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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