A simple pendulum of time period 1s and length l is hung from a fixed support at 0. Such that the bob is at a distance H vertically above A on the ground (figure) the amplitude is `theta_(0)` the string snaps at `0=theta_(0)//2.` Find the time taken by the bob to hit the ground. Also find distance from A where bob hits the ground. Assume `theta_(0)` to be small, so that `sintheta_(0)~~theta_(0)and costheta_(0)~~1.`

Explanation

Consider the diagram.

Let us assume `t=0" when"theta=theta_(0) "then" theta =theta_(0) cosomegat`Given a seconds pendulum `omega=2pirArrtheta=theta_(0)cos2pit" "....(i)`
At time `t_(1)" let " theta=0_(0)//2`
`:.cos2pit_(1)=1//2rArrt_(1)=(1)/(6)" " [:'cos2pit_(1)=cos""(pi)/(3)rArr2pit_(1)=(pi)/(3)]`
`(d theta)/(dt)=-( theta_(0)2pi)sin2pit" " ["from Eq.(i)"]`
At `t=t_(1)=(1)/(6)`
`(d theta)/(dt)=-theta_(0)2pisin.(2pi)/(6)=-sqrt(3) pi theta_(0)`
Negative sign shows that it is going left.
Thus, the linear velocity is
`u=-sqrt(3)pi theta_(0)l" perperndicular ot the string."`
The vertical component is
`u_(y)=-sqrt(3)pi theta_(0)l" "sin( theta_(0)//2) `
and the horizontal component is
`u_(x)=-sqrt(3)pi theta_(0)lcos( theta_(0)//2)`
At the time it snaps, the vertical height is
`H'=H+l(1-cos( theta_(0)//2))" ".....(ii)`
Let the time required for fall be t, then
`H'=u_(y)t+(1//2)"gt"^(2)" "("notice g also in the negative direction")`
or `(1)/(2)"gt"^(2)+sqrt3pitheta_(0)lsin""theta_(0)/(2)-H'=0`
`-sqrt3pitheta_(0)lsin""(theta_(0))/(2)+-sqrt(3pitheta_(0)^(2)l^(2)sin^(2)""theta_(0)/(2)+2gH')`
`:. t=(sqrt(3)pitheta_(0)lsin.(theta_(0))/(2)+-sqrt(3pitheta_(0)^(2)l^(2)sin^(2) .(theta_(0))/(2))+2gh')/(g)`
`=(sqrt(3)pil(theta_(0)^(2))/(2)+-sqrt(3pi^(2)((theta_(0)^(4))/(4))l^(2)+2gH'))/(g)[ :. sin.(theta_(0))/(2)=(theta_(0))/(2) "for small ange"]`
Given that `theta_(0)` is small, hence neglecting terms ot order `theta_(0)^(2)` and higher
`t=(sqrt(2H)/(g)) " "` [from Eq . .. (iii)]
Now `H'~~H+l(1-1) " " [ :.costheta _(0)//2~~1]`
`rArrt=sqrt((2H)/(g))" " ` [ from Eq. (ii) ]
The distance travelled in the x-direction is u `underset(x)t` to the left of where the bob is snapped
`X=Ux=tsqrt(3)pitheta_(0)lCos ((theta_(0))/(2))sqrt((2H_(s))/(g))`
As `theta_(0)` is small `rArrcos((theta_(0))/(2))~~1` `X=sqrt(3)pitheta_(0)lsqrt((6H)/(g))theta_(0)pil`
At the time of snapping, the bob was a horizontal distance of `l sin (theta_(0)//2)~~l(theta_(0))/(2)` from A
Thus, the distance of bob from A where it meets the ground is
`(l theta_(0))/(2)-X=(l theta_(0))/(2)-sqrt((6H)/(g))theta_(0)lpi`
`=theta_(0)l((1)/(2)-pisqrt((6H)/(g)))`