Question

One end of a spring of natural length h and spring constant 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 h (following figure). Initially, the spring makes an angle of 37° with the vertical when the system is released from rest. Find the speed of the ring when the spring becomes vertical.

Answer 1

\[\theta = 37^\circ,\text{  l = natural length h } \]

Let the velocity be  \['\nu'\] .

 

\[\cos 37^\circ= \frac{\text{BC}}{\text{AC}} = 0 . 8 = \frac{4}{5}\]
\[ A_C = \left( \text{ h + x }\right) = \frac{5\text{h}}{4}\]

Applying the law of conservation of energy,

\[\frac{1}{2}\text{kx}^2 = \frac{1}{2}\text{m } \nu^2 \]

\[ \Rightarrow \nu = \text{x}\sqrt{\left( \frac{\text{k}}{\text{m}} \right)} = \frac{\text{h}}{4} \sqrt{\left( \frac{\text{k}}{\text{m}} \right)}\]