In the figure shown, blocks `P` and `Q` are in contact but do not stick to each other. The lower face of `P` behaves as a plane mirror. The springs are in their natural lengths. The system is released from rest. Then the distance between `P` and `Q` when `Q` is at the lowest point first time will be

Options

A. `(2mg)/(K)`
B. `(mg)/(K)`
C. `(4mg)/(K)`
D. 0

Answer

A

Explanation

Both blocks loose contact immediately after the release
`T_(P) = 2pi sqrt((m)/(4K)) .T_(Q) = 2 pi sqrt((m)/(K))`
`:. T_(Q) = 2T_(P)`
Q comes at lowest position at time `(T_(Q))/(2)` travelling a distance `(2mg)/(K)` downwards.
In time `(T_(Q))/(2)`, i.e. time period of `P(T_(P))` the block `P` come back to original position
`:.` Time distance P and Q is `(2 mg)/(K)`