/(2)` `a^(2)- x^(2) = (a^2)/(4)` `therefore a^(2)- (a^2)/(4)= x^(2)`. `therefore x= (sqrt(3))/(2)a` Now, `x= a sin (omega t+(pi)/(6))` `(sqrt(3))/(2)a= a sin (omega t+(pi)/(6))` `therefore (sqrt(3))/(2) = sin (omega t+(pi)/(6))` `therefore omega t+(pi)/(6)= (pi)/(3)` `therefore omega t= (pi)/(3)-(pi)/(6)` `therefore omega t= (pi)/(6)` `therefore t= (pi)/(6)xx (T)/(2pi)` `therefore t= (T)/(12)`.