question_answer

A block of mass 2 M is attached to a massless spring with spring-constant K. This block is connected to two other blocks of masses M and 2 M using two massless pulleys and strings. The accelerations of the blocks are \[{{a}_{1}},\]\[{{a}_{2}}\]and \[{{a}_{3}}\] as shown in the figure. The system is released from rest with the spring in its unstretched state. The maximum extension of the spring is \[{{x}_{0}}.\] Which of the following option(s) is/are correct? [g is the acceleration due to gravity. Neglect friction]

A) \[{{x}_{0}}=\frac{4\text{Mg}}{k}\]

B) When spring achieves an extension of \[\frac{{{x}^{0}}}{2}\]for the first time, the speed of the block connected to the spring is \[3g\sqrt{\frac{\text{M}}{5k}}\]

C) At an extension of \[\frac{{{x}_{0}}}{4}\] of the spring, the magnitude of acceleration of the block connected to the spring is \[\frac{3g}{10}\]

D) \[{{a}_{2}}\,-\,{{a}_{1}}\,=\,{{a}_{1}}\,-\,{{a}_{3}}\]

Correct Answer: D

Solution :

We know, \[2{{a}_{1}}={{a}_{2}}+{{a}_{3}}\]

or         \[{{a}_{2}}-{{a}_{1}}={{a}_{1}}\,-\,{{a}_{3}}\]

and       \[\frac{\text{T}}{{{g}_{app}}}=\frac{8m}{3}\]

\[{{x}_{0}}=\,\frac{16mg}{3k}\]

amplitude of SHM in \[\frac{{{x}_{0}}}{2}\]

\[{{\text{U}}_{\text{max}}}=\omega \text{A}=\left( \sqrt{\frac{\text{K}}{2m+8m/3}} \right).\left( \frac{x}{2} \right)\]

\[=\frac{{{x}_{0}}}{2}\sqrt{\frac{3K}{14m}}\]\[=\frac{8mg}{3K}\sqrt{\frac{3K}{14m}}\]\[=g\sqrt{\frac{64\times 3m}{9\times 14k}}\]

ACC at \[\frac{{{x}_{0}}}{4}\]

\[a={{\omega }^{2}}x=\frac{3k}{14m}.\frac{{{x}_{0}}}{4}\]\[=\,\frac{3\text{K}}{14m}\times \frac{4ms}{3k}\]\[=\frac{4g}{14}=\frac{2}{7}g.\]