Question

Two blocks of mass $m_1$ and $m_2$ connected by a string rest on the smooth horizontal plane as shown. A constant forced F starts acting on the block $m_2$:

• A. Length of the spring increases continuously if $m_1>m_{1}2$
• B. Blocks start performing SHM at about the centre of mass with uniform amplitude
• C. Blocks start performing SHM at about the centre of mass with increasing amplitude
• D. Acceleration of $m_2$ is maximum at initial

Answer 1

The correct option is B Blocks start performing SHM at about the centre of mass with uniform amplitude

The only external horizontal force acting on the system of the two blocks and the spring is $F$. Therefore, the acceleration of the centre of mass of the system is equal to $\frac{F}{m_1+m^2}$.
Hence, centre of mass of the system moves with a constant acceleration. Initially, there is no tension in the spring, therefore at initial moment $m_2$ has an acceleration $\frac{F}{m_2}$ and it starts to move to the right. Due to its motion, the spring elongates and tension is developed. Therefore, acceleration of $m_2$ decreases while that of $m_1$ increases from zero initial value.
The block start to perform SHM about their centre of mass and the centre of mass moves with the acceleration calculated above. Hence, option (b) is correct.
Since the blocks perform SHM about centre of mass, therefore the length of the spring varies periodically. Hence option (a) is wrong.
Since the magnitude of the force $F$ remains constant, therefore the amplitude of oscillations also remains constant, therefore the amplitude of oscillations also remains constant. So, option (c) is also wrong.
Acceleration of $m_2$ is maximum at the instant when the spring is in its minimum possible length, which is equal to its natural length. Hence, at initial moments, acceleration of $m_2$ is maximum possible.
The spring is in its natural length, not only at the initial moment but at time $t=T,2T,3T,.....$ also, where $T$ is the period of oscillation. Hence, option (d) is wrong.