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A ball is suspended by a thread of lengt...

A ball is suspended by a thread of length l at the point O on an incline wall as shown. The inclination of the wall with the vertical is (a)the thread is displacement through a small angle away from the vertical and (b) the ball is released. Find the period of oscillation of pendulum. Consider both cases
a. `alpha gt beta`
b. `alpha lt beta`
Assuming that any impact between the wall and the ball is elastic.

Text Solution

Verified by Experts

If `alpha gt beta`, the ball does not collide which the wall and it performed full oscillation like simple pendulum. Thus,
`Period = 2 pi sqrt((l)/(g))`

b. If `alpha lt beta`, the ball collide with the wall and rebounds with same speed the motion of ball from `A and Q` is one part of a simple pendulum time Period of ball`= 2 (t_(AQ)`
Consider A as the starting point `(t = 0)`, equition of motion is `x(t) = A cos omega t`.
`x (t) = l beta cos omega t`, because amplitude `= A = l beta`
Time from A to Q is the time t when x becomes - l alpha`
`implies - l alpha - l beta cos omega t implies t = t_(AQ)` = l// omega cos^(-1) ((-alpha)/(beta))`
The return path from `Q and A` will involve the same time interval. Hence time period of ball `= 2 t_(AQ)`
`= (2)/(omega) cos^(-1) (-(alpha)/(beta)) = 2 sqrt((t)/(g)) cos^(-1) ((-alpha)/(beta))`
`= 2 pi sqrt((l)/(g)) - 2 sqrt ((l)/(g)) cos ^(-1) ((alpha)/(beta))`
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