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 $37{}^{\circ }$ with the vertical when the system is released from the rest. The speed of the ring, when the spring becomes vertical will be

• A. $\sqrt{\left(\frac{3k}{m}\right)h}$
• B. $\frac{h}{2}\sqrt{\frac{3k}{m}}$
• C. $\frac{h}{4}\sqrt{\frac{k}{m}}$
• D. $\frac{5h}{4}\sqrt{\frac{3k}{2m}}$

Answer 1

Answer: (C)

Consider the situational diagram
Here, $\theta =37{}^{\circ },$ $l=h=$ natural length of spring Also, let the velocity be v. From the diagram $\cos 37{}^{\circ }=\frac{BC}{AC}\Rightarrow \frac{4}{5}=0.8=\frac{h}{h+x}$ $\therefore$ $AC=h+x=\frac{5h}{4}\Rightarrow x=\frac{h}{4}$ Applying work-energy theorem, $\frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2\Rightarrow v=x\sqrt{\frac{k}{m}}$ $\Rightarrow$ $v=\frac{h}{4}\sqrt{\frac{k}{m}}$