# response of unit step input in harmonically oscillating system

As far as I've understood or misunderstood in constant coefficient second order differential equation $$\frac{d^2y}{dt^2} + b \frac{dy}{dt} +cy = ef(t)$$ $b$, $c$ being constants, $f(t)$ the input to the system, $y$ being response of the system.

Let, $$\frac{d^2y}{dt^2} +\omega_0^2y = u(t)k$$ such that $k$ is non zero be a system and $u(t)$ a unit step function.

How to physically visualize this system(or input to the system), isn't the $u(t)$ like applying constant force to the system? Will it not bring the system to halt after certain time? Like if we keep poking a mass-spring system with constant force only in one direction??

But the solution seems different. Please Help me to clear this simple misconception.

Thanks you!!

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 Thanks @Rahul Narain ... how do you put that ugly equation beautifully like that?? – experimentX Jun 7 '12 at 20:12 Use two dollar signs ($$...$$) to create "display math" that is centered and has larger fractions and super/subscripts. – Rahul Narain Jun 7 '12 at 20:14

Yes, after time $t = 0$, the step function $ku(t)$ is like applying a constant force $k$ to the system, but no, it will not bring the system to a halt. In fact, it merely shifts the equilibrium position of the system, but the system will continue to oscillate about the new equilibrium position just as before. Since we only care about $t \ge 0$, let's just assume the force is a constant $k$. Observe that if $\newcommand{\d}{\mathrm d}$ $$\frac{\d^2y}{\d t^2} + \omega^2y = k,$$ this is equivalent to $$\frac{\d^2y}{\d t^2} + \omega^2\left(y - \frac k{\omega^2}\right) = 0,$$ and if you let $\tilde y = y - k/\omega^2$, you get back the equation of the simple harmonic oscillator centered at $0$, $$\frac{\d^2\tilde y}{\d t^2} + \omega^2\tilde y = 0.$$ So the system with a constant force behaves exactly like the unforced system, only shifted by $k/\omega^2$.