Answer to Question #1837 in Mechanics | Relativity for nirmala

Question

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.

Answer 1

Let's find the solutions of the homogeneous differential equation:
d2x / dt2 + 4 dx/ dt +8x = 0

The characteristic equation :
λ² + 4 λ +8 = 0
λ₁ = -2 +i2; λ₂ = -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