    Question

# 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 sLet 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(l−x)]−[4mg(l−x)+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=2π√5l3g From the data given in the question, l=1504π2 m and g=10 m/s2 ⇒T=2π√5×1503×10×4π2=5 s Thus, option (a) is the correct answer.  Suggest Corrections  0      Explore more