A particle executes simple harmonic motion with an amplitude of 24 cm . At t=0 , the displacement of the particle is 12 3 cm and it is moving towards the positive extreme position. If the time taken by the particle to reach the positive extreme position from this initial position is 1 s , the magnitude of its velocity at t=0 is n cm/s . The value of n is

Step-by-step solution

Let the equation of motion be x(t) = A ( t + ) . Given amplitude A = 24 cm . At t=0 , x = 12 3 cm and velocity is positive. 24 ( ) = 12 3 ( ) = 3 2 Since the velocity is positive, ( ) > 0 , which gives the initial phase = 3 . The phase at the positive extreme position is 2 . The phase difference to reach the extreme position is = 2 - 3 = 6 . The time taken to cover this phase difference is 1 s . 1 = 6 = 6 rad/s The velocity of the particle at t=0 is given by: v = A ( ) v = 24 ( 6 ) ( 3 ) v = 4 1 2 = 2 cm/s Comparin