Question

A pendulum has time period T for small oscillations.An obstacle p is situated below the point of suspension from a top at a distance 3l/4 .The pendulum is at released from rest . Throughout the motion thr moving string makes an anglr with vertical.Find thr time after which the penndulum returns back to its initial position?

Answer 1

The pendulum will return to its original position after 3/4th of its previous time period.

According to the question, the total time period will be sum of half of the time period of both the oscillations.

T = T₁/2 + T₂/2

For the larger oscillation.

T₁ = 2π√(l/g)

For the smaller oscillations:

l₂ = l - 3l/4 = l/4

Hence ,

T₂ = 2π√(l/4g)

Hence the total time period,

T = T₁/2 + T₂/2

= π√(l/g) + π√(l/4g)

= π(√(l/g) + √(l/g)/2)

= 3π/2[ √(l/g)]

= 3/4[2π√l/g]

= 3T/4

Hence the pendulum will return to its original position after 3/4th of its previous time period.