Question #1837

The equation of motion of a particle moving along the x-axis is given by:
d[sup]2[/sup]x/dt[sup]2[/sup]+4dx/dt+8x=20cos2t
Calculate the amplitude, period and frequency of the oscillation after a long time has elapsed.

Expert's answer

Let's find the solutions of the homogeneous differential equation:

d2x / dt2 + 4 dx/ dt +8x = 0

The characteristic equation :

λ^{2} + 4 λ +8 = 0

λ_{1} = -2 +i2; λ_{2} = -2 - i2

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

d2x / dt2 + 4 dx/ dt +8x = 0

The characteristic equation :

λ

λ

The CF of the equation is:

x = e

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

At t -> infinity x(t) -> e

Thus the amplitude is √5, the period is T = π, the frequncy is f = 1/T = 1/π

## Comments

Assignment Expert10.03.11, 17:39When 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.

nirmala10.03.11, 08:17sir<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

Assignment Expert09.03.11, 16:03Check the correct answer above please.

nirmala08.03.11, 12:51The 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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