Question #103278
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
1
Expert's answer
2020-02-24T10:04:51-0500

The equation of motion of a particle moving along the x-axis is given by


d2xdt2+4dxdt+8x=20cos2t.\frac{d²x}{dt²} + 4\frac{dx}{dt} + 8x = 20\cos2t.

The solution


x=Acos2t+Bsin2t.x=A\cos2t+B\sin2t.


We obtain


4Acos2t4Bsin2t8Asin2t+8Bcos2t-4A\cos2t-4B\sin2t - 8A\sin2t+8B\cos2t+8Acos2t+8Bsin2t=20cos2t.+ 8A\cos2t+8B\sin2t = 20\cos2t.

Hence


4A+8B=20,8A+4B=0.4A+8B=20,\quad -8A+4B=0.B=2A,  A=1.B=2A, \; A=1.x=cos2t+2sin2t=5cos(2t+arctan2).x=\cos2t+2\sin2t=\sqrt{5}\cos(2t+\arctan2).

Therefore:

amplitude a=5;a=\sqrt{5};

period T=2πω=π;T=\frac{2\pi}{\omega}=\pi;

frequency f=1/T=1/π.f=1/T=1/\pi.


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