Question

How long after the beginning of moon is the displacement of a harmonically oscillating particle equal to
onh am
period is 24 and particle starts
from rest
​

Answer 1

Answer:

Displacement x of SHM, x = a sin( ωt ) .....................(1)

where a is amplitude, ω is angular frequency which is given by ω = 2π/T where T is period of SHM and t is time

if displacement x is half of amplitude, then time taken is calculated using eqn.(1) as given below

(a/2) = a sin [ (2π/ 24) t ]

hence ( π/12 )t = sin-1(1/2) = π/6 or t = 2 s

hope it's helpful to you