A simple pendulum of length l swimings from a small angle `theta` . Its swinging is constrained by the smooth inclined planes OP and PC. Assuming elastic collision of the bob with the plane PC, find
angular amplitude for the motion of the bob in the left hand side of its mean possition.
.

Explanation

a. Let us assume that the bob swings a maximum a maximum angle `beta` in the left hand side of the mean position in the absence of the position PC of the wall.

Conserving energy at A and D, we have
`U_(A) + K_(A) = U_(D) + K_(D)`
where `U_(A) = mgh_(A). U_(B) = mgh_(B)`
and `K_(A) = K_(D) = 0`
because the bob compes to rest at the exterme positions.
Then, we have `h_(A) = h_(D)`
Substituting `h_(A) = OA (1 - cos theta)`
and `h_(0) = PA (1 - cos theta)` we have
`OA (1 - cos theta) = PD (1 - cos beta)`
Substituting `OA = l. PD = (l - d)`, we have
`beta = (sqrt((l)/(l - h))) theta`
b. Since the bob from A to B
`t_(AB) = (T)/(4)`
where `T = 2 pi sqrt((l)/(g))`
Then `t_(AB) = (pi)/(2) sqrt((l)/(g))`
The time of miotion from B to C can be found by subsituting
`t = t_(BC). theta = 0`
and `theta = beta sqrt((l//(l - h)) theta)`
in the equation `theta = theta_(0) sin omega t`.
where `omega = sqrt(g//(l - h))`
for the swinging position of the string from B to C. This gives
`theta = sqrt((l)/(l - h)) theta sin sqrt((g)/(l - h)) t_(BC)`
Then, we have
` t_(BC) = sqrt((l - h)/(g)) sin ^(-1) sqrt((l - h)/(l))`
Finally, substituting `t_(AB) and t_(BC)` in the expression of time . `T = 2 (t_(AB) + t_(BC))` , we have
`T = pi sqrt((l)/(g)) + sqrt((l - h)/(l)) sin^(-1) sqrt((l - h)/(l))`