# A simple pendulum of time period 1s and length l is hung from a fixed support at O

246 views
in Physics
edited

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 (Fig. 14.11).The amplitude is θ0. The string snaps at θ = θ0/2. Find the time taken by the bob to hit the ground. Also find distance from A where bob hits the ground. Assume θ0 to be small so that sinθ0 = θand cosθ0 = 1.

by (13.6k points)
selected

Assume that t = 0 when θ = θ2. Then,

θ = θ2 cos ωt

Given a seconds pendulum ω = 2π

At time t1, let θ = θ0/2

cos 2πt1 = 1/2  t1=1/6

θ = θ02π sin 2πt [θ = dθ/dt]

At t1= 1/6

θ = -θ02π sin 2π/6 = - √3πθ0

Thus the linear velocity is u = –3πθ0l perpendicular to the string.

The vertical component is uy = -3πθ0l sinθ0 and the horizontal component is uy = -3πθ0l cosθ0

At the time it snaps, the vertical height is H' = H+l(1 - cos(θ0/2))

Let the time required for fall be t, then H' uyt + gt2 (notice g is also in the negative direction)

Or, (1/2)gt2 + 3πθ0l sinθ0t - H' = o

Neglecting terms of order θ20 and heigher

t≃  √2H'/g.

Now H' ≃ H + l(1–1) = H∴ t≃ √2H/g

The distance travelled in the x direction is uxt to the left of where it snapped.

X = 3πθ0lcosθ0√2H/g

To order of θ0.

X =3πθ0l√2H/g = √(6H/g)θ0l.

At the time of snapping, the bob was l sinθ0 ≃ lθ0 distance from A.

Thus, the distance from A is

0 - √(6H/g)lθ0 = lθ0 (1 - √(6H/g).