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 be slide without friction in a vertical rod fixed at a distance d from the wall. Initially the spring makes an angles of ${37}^0$ 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 [sin${37}^0$ = 3/5]

• A. $\frac{h}{12}\sqrt{\frac{K}{m}}$
• B. $\frac{h}{6}\sqrt{\frac{K}{m}}$
  
• C. $\frac{h}{4}\sqrt{\frac{K}{m}}$
• D. None of these

Answer 1

The correct option is C $\frac{h}{4}\sqrt{\frac{K}{m}}$

Given, $\theta ={37}^0$

Here, l=h= natural length.

Let velocity when the spring is vertical be 'v'.

$cos{37}^0$= $\frac{BC}{AC}=0.8=\frac{4}{5}$

AC= (h+x) Where x=$\frac{5h}{4}(as,BC=h)$

So, $\frac{5h}{4}-hx=\frac{h}{4}$

From work energy principle, $\frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2$

$v=x\sqrt{\frac{K}{m}}=\frac{h}{4}\sqrt{\frac{K}{m}}$