When a long time elapsed we take t -> ∞, therefore& e<sup>-2t</sup> -> 0 and the whole term e<sup>-2t </sup>(Acos 2t +B sin 2t) becomes infinitely small.
According to trigonometrical identity:& asin(x) + bcos(x) = √(a2 + b2) sin(x + arctan(b/a)) :
& 2sin(2t) + cos(2t) =√(2<sup>2</sup>+1<sup>2</sup>) sin (2t + arctan(1/2))
The period is equal to 2π /(the coefficient at <i>t</i>) = 2π/2 = π.
sir<br>Please help me, i have not understood this step,<br>why we have taken t->t;infinite and how <br>& e-2t (Acos 2t +B sin 2t) + 2sin(2t) + cos(2t) & <br>converted to <br>= A + √5sin(2t + arctan(0.5))<br><br>At t -> infinity& x(t) ->& e-2t (Acos 2t +B sin 2t) + 2sin(2t) + cos(2t) = A + √5sin(2t + arctan(0.5))<br>how we got the amplitude = √5, and the period is T = π, the frequency is f = 1/T& = 1/π
Check the correct answer above please.
The equation of motion of a particle moving along the x-axis is given by:
Calculate the amplitude, period and frequency of the oscillation after a long time has elapsed