question_answer

The spring block system as shown in figure is in equilibrium. The string connecting blocks A and B is cut. The mass of all the three blocks is m and spring constant of both the spring is k. The amplitude of resulting oscillation of block A is (string massless)

A) \[\frac{mg}{k}\]            

B) \[\frac{2mg}{k}\]

C) \[\frac{3mg}{k}\]                      

D) \[\frac{4mg}{k}\]

Correct Answer: A

Solution :

[B] Just after cutting the string extension in spring \[=\frac{3mg}{k}\] The extension in the spring when block is in mean position \[=\frac{mg}{k}\] \[\therefore \] Amplitude of oscillation \[A=\frac{3mg}{k}-\frac{mg}{k}=\frac{2mg}{k}.\]