a particle is in SHM with time period 3sec. calculate the displacement of the particle from mean position after t=0.5sec is
Asked by vvelesha
13th September 2018,
6:45 PM
Answered by Expert
Answer:
displacement x is given by, x = a×sin ωt = a×sin (2π/T)t = a×sin [ (2π/3)×0.5 ] = a×sin(π/3) = (√3/2)×a
where a is amplitude, ω is angular velocity i.e., ω = 2π/T and T is period of oscillation.
hence at t=0.5 sec, particle is at 0.866 times of amplitude from mean position.
Answered by Expert
13th September 2018,
7:33 PM
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