Question

One end of a light spring of natural length $d$ and spring constant $K$ is fixed on a rigid wall and the other is fixed to a smooth ring of mass $m$ which can slide without friction in a vertical rod fixed at a distance $d$ from the wall. Initially the spring makes an angles of $37^{\circ}$ with the horizontal as shown in figure. When the system is released from rest, find the speed (in $m / s$ ) of the ring when the spring becomes horizontal.
$\left[\sin 37^{\circ}=3 / 5, \frac{g}{d}=2\right.$ unit and $\frac{k}{m}=16$ unit]

Answer 1

From energy conservation
At point $A$ & $B E _{ A }= E _{ B }$
$\frac{1}{2} kx ^{2}+ mgh =\frac{1}{2} mv ^{2}$
Here $ x=\ell-d=\frac{5 d}{4}-d=d / 4$
$h = d \tan 37=\frac{3}{4} d$
$\therefore \frac{1}{2} k \times \frac{ d ^{2}}{16}+\frac{3\, mgd }{2}= mv ^{2}$
$v=\sqrt{\frac{k d^{2}}{16 m}+\frac{3 g d}{2}}$
$v=d \sqrt{\frac{k}{16 m}+\frac{3 g}{2 d}}$