Question

Two blocks each of mass $m$ are connected with springs of force constant $k$. Initially springs are relaxed. Mass $A$ is displaced to left and $B$ is displaced towards right by same amount and released then time period of oscillation of any one block (Assume collision to be perfectly elastic)

• A. $2\pi \sqrt{\frac{m}{k}}$
• B. $2\pi \sqrt{\frac{m}{2k}}$
• C. $\pi \sqrt{\frac{m}{k}}$
• D. $\pi \sqrt{\frac{m}{2k}}$

Answer 1

The correct option is C $\pi \sqrt{\frac{m}{k}}$

When the block come back they will equal velocities but in opposite direction. So they will stop after collision. During their motion, they only can complete their half oscillation. So time period of any one block is half of its original time period. Original time period of any block $T=2\pi \sqrt{\frac{m}{k}}$

So, the new time period $\frac{T}{2}=\pi \sqrt{\frac{m}{k}}$