$m_{1}$ and $m_{2}$ are connected with a light inextensible string with $m_{1}$ lying on smooth table and $m_{2}$ hanging as shown in figure. $m_{1}$ is also connected to a light spring which is initially unstretched and the system is released from rest –

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maximum displacement of $m_{1}$ will be $\frac{2 m_{2} g}{k}$

system performs SHM with angular frequency given by $\sqrt{\frac{k (m_{1} + m_{2})}{m_{1} m_{2}}}$

system performs SHM with angular frequency given by $\sqrt{\frac{k}{m_{1} + m_{2}}}$

tension in string will be $0$ when the system is released.

answer is B, D.

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Detailed Solution

The equivalent system is shown in figure

$∴ ω = \sqrt{\frac{k}{(m_{1} + m_{2})}} L e t m a x i m u m d i s p l a c e m e n t o f m_{1} b e x (a s s u m i n g i t i s s t i l l o n t h e t a b k e t o p) T h e b l o c k s m o m e n t a r i l y c o m e t o r e s t a t t h i s i n s \tan t . c o n s e r v i n g e n e r g y, \frac{1}{2} k x^{2} + (- m_{2} g x) = 0 \Rightarrow x = \frac{2 m_{2} g}{k}$ .