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). 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.

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is D

None of these

$\theta ={37}^{\circ }$; I = h = natural length

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

Cos ${37}^{\circ }$ = $\frac{BC}{AC}$ = 0.8 = $\frac{4}{5}$

Ac = (h + x) = $\frac{5h}{4}$ (because BC = h)

So,

$x=\frac{5h}{4}-h=\frac{h}{4}\frac{1}{2}k{x}^2=\frac{1}{2}m{v}^{2}v=x\sqrt{\frac{k}{m}}=\frac{h}{4}\sqrt{\frac{k}{m}}$