question_answer

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)  \[\frac{5h}{4}\sqrt{\frac{3k}{2m}}\]

Correct Answer: C

Solution :

                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, \[\frac{1}{2}k\,{{x}^{2}}=\frac{1}{2}m{{v}^{2}}\Rightarrow v=x\sqrt{\frac{k}{m}}\] \[\Rightarrow \]               \[v=\frac{h}{4}\sqrt{\frac{k}{m}}\]