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

Here, `theta_(0)` is the angular amplitude of simple pendulum and time period, `T=1s`. The angular displacement of the bob of the pendulum at any inisant t is given by
`theta=theta_(0)sinomegat=theta_(0)sin((2pi)/(T))t`
`=theta_(0)sin((2pi)/(1))t=theta_(0)sin2pit` ..(i)
`( :' T=1s.)`
At time `t_(1)`, let `theta=theta_(0)//2` then from (i), we have
`(theta_(0))/(2)=theta_(0)sin 2pit_(1)` or `2pit_(1)=(pi)/(6) or t_(1)=(1)/(12)s`
Differentiating (i) w.r.t. t, we have
`(d theta)/(dt)=2pitheta_(0)cos2pit`
When `t=t_(1)=(1)/(12),` then `(d theta)/(dt)=2pithet_(0)cos2pixx(1)/(12)=2pitheta_(0)xx(sqrt(3))/(2)=sqrt(3)pitheta_(0)`
Angular velocity of the bob of simple pendulum is `omega=(d theta)/(dt)=sqrt(3)pitheta_(0)`
velocity of the bob of simple pendulum, `v=romega=lomega=l(sqrt(3)pitheta_(0))`
It is acting perpendicular to string.
The vertical component velocity of bob, `v_(y)=vsin(theta_(0)//2)=l sqrt(3)pitheta_(0)sin(theta_(0)//2)`
Horizontal component velocity of bob, `v_(x)=vcos(theta_(0)//2)=lsqrt(3)pitheta_(0)cos(theta_(0)//2)`
The height of the bob when the pendulum snaps is , `H^(')+H+lcos(theta_(0)//2)=H+l[1-cos(theta_(0)//2)]`
If `t'` is the time taken by the bob to fall through height `H(')` , then using the relation , `S=ut+(1)/(2)at^(2)`,
we have `s=H^('), u^(')=v_(y)=lsqrt(3)pitheta_(0)sin(theta_(0)//2), t=?`
`:. H^(')=[sqrt(3)pitheta_(0)lsin(theta_(0)//2)]t+(1)/(2)g t^(2)`
or `t^(2)+[(2sqrt(3)pitheta_(0)lsin(theta_(0)//2))/(g)]t-(2H^('))/(g)=0`
`:. t=(-[(2sqrt(3)theta_(0)lsin(theta_(0)//2))/(g)]+-sqrt([(12pi^(2)theta_(0)^(2)l^(2)sin^(2)(theta_(0)//2))/(g^(2))]+4xx2H^(')lg))/(2)`
`=(-[sqrt(3)pitheta_(0)lsin(theta_(0)//2)]+-sqrt(3pi^(2)theta_(0)^(2)l^(2)sin^(2)(theta_(0)//2)+2H^(')g))/(g)`
`=(-sqrt(3)pi(theta_(0)//2)+- sqrt(3pi^(2)theta_(0)^(2)(theta_(0)^(2)//4)+2H^(')g))/(g)`
As `theta_(0)` is small so the terms with `theta_(0)^(2)` beign very very small can be neglected.
`:. t=(sqrt(2H^(')g))/(g)=sqrt((2H^('))/(g)`
Now, `H^(')=H+l-lxxl=H` `[:'cos(theta_(0)//2)=1]`
` :. t=sqrt((2H^('))/(g))=sqrt((2H)/(g))`
The horizontal distance travelled by bob after snapping
`x=v_(x)t=[sqrt(3)lpitheta_(0)cos(theta_(0)//2)]sqrt((2H)/(g))=sqrt(3)lpitheta_(0)xx1xxsqrt(2H//g)=sqrt((6H)/(g))xxtheta_(0)lpi`j
At the time of snapping, the horizontal distance of the bob from point A is `=lsin(theta_(0)//2)=l theta_(0)//2`
Thus the distance of bob from A where it meets the ground is
`=(l theta_(0))/(2)-sqrt((6H)/(g))l pi theta_(0)=l theta_(0)[(1)/(2)-pisqrt((6H)/(g))`