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प्रश्न
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.
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उत्तर
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:
\[\left( \frac{1}{2} \right)k x_0^2 = \left( \frac{1}{2} \right)k( x_1 + x_2 - x_0 )^2 \]
\[ \Rightarrow x_0 = x_1 + x_2 - x_0 \]
\[ \Rightarrow x_1 + x_2 = 2 x_0 . . . (2)\]
\[\therefore x_1 = 2 x_0 - x_2 \]
\[\text{ Putting this value in equation }\left( 1 \right),\text{ we get }: \]
\[ m_1 (2 x_0 - x_0 ) = m_2 x_2 \]
\[ \Rightarrow 2 m_1 x_0 - m_1 x_2 = m_2 x_2 \]
\[ \Rightarrow x_2 = \frac{2 m_2}{m_1 + m_2} x_0 \]
\[\text{ Similarly, x}_1 = \left( \frac{2 m_2}{m_1 + m_2} \right) x_0\]
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