question_answer

One end of a spring of force constant \[k\] is fixed to a vertical wall and the other to a block of mass \[m\] resting on a smooth horizontal surface. There is another wall at a distance \[{{x}_{0}}\] from the block. The spring is then compressed by \[2{{x}_{0}}\] and released. The time taken to strike the wall is

A) \[\frac{1}{6}\pi \sqrt{\frac{k}{m}}\]      

B) \[\sqrt{\frac{k}{m}}\]

C) \[\frac{2\pi }{3}\sqrt{\frac{k}{m}}\]      

D) \[\frac{\pi }{4}\sqrt{\frac{k}{m}}\]

Correct Answer: C

Solution :

[c] The total time from A to C \[{{t}_{AC}}={{t}_{AB}}+t=(T/4)+{{t}_{BC}}\] Where T=time period of oscillation offspring mass system \[{{t}_{BC}}\]can be obtained form, \[BC=A\,B\sin {{(2\pi /T)}^{{{t}_{BC}}}}\] Putting \[\frac{BC}{AB}=\frac{1}{2}\]we obtain \[{{t}_{BC}}=\frac{T}{12}\] \[\Rightarrow {{t}_{AC}}=\frac{T}{4}+\frac{T}{12}=\frac{2\pi }{3}\sqrt{\frac{m}{k}}\].