One end of a spring of natural length $ℓ_{0} = 0.1 m$ and spring constant $k = 80$ N/m is fixed to the ground and other end is fitted with a smooth ring of mass $m = 2 g m$ , which is allowed to slide on a horizontal rod fixed at a height h $= 0.1 m$ . Initially, the spring makes an angle of $37^{o}$ with vertical when the system is released from rest.
When the spring becomes vertical, speed of ring is ' $v$ ' m/s, find ' $v$ '. $(cos 37^{\circ} = \frac{4}{5})$

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Solution

Loss in elastic potential energy $=$ gain in kinetic energy
$\Rightarrow \frac{1}{2} k (l - l_{0})^{2} = \frac{1}{2} m v^{2}$
$\Rightarrow v = (l - l_{0}) \sqrt{\frac{k}{m}}$
$= (\frac{l_{0}}{c o s 37^{0}} - l_{0}) \sqrt{\frac{k}{m}}$
$= l_{0} [\frac{1}{c o s 37^{0}} - 1] \sqrt{\frac{k}{m}}$
$= (0.1) [\frac{5}{4} - 1] \sqrt{\frac{80}{2 \times 10^{- 3}}}$
$= 0.1 \times \frac{1}{4} \times 2 \times 10^{2}$
$= 5 m / s$

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