Question #286416

An Over Damped harmonic oscillator satisfied the equation (d²x/dt²)+(10dx/dt)+16x=0. At time t=0 the particle is projected from the point x=1 towards the origin with the speed "U". Find x(t) in subsequent motion.

Expert's answer

The equation of motion of overdamped oscillator is given by

d2xdt2+10dxdt+16x=0\frac{d²x}{dt²}+10\frac{dx}{dt}+16x=0

Let

x=eλtx=e^{\lambda t}

The characteristic equation:

λ2+10λ+16=0\lambda^2+10\lambda+16=0

Roots:

λ1=−8,λ2=−2\lambda_1=-8,\quad \lambda_2=-2

Solution:

x(t)=C1e−8t+C2e−2tx(t)=C_1e^{-8t}+C_2e^{-2t}

The initial conditions

x(0)=1,x′(0)=−ux(0)=1,\quad x'(0)=-u

give

x(0)=C1+C2=1x′(0)=−8C1−2C2=−ux(0)=C_1+C_2=1\\ x'(0)=-8C_1-2C_2=-u

Finally, we get

x(t)=13(u/2−1)e−8t−13(u/2−4)e−2tx(t)=\frac{1}{3}\left(u/2-1\right)e^{-8t}-\frac{1}{3}\left(u/2-4\right)e^{-2t}


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