Question

The bob of a swinging seconds pendulum (one whose time period is 2 s) has a small
speed v, at its lowest point. Its height from this lowest point 2.25 s after passing through
it is given by

Answer 1

Answer:

This is the answer of your question.

Answer 2

Answer: v02/4g

Explanation:

Given: Time period of swinging seconds pendulum is 2s.

It means it completes its one revolution in 2s.

So the bob will reach its lowest point after 2s.

As it is travelling further for 1/4 sec. i.e. t = 2.25s

Hence, v = Aω cos ωt = v0 cos π/4

= v0/√2 { since Aω = v0 = vmax}

∵ displacement x = A sin ωt

Velocity, v = dx/dt = Aω cos ωt

From law of conservation of energy

½ mv02 = ½ m(v02/2)+mgh

½ v02 - ½ (v02/2) = gh

(1/2)v02(1-1/2) = gh

v02/4 = gh

⇒ h = v02/4

source: wbjee 2020 question 1