1. Speed Bump Dynamics

so far i have found the problem for when the tire is before the bump and during but i am having trouble with finding after it always turns out to be equal to the before so im looking for someone to check my work. to save time i scanned the problem and all of my work and put it on photobucket so any help would be great.
http://i1199.photobucket.com/albums/...238problem.jpg
http://i1199.photobucket.com/albums/.../mathstuff.jpg
http://i1199.photobucket.com/albums/...mathstuff0.jpg
http://i1199.photobucket.com/albums/...mathstuff2.jpg

thanks for the help

2. Your choice of a particular function for the method of undetermined coefficients is incorrect. You're going to have to have the function

$\displaystyle y_{p}=A\cos\left(\dfrac{\pi V t}{L}\right)+B\sin\left(\dfrac{\pi V t}{L}\right).$

In solving the DE for the region

$\displaystyle -\dfrac{L}{2V}<t<\dfrac{L}{2V},$

you're not applying the initial conditions at the correct location. It needs to be at

$\displaystyle -\dfrac{L}{2V},$ not $\displaystyle -\dfrac{\pi}{2V}.$

Try carrying those corrections through and see what happens.

Very interesting and fun problem, by the way! Thanks for posting!

3. ok

I know I didn't show the canceling out when I solved the problem but it states that L=pi so if you were to plug that into your suggestion it would give you the same thing unless canceled out wrong

4. Oh, gotcha. Ok, time to peruse your solution further, then. Hang on a bit.

5. In the mathstuff0.jpg file, where you're applying the initial condition to the derivative, I would simplify a bit:

$\displaystyle y'\left(-\dfrac{\pi}{2V}\right)=0=-C_{1}\left(\sin\left(-\dfrac{\pi}{2V}\right)+\dfrac{\cos^{2}\left(-\dfrac{\pi}{2V}\right)}{\sin\left(-\dfrac{\pi}{2V}\right)}\right)+\dfrac{V}{1-V^{2}}$

$\displaystyle =-C_{1}\left(\dfrac{\sin^{2}\left(-\dfrac{\pi}{2V}\right)+\cos^{2}\left(-\dfrac{\pi}{2V}\right)}{\sin\left(-\dfrac{\pi}{2V}\right)}\right)+\dfrac{V}{1-V^{2}}$

$\displaystyle =-C_{1}\left(\dfrac{1}{\sin\left(-\dfrac{\pi}{2V}\right)}\right)+\dfrac{V}{1-V^{2}}\dots$

I think your simplifications are going awry somewhere, because the solution for this region that I get is

$\displaystyle y(t)=\dfrac{\cos(V t)-V\sin\left(t+\dfrac{\pi}{2V}\right)}{1-V^{2}}.$

Also note that, once you've got this solution, you'll need to obtain the initial conditions for the far right region by plugging in $\displaystyle \pi/(2V)$ into this solution. That's why you're not going to get zero for the far right region: you don't have zero initial conditions (well, most likely you don't).

Make sense?

6. got it

I eventually solved it just took some time and determination and one thing I had wrong in my work was a negative sign was off in C1 so thanks again for the help

7. You're welcome for whatever help I gave!