Answer:
At t = t1 and θ= θ0/a
We have, T = 1 sec
θ0/2 = θ0 cos 2π/T t1
d θ/dt = – θ0 2π sin 2πt
At t = 1/6, that is when θ = θ0/2
d θ/dt = – θ0π√3
v/l = – θ0π√3
v = -θ0π√3l
θ0l [1/2 – π√6H/g] gives us the distance between the bob from A.
A simple pendulum of time period 1s and length l is hung from a fixed support at O, such that the bob is at a distance H vertically above A on the ground. The amplitude is θo. The string snaps at θ = θ0/2. Find the time taken by the bob to hit the ground. Also, find the distance from A where bob hits the ground. Assume θ0 to be small so that sin θo, θo and cos θ0 1.
A simple pendulum of time period 1s and length l is hung from a fixed support at O, such that the bob is at a distance H vertically above A on the ground. The amplitude is θo. The string snaps at θ = θ0/2. Find the time taken by the bob to hit the ground. Also, find the distance from A where bob hits the ground. Assume θ0 to be small so that sin θo, θo and cos θ0 1.