Question

A point particle, mass of m, starts at the top of a (frictionless) ramp of width 'a'. There is a mirror ramp next to this one which it in turn slides up. Using potential and kinetic energy, work out the period of oscillation to get back to the same point (as a function of 'a')

Answer 1

Do you know how to start a problem like this?

Answer 2

Yes, I've got an answer which I suspect is wrong

Answer 3

Oh, yes, you're also given the angle of inclination

Answer 4

Alright, well, here's mine:

Answer 5

What I had:
At the top:
\[V=V _{0}\]\[T=0\]
\[V _{0}=mgh=mg atan \theta\]
Generally:
\[V =mg atan \theta-mg xtan \theta =V _{0}-mg xtan \theta\]
\[E=T+V=0+V _{0}\]\[T=mgxtan \theta\]
\[v=\sqrt{\frac{2mgxtan \theta}{m}}\]
Finding the average speed:
\[\frac{\int\limits_{0}^{a}k \sqrt{x}}{a}\]
\[\frac{2ka^\frac{3}{2}}{3a}\]
\[\frac{2ka^\frac{1}{2}}{3}\]
Now use t=distance(a)/speed and multiply by 2.

Answer 6

*Integrating wrt dx

Answer 7

Let me type this up and then put it in, I'll be a minute...

Answer 8

I may be a while in replying to your coming answer

Answer 9

\[ T = \int dt = \int \frac{ds}{v} = \sec{\theta}\int \frac{dx}{v} \]
let x = -a at the left edge, and x = 0 at the right edge (for simplicity). The time it takes to get to the bottom is
\[T =\sec(\theta) \int_{-a}^0 \frac{dx}{\sqrt{2g(a\tan(\theta) + x\tan(\theta))}} = \frac{\sec{\theta}}{\sqrt{2g\tan(\theta)}}\int_{-a}^0 \frac{dx}{\sqrt{a+x}} = \sec(\theta)\sqrt{\frac{2a}{g\tan(\theta)}} \]
The full answer is just four times this. We can check to see if this makes sense by expressing a in terms of the height and the angle:
\[a = \frac{h}{\tan(\theta}\]
\[T_{1/4} = \sec(\theta) \sqrt{ \frac{2h}{g\tan^2(\theta)}} = \frac{1}{\sin(\theta)} \sqrt{\frac{2h}{g}}\]

As theta goes to pi/2, this expression becomes the time it takes for an object at rest to fall a height h, which is just what we expect. In conclusion, I believe that the expression you're looking for is
\[ T = 4\sec(\theta) \sqrt{\frac{2a}{g\tan(\theta)}} \]

Answer 10

Thanks for that- I messed up with not integrating with respect to displacement