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 (figure 8-E13). 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.

Figure

Answer 1

$\mathrm{\theta }=37°,l=\mathrm{natural}\mathrm{length}h$
Let the velocity be $\text{'}\nu \text{'}$.



$$\begin{gathered} \cos 37°=\frac{\mathrm{BC}}{\mathrm{AC}}=0.8=\frac{4}{5} \\ {\mathrm{A}}_{\mathrm{C}}=\left(h+x\right)=\frac{5h}{4} \end{gathered}$$

Applying the law of conservation of energy,

$$\begin{gathered} \frac{1}{2}k{x}^2=\frac{1}{2}m{\nu }^2 \\ \Rightarrow \nu =x\sqrt{\left(\frac{k}{m}\right)}=\frac{h}{4}\sqrt{\left(\frac{k}{m}\right)} \end{gathered}$$