The CF of the equation is: x = e-2t (Acos 2t +B sin 2t) We try a particular integral of the form Csin(2t)+Dcos(2t) d2x/dt2 = -4Csin(2t) - 4D cos(2t); dx/dt = 2C cos(2t) - 2D sin(2t) -4Csin(2t) - 4D cos(2t) + 4(2C cos(2t) - 2D sin(2t)) + 8(Csin(2t)+Dcos(2t))= 20 cos(2t) cos(2t) (8C - 4D+8D) - sin(2t) (8D + 4C - 8C) = 20 cos(2t) We obtained the system of equations for C and D: 8C - 4D + 8D = 20 8C - 8D - 4C = 0 ;
2C + D = 5 C = 2D; D + 4D = 5; D = 1, C = 2 Particular integral is PI = 2sin(2t) + cos(2t). Hence x(t) = e-2t (Acos 2t +B sin 2t) + 2sin(2t) - cos(2t).
At t -> infinity x(t) -> e-2t (Acos 2t +B sin 2t) + 2sin(2t) + cos(2t) = √5sin(2t + arctan(0.5)) Thus the amplitude is √5, the period is T = π, the frequncy is f = 1/T = 1/π
When a long time elapsed we take t -> ∞, therefore& e-2t -> 0 and
the whole term e-2t (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) =√(22+12) sin (2t +
arctan(1/2)) The period is equal to 2Ï€ /(the coefficient at _t_) =
2π/2 = π.
nirmala
10.03.11, 08:17
sir Please help me, i have not understood this step, why we have taken
t->t;infinite and how & e-2t (Acos 2t +B sin 2t) + 2sin(2t) + cos(2t)
& converted to = A + √5sin(2t + arctan(0.5)) At t -> infinity& x(t)
->& e-2t (Acos 2t +B sin 2t) + 2sin(2t) + cos(2t) = A + √5sin(2t +
arctan(0.5)) how we got the amplitude = √5, and the period is T =
Ï€, the frequency is f = 1/T& = 1/Ï€
Assignment Expert
09.03.11, 16:03
Check the correct answer above please.
nirmala
08.03.11, 12:51
The equation of motion of a particle moving along the x-axis is given
by: d2x/dt2+4dx/dt+8x=20cos2t Calculate the amplitude, period and
frequency of the oscillation after a long time has elapsed
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When a long time elapsed we take t -> ∞, therefore& e-2t -> 0 and the whole term e-2t (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) =√(22+12) sin (2t + arctan(1/2)) The period is equal to 2π /(the coefficient at _t_) = 2π/2 = π.
sir Please help me, i have not understood this step, why we have taken t->t;infinite and how & e-2t (Acos 2t +B sin 2t) + 2sin(2t) + cos(2t) & converted to = A + √5sin(2t + arctan(0.5)) At t -> infinity& x(t) ->& e-2t (Acos 2t +B sin 2t) + 2sin(2t) + cos(2t) = A + √5sin(2t + arctan(0.5)) 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: d2x/dt2+4dx/dt+8x=20cos2t Calculate the amplitude, period and frequency of the oscillation after a long time has elapsed
Leave a comment