MHT CET20255 May 2025Evening ShiftPhysicsOscillationsActual
A point particle of mass 200 gram is executing S.H.M. of amplitude 0.2 m . When the particle passes through the mean position, its kinetic energy is 16 10⁻³ ~J . The equation of motion of this particle is (Initial phase of oscillation =0^ )
Options
- AY=0.2 (4 t)
- BY=0.2 ( t 4 )
- CY=0.2 ( t 2 )
- DY=0.2 (2 t)
Correct answer
D. Y=0.2 (2 t)
Step-by-step solution
Maximum kinetic energy in SHM relates to amplitude and angular frequency through the equation KE_ max = 1 2 m (A )^2 . Substituting known values m = 0.2 kg , A = 0.2 m , and KE_ max = 16 10⁻³ J : 16 10⁻³ = 1 2 (0.2) (0.2 )^2 Solving for ^2 : ^2 = 16 10⁻³ 0.1 0.04 = 4 Thus = 2 rad/s . The displacement equation becomes Y = 0.2 (2t) , which corresponds to option D .