A ball is suspended by a thread of length L at the point O on a wall which is inclined to the vertical by α. The thread with the ball is displaced by a small angle β away from the vertical and also away from the wall. If the ball is released, the period of oscillation of the pendulum when β > α will be?

The motion of the string can be represented as an angular displacement from the mean position, given as:
θ = βsin(ωt)
Time taken from displacement from mean position to β and back to mean position = t₁ = T/2
where T is the time period of the oscillation without the wall barrier.
Thus t₁ = 1/2 × 2π√Lg = π√Lg
Time taken to displace from mean position to α can be found by:
α = βsin(ωt) ⇒ t = 1/ωsin⁻¹(α/β) = √Lg sin⁻¹(α/β)
Thus time taken to displace from mean position to α and back to mean position = t₂ = 2√Lg sin⁻¹(α/β)
Thus the time period of the oscillation = t₁ + t₂ = √Lg[π + 2sin⁻¹(α/β)]