A spring of force constant k =300 ~N / m connects two blocks having masses 2 kg and 3 kg , lying on a smooth horizontal plane. If the spring block system is released from a stretched position, the number of complete oscillations in 1 minute is 6 n . Find the value of n . Take = 10 .

Step-by-step solution

The total momentum of the system in the horizontal direction is conserved. We draw the F.B.D., assuming the displacement of the block to be x ₁ and x ₂ in opposite directions, and the total extension x is given by, x=x₁+x₂ and m₁ x₁=m₂ x₂ m ₁ d ^2 x ₁ dt ^2 =- k ( x ₁+ x ₂ ) and m₂ d^2 x₂ d t^2 =-k (x₁+x₂ ) after suitably manipulating the equations, we get, d^2 x₂ d t^2 =- -k (m₁+m₂ ) m₁ m₂ x i.e. the frequency = 1 2 k (m₁+m₂ ) m₁ m₂ = 1 2 300(2+3) 2 3 =2.5 ~Hz number of complete oscillations in 1 minute = 60 2.5 =