2
$\begingroup$

A tutorial (page 17) on ODE scaling takes the equation $m\ddot{u}+c\dot{u}+ku=F_0sin(\omega t)$, with $u$ a function of $t$, and substitutes $\eta=u/a$, $\tau=t/b$ to get:

$$\frac{ma}{b^2}\ddot{\eta}+\frac{ca}{b}\dot{\eta}+ka\eta=F_0sin(\omega\tau)$$

Now, when I differentiate $\eta=u/a$, I get $\dot{\eta}=t\dot{u}/a=\tau b\dot{u}/a$. Substituting, the final result is:

$$\frac{ma}{\tau^2b^2}\ddot{\eta}+\frac{ca}{\tau b}\dot{\eta}+ka\eta=F_0sin(\omega\tau b)$$

That's not the same. What am I missing?

$\endgroup$
2
$\begingroup$

Gerry has already explained the differences on the left-hand side. On the right-hand side, this is a typo in the book. The next equation has $\sin\omega'\tau$ instead of $\sin\omega\tau$, with $\omega'=\omega\sqrt{m/k}$ and $b=\sqrt{m/k}$, so this is $\sin\omega\tau b$ as you expected. So it should be either $\sin\omega t$ or $\sin\omega'\tau$ on the right-hand side.

$\endgroup$
2
$\begingroup$

Differentiating $\eta=u/a$ gives $\eta'=u'/a$, not $\eta'=tu'/a$, no?

$\endgroup$
  • $\begingroup$ Heh. Of course, you are right. I would be much obliged, if you coud explain how to get their result. $\endgroup$ – Don Reba Aug 24 '11 at 8:53
  • $\begingroup$ @Don, I'm a little puzzled myself as to how they get $\omega\tau$ on the right side, where your $\omega\tau b$ looks better to me. Sorry, don't have the time to work through it now, but I'm sure someone will be along to help out. $\endgroup$ – Gerry Myerson Aug 24 '11 at 9:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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