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 ${37}^{\circ }$ with the vertical when the system is released from rest. Find the speed of the ring when the spring becomes vertical.

Answer 1

$\theta ={37}^{\circ },l=h$natural length

Let the velocity be'v'

$cos{37}^{\circ }=\frac{BC}{AC}=0.8=\frac{4}{6}$

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

Applying work energy principle,

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

$\Rightarrow v=x\sqrt{\left(\frac{k}{m}\right)}=\frac{h}{4}\sqrt{\left(\frac{k}{m}\right)}$