Q.

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 :Question Image

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a

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

b

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

c

length of spring increases continuously if m1 > m2

d

acceleration of m2 is maximum at initial moment of time only

answer is C.

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Detailed Solution

acom = \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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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 :