What is the period of a pendulum bouncing off an inclined wall?

[Demakras]

Homework Statement

A pendulum hangs from an inclined wall. Suppose that this pendulum is released at an initial angle of 10 degrees and it bounces off the wall elastically when it reaches an angle of -5 degrees. What is the period of this pendulum?
View attachment pendulum.bmp

Homework Equations

T=2π√(l/g)

The Attempt at a Solution

A=10
θ=Acos(wt)
a=-A(w^2)cos(wt)
a=-(g/L)(-5)=-A(g/L)cos(wt)
(-5)=Acos(wt)
arccos((-5)/A)=wt
time to reach the wall t=(arccos((-5)/A))(L/g)=(1/3)π√L/g)
and then, I don't understand how it moves back up.

 

[AlexChandler]

Just thinking about it, the amount of time the pendulum spends traveling to the wall will be identical to the time it takes to travel back. The situation makes it so that the total distance traveled is 3/4 of what it would be if the wall were not there. The wall only affects the pendulum by cutting out 1/4 of the time it needs to travel to complete one period. Therefore I would suggest that the answer should be 3/4 of the period that the pendulum would have if the wall were not there.
Another way to think about it is (actually really the same) that you can just calculate the amount of time it takes to travel to the wall (which is 3/8 of the pendulum's normal period) and then multiply this by two, because the path back to the start point will take the same amount of time. then you get 6/8 (or 3/4) of the pendulum's normal period.

 

[Super_Leunam]

I'd say AlexChandler's solution is flawed. He is assuming that the bob will take 1/4 of a period to travel 1/4 of a round trip which is how most people would tend to think off guard. The motion is not uniform but is well known to be non-elementary. Fortunately, the small angle approximation applies here and a SHM ensues. We can use the analogy of SHM witha uniform circular one and see that the bob bounces off when its phase is such that the angular "position" is negative half of the amplitude. This happens at a phase of 2pi/3. It jumps off immediately to 4pi/3 (because of the elasticity of the collision with the wall) with the same angular frequency and finishes a round trip going through a total phase of 4pi/3...
w*T=4pi/3 and the period is therefore 4pi/(3w) T=4pi/3 * sqrt(L/g)