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Two uniform ropes having linear mass densities m and 4m are joined to form a closed loop. The loop is hanging over a fixed frictionless small pulley with the lighter rope above and denser rope below, as shown in the figure. (Figure represents equilibrium position). Now if point A (joint) is slightly displaced in downward direction and released, it is found that the loop performs SHM. Find the time period of oscillation.
[Take l=1504π2 m,g=10 m/s2]


A
5 s
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B
15 s
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C
10 s
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D
25 s
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Solution

The correct option is A 5 s
Let point A be displaced in the downward direction by x.
The imbalance of weights on the two sides would give rise to the restoring force.
Net imbalance of force =mtotal a, where a respresents the acceleration of the motion.


Net imbalance in force
ΔF=[4mg(l+x)+mg(lx)][4mg(lx)+mg(l+x)]
=6mgx [downward -ve]
ΔF=mtotal×a
6mgx=(4m×2l+m×2l)(a)
a=(3g5l)x
Comparing this with a=ω2x, we get
ω=3g5l
Time period of oscillation T=2πω=2π5l3g
From the data given in the question,
l=1504π2 m and g=10 m/s2
T=2π5×1503×10×4π2=5 s
Thus, option (a) is the correct answer.

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