Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!!

share|improve this question
    
Thanks @Rahul Narain ... how do you put that ugly equation beautifully like that?? –  Santosh Linkha 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 Jun 7 '12 at 20:14
add comment

1 Answer

up vote 2 down vote accepted

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 $k/\omega^2$.

Perhaps you're imagining the constant force to be like holding the oscillator and pushing it to one side. But when you do that in real life, you're also opposing the relative motion of the oscillator with respect to your hand, and that is what damps out the motion of the system. A constant force is not like that; it continues to push in one direction, no matter whether the oscillator is above or below its new equilibrium, no matter whether it is moving towards or away from it.

Or just think of a spring held up at one end, with a weight at the other end being pulled down by gravity. It's not the gravitational force that makes it eventually come to a stop, it's the friction in the spring itself.

share|improve this answer
    
thanks :) ... it helped to clear a lot of misconception ... i am also assuming that the equilibrium point will be top point for oscillation –  Santosh Linkha Jun 8 '12 at 9:17
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.