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 attached to a smooth ring of mass $m$ which can slide without friction on a vertical rod fixed at a distance $d$ from the wall. Initially the spring makes an angle of ${37}^{\circ }$with the horizontal as shown in figure. When the system is released from rest, find the speed of the ring when the spring becomes horizontal.

• A. $d\sqrt{\frac{3g}{2d}+\frac{k}{16m}}$
• B. $d\sqrt{\frac{3g}{d}+\frac{k}{16m}}$
• C. $d\sqrt{\frac{3g}{2d}+\frac{k}{4m}}$
• D. $d\sqrt{\frac{3g}{2d}+\frac{k}{m}}$

Answer 1

The correct option is A $d\sqrt{\frac{3g}{2d}+\frac{k}{16m}}$


If $l$ is the stretched length of the spring, then from figure,

$\frac{d}{l}=\cos {37}^{\circ }=\frac{4}{5}\Rightarrow l=\frac{5}{4}d$

So, the stretch in spring,

$y=l-d=\frac{5}{4}d-d=\frac{d}{4}$

and $h=l\sin {37}^{\circ }=\frac{5}{4}d\times \frac{3}{5}=\frac{3}{4}d$

Now, taking point $B$ as reference level and applying law of conservation of mechanical energy between $A$ and $B$,

$M.E_A=M.E_B$

$\Rightarrow mgh+\frac{1}{2}k{y}^2=\frac{1}{2}m{v}^2$

As for point $B$, $h=0$ and $y=0$ and for point $A$ , $h=\frac{3}{4}d$ and $y=\frac{1}{4}d$

$\Rightarrow \frac{3}{4}mgd+\frac{1}{2}k{\left(\frac{d}{4}\right)}^2=\frac{1}{2}m{v}^2$

$\therefore v=d\sqrt{\frac{3g}{2d}+\frac{k}{16m}}$

Hence, option (a) is the correct answer.