Two identical ideal springs of spring constant $1000 \mathrm{~N} / \mathrm{m}$ are connected by an i

Question

Two identical ideal springs of spring constant $1000 \mathrm{~N} / \mathrm{m}$ are connected by an ideal pulley as shown and system is arranged in vertical plane. At equilibrium $	heta$ is $60^{\circ}$ and masses $m_1$ and $m_2$ are 2 kg and 3 kg respectively. The elongation in each spring when $	heta$ is $60^{\circ}$ is p cm . The value of $\frac{\sqrt{3} \mathrm{p}}{1.6}$ is :

Professor PC Thomas Classes · Answer

$\begin{aligned} & 2 \mathrm{kx} \sin 	heta=2 \mathrm{~T}=\left(\frac{4 \mathrm{~m}, \mathrm{~m}_2}{\mathrm{~m}_1+\mathrm{m}_2}ight) \mathrm{g} \ & \mathrm{kx} \frac{\sqrt{3}}{2}=\frac{2 	imes 2 	imes 3}{5} 	imes 10 \ & 1000 	imes \frac{\mathrm{x}}{2} 	imes \sqrt{3}=8 	imes 3 \ & \frac{\mathrm{x}}{2}=\frac{\sqrt{3}}{125} 	imes 100 \mathrm{~cm} \ & \frac{\mathrm{x}}{2}=\frac{\sqrt{3}}{5} 	imes 4 \ & \frac{\mathrm{P}}{2}=\frac{\mathrm{x}}{2}=\sqrt{3} 	imes 0.8 \ & \frac{\sqrt{3} \mathrm{P}}{1.6}=3\end{aligned}$