Two blocks connected by a spring rest on a smooth horizontal plane as shown in figure A constant force F starts acting on the block m2, then :

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blocks start performing SHM about centre of mass the system which moves rectili-nearly with constant acceleration

blocks start performing SHM about centre of mass the system with increasing amplitude

length of spring increases continuously if m₁ > m₂

acceleration of m2 is maximum at initial moment of time only

answer is C.

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Detailed Solution

a_com = $\large \frac{F}{{\left( {{m_1} + {m_2}} \right)}}$

$\large \therefore \,{m_1}\frac{{{d^2}{x_1}}}{{d{t^2}}}\, = \,{m_1}\left( {\frac{F}{{{m_1} + {m_2}}}} \right) - kx.......\left( 1 \right)$

$\large {m_2}\frac{{{d^2}{x_2}}}{{d{t^2}}}\, = \,F - kx - \frac{{{m_2}F}}{{\left( {{m_1} + {m_2}} \right)}}.....\left( 2 \right)\,$

x₁ +x₂ = x $\large \Rightarrow \,\frac{{{d^2}{x_1}}}{{d{t^2}}} + \frac{{{d^2}{x_2}}}{{d{t^2}}} = \frac{{{d^2}x}}{{d{t^2}}}.....\left( 3 \right)$

combining (1), (2) and (3), we can write,

$\large \frac{{{d^2}x}}{{d{t^2}}}\, = \, - \frac{{\left( {{m_1} + {m_2}} \right)k}}{{{m_1}{m_2}}}\left[ {x\, = \,\frac{{{m_1}F}}{{{m_1} + {m_2}}}} \right]$

This equation represents SHM.

$\large \therefore \,x\, = \,\left( {\frac{{{m_1}F}}{{{m_1} + {m_2}}}} \right) + A\cos \left( {\omega t + \phi } \right)$

where $\large \omega \, = \,\sqrt {\frac{{\left( {{m_1} + {m_2}} \right)k}}{{{m_1}{m_2}}}}$

At t=0, x = 0 and $\large \frac{{dx}}{{dt}}\, = \,0$

$\large \therefore \phi = 0\,\,and\,\,A\, = \, - mF\left( {{m_1} + {m_2}} \right)$

so option -(3) is right.

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