Question

Two mass m₁ and m₂ are connected by a spring of spring constant k and are placed on a frictionless horizontal surface. Initially the spring is stretched through a distance x₀ when the system is released from rest. Find the distance moved by the two masses before they again come to rest. 

Answer 1

It is given that two blocks of masses m₁ and m₂ are connected with a spring having spring constant k.
Initially the spring is stretched by a distance x₀.

For the block to come to rest again, 
Let the distance travelled by m₁ be x₁ (towards right), and that travelled by m₂ be x₂ towards left.

As no external force acts in horizontal direction, we can write:
m₁x₁ = m₂x₂    ...(1)

As the energy is conserved in the spring, we get: 

\[\left( \frac{1}{2} \right)k x_0^2 = \left( \frac{1}{2} \right)k( x_1 + x_2 - x_0 )^2 \]

\[ \Rightarrow x_0 = x_1 + x_2 - x_0 \]

\[ \Rightarrow x_1 + x_2 = 2 x_0 . . . (2)\]

\[\therefore x_1 = 2 x_0 - x_2 \]

\[\text{ Putting this value in equation }\left( 1 \right),\text{  we get }: \]

\[ m_1 (2 x_0 - x_0 ) = m_2 x_2 \]

\[ \Rightarrow 2 m_1 x_0 - m_1 x_2 = m_2 x_2 \]

\[ \Rightarrow x_2 = \frac{2 m_2}{m_1 + m_2} x_0 \]

\[\text{ Similarly, x}_1 = \left( \frac{2 m_2}{m_1 + m_2} \right) x_0\]