A ball is suspended by a thread of length $l$ at the point O on an inclined wall as shown. The inclination of the wall with the vertical is $α$ . The thread is displaced through a small angle $β (> α)$ away from the vertical and the ball is released. The period of oscillation of pendulum( the collision between the wall and the ball is elastic) is

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$2 π \sqrt{\frac{l \sin α}{g \cos β}}$

$2 π \sqrt{\frac{l}{g}} + 2 \sqrt{\frac{l}{g}} \cdot \cos^{- 1} (\frac{α}{β})$

$2 π \sqrt{\frac{l}{g}} - 2 \sqrt{\frac{l}{g}} \cdot \cos^{- 1} (\frac{α}{β})$

$2 π \sqrt{\frac{l \sin α}{g}}$

answer is D.

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Detailed Solution

If $α < β,$ , the ball collides with the wall and rebounds with same speed. The motion of ball from A to Q is one part of a simple pendulum. Time period of ball
$= 2 (t_{A Q})$ . Consider A as the starting point (t = 0) .
Equation of motion is x(t) = $A \cos ω t$ $x (t) = ℓ β \cos ω t$ , because amplitude $= A = ℓ β$ time from A to Q is the time t when x becomes $- ℓ α . \Rightarrow - ℓ α = ℓ β \cos ω t$

$\Rightarrow t = t_{A Q} = 1 / ω \cos^{- 1} (\frac{- α}{β})$

The return path from Q to A will involve the same time interval. Hence time period of ball $= 2 t_{A Q}$ .

$\frac{2}{ω} \cos^{- 1} (- \frac{α}{β}) = 2 \sqrt{\frac{l}{g}} \cos^{- 1} (\frac{- α}{β}) = 2 π \sqrt{\frac{ℓ}{g}} - 2 \sqrt{\frac{ℓ}{g}} \cos^{- 1} (\frac{α}{β}) .$