Submitted by Nomreg on Thu, 02/11/2016 - 07:51
Forums:
Hello,
I've been having a go at this question however can't get the first D.E out. I started off by saying $m(\frac {d^2y}{dt^2}) = m(g-f) - \frac {\lambda (a + y)}{l}$, with $\frac{\lambda a}{l}=mg$, however I'm not certain I've put $f$ in the right place? Even so when I expand the above I get very close to the answer except I'm missing the $g$ term. So i was wondering whether the question meant $y$ is the extension of the elastic band from it's natural length instead of equilibrium length, as this would make my equation work out?
Hi there, I tried this
Hi there, I tried this question and ran into the same problem as you so I think it's reasonable to assume $y$ is the extension of the string from it's natural length.
Your equations are correct apart from the part already mentioned but it might be easier to see why you've inserted $f$ correctly by considering the acceleration (downwards) of the particle at the end of the string. This is just $\ddot{y}+f$ so, by Newton's second law, we have $$m(\ddot{y}+f)=mg-\frac{\lambda y}{l},$$
Hope that helps.
Yes that's great thanks! I
Yes that's great thanks! I originally wrote your final equation but then confused myself because the $mg$ wasn't causing it to accelerate at $f$. I'll give the rest of the question a go now. Thanks again.