    Question

# Two simple pendulums A and B having length l and l4 respectively are released from the positions shown in the figure. If the time after release when the two strings become parallel for the first time is given by 2πx√lg s, then the value of x will be [Assume angle θ is very small and motion starts at t=0] A
3.00
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B
3
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C
3.0
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Solution

## Let at any time t after release, the angular position of the strings be θ1 and θ2 w.r.t XX′. We know that, angular frequency ω=√gl So, ω1=√gl/4=2√gl and ω2=√gl Now, θ1=θcos(ω1t) ⇒θ1=θcos(2√glt) .......(1) And, θ2=−θcos(ω2t) ⇒θ2=θcos(−√glt) ..........(2) For the strings to become parallel, the displacement angles should be the same. So, cos(2√glt)=cos(−√glt) [from (1) and (2)] ⇒2√glt=2nπ±√glt For n=0,t=0 [trivial] For n=1,t=2π3√lg(minimum) and t=2π√lg So, after t=2π3√lg, two strings will become parallel for the first time. t=2πx√lg [given] Hence, x=3  Suggest Corrections  0      Explore more