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.
$[\text{Take}l=\frac{150}{4{\pi }^2}\text{m},g=10{\text{m/s}}^2]$

• A. $5\text{s}$
• B. $15\text{s}$
• C. $10\text{s}$
• D. $25\text{s}$

Answer 1

The correct option is A $5\text{s}$

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 $=m_{\text{total}}a,$ where $a$ respresents the acceleration of the motion.


Net imbalance in force
$\Delta F=-\left[4mg\right(l+x)+mg(l-x\left)\right]-\left[4mg\right(l-x)+mg(l+x\left)\right]$
$=-6mgx$ [downward -ve]
$\Delta F=m_{total}\times a$
$\Rightarrow -6mgx=(4m\times 2l+m\times 2l)\left(a\right)$
$\Rightarrow a=-\left(\frac{3g}{5l}\right)x$
Comparing this with $a=-{\omega }^{2}x$, we get
$\omega =\sqrt{\frac{3g}{5l}}$
Time period of oscillation $T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{5l}{3g}}$
From the data given in the question,
$l=\frac{150}{4{\pi }^2}\text{m}$ and $g=10{\text{m/s}}^2$
$\Rightarrow T=2\pi \sqrt{\frac{5\times 150}{3\times 10\times 4{\pi }^2}}=5\text{s}$
Thus, option (a) is the correct answer.