A particle executing linear S.H.M has period 3 second and amplitude 6 ~cm . The time required by it to travel a distance of 3 ~cm from positive extreme position is [ 30^ = 60^ = 1 2 , 60^ = 30^ = 3 2 ]

Step-by-step solution

Lets take simplest equation of SHM, y=A ( t) Given, A=6 ~cm , ~T =3 sec = 2 T = 2 3 ⁻¹ . Time when particle reaches extreme position P is 1.5 sec . Let time to reach Q is . aligned & y =3 ~cm =6 ~cm ( 2 + ) & ( 2 + )= 1 2 & ( )= 1 2 & = 3 = (3) 3(2 ) =0.5 ~s aligned