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 $h$. Initially, the spring makes an angle of $37^\circ$ with the vertical when the system is released from rest. The speed of the ring when the spring becomes vertical is (take $\cos 37^\circ = \dfrac{1}{{1.25}}$)
(A) $\sqrt {\dfrac{k}{m}}$
(B) $\dfrac{h}{2}\sqrt {\dfrac{k}{m}}$
(C) $\dfrac{h}{3}\sqrt {\dfrac{k}{m}}$
(D) $\dfrac{h}{4}\sqrt {\dfrac{k}{m}}$

Answer 1

Here, you need to first picturise the whole situation given to you. Consider a vertical plane in front of you. On that plane, there is a horizontal rod fixed at a height $h$ above the ground. There is a smooth ring which can slide on this rod and this ring is attached to a spring. Now, the spring is attached to the ground. So, you can imagine the motion of the ring and the change that the spring will go through, all happens in that vertical plane only. What you can do is think of the kinetic and potential energy of this system and try to derive the value of velocity of the ring.

Complete step by step answer:
The situation is shown in the above figure. When the spring makes an angle of $37^\circ$, the total length of the spring is $h + x$, where $h$ is the natural length of the spring and $x$ is the elongation produced in the spring. When the spring is vertical, the elongation $x = 0$, as the natural length of the spring and height of the ring from the ground is the same, that is $h$.Now, if you observe carefully, there is no external force acting on the system. By system we mean the ground, ring, spring and the rod. So, the mechanical energy of the system will be conserved.
$M.{E_i} = M.{E_f}$
Initial conditions are $u = 0,$height of ring from ground $= h$, elongation ${x_f} = x$
For elongation, we can apply trigonometry.
\cos 37^\circ = \dfrac{h}{{h + x}} \\\ \Rightarrow\dfrac{1}{{1.25}} = \dfrac{h}{{h + x}} \\\ \Rightarrow\dfrac{1}{{\dfrac{5}{4}}} = \dfrac{h}{{h + x}} \\\ \Rightarrow\dfrac{4}{5} = \dfrac{h}{{h + x}} \\\ \Rightarrow x = \dfrac{h}{4} \\\
Final conditions are height of ring from ground $= h$, final velocity $= v$, elongation ${x_i} = 0$
$M.{E_i} = \dfrac{1}{2}m{u^2} + mgh + \dfrac{1}{2}k{x_i}^2$ and $M.{E_f} = \dfrac{1}{2}m{v^2} + mgh + \dfrac{1}{2}k{x_f}^2$
\Rightarrow\dfrac{1}{2}m{u^2} + mgh + \dfrac{1}{2}k{x_i}^2 = \dfrac{1}{2}m{v^2} + mgh + \dfrac{1}{2}k{x_f}^2 \\\ \Rightarrow\dfrac{1}{2}m{\left( 0 \right)^2} + mgh + \dfrac{1}{2}k{x^2} = \dfrac{1}{2}m{v^2} + mgh + \dfrac{1}{2}k{\left( 0 \right)^2} \\\ \Rightarrow v = x\sqrt {\dfrac{k}{m}} \\\ \therefore v = \dfrac{h}{4}\sqrt {\dfrac{k}{m}} \\\
Therefore, the speed of the ring when the spring becomes vertical is $\dfrac{h}{4}\sqrt {\dfrac{k}{m}}$.

Hence, option D is correct.

Note: In this type of questions, you need to picturise the whole situation given to you and then analyse the motion at initial stage and the final stage. Here, remember that the mechanical energy is conserved because there is no external force acting on the ring. Also, you need to keep track of trigonometric properties, it will help you solve many problems. Keep in mind the formulae and forms of quantities such as potential energy, kinetic energy and potential energy stored in the spring.