Solve the following :


One end of a spring of natural length $\mathrm{h}$ and spring constant $\mathrm{k}$ is fixed at the ground and the other is fitted with a smooth ring of mass $m$ which is allowed to slide on a horizontal rod fixed at a height $\mathrm{h}$ (figure 8-E13). Initially, the spring makes an angle of $37^{\circ}$ with the vertical when the system is released from rest. Find the speed of the ring when the spring becomes vertical.


$\cos \theta=\cos 37^{\circ}=\mathrm{BC} / \mathrm{AC}=4 / 5$

$A C=(h+x)=\frac{5 h}{4}$

and $x=A C-h=\frac{5 h}{4}-h=\frac{h}{4}$

By work energy principle,

$\frac{1}{2} k x^{2}=\frac{1}{2} \mathrm{~m} \mathrm{}^{2}$

$\mathrm{v}=\frac{h}{4} \sqrt{\frac{k}{m}}$

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