Given system is in equilibrium. All surfaces are smooth. Spring is ideal and blocks are sticked at the ends of spring Now `F_(0)` is removed Average normal contact force between wall and mass `m_(1)` upto the time spring attains its natural length for the first time in newton is (given that `F_(0)=2pi` Newton)

Answer

4

Explanation

if `J` is the impulse imparted to the `m_(2)` by spring after removal of `F_(0)` (up to the spring attains its natural length for the first time.)
`(J^(2))/(2m_(2))=(1)/(2)K((F_(0))/(K))^(2)=(F_(0)^(2))/(2K)`
`J=F_(0)sqrt((m_(2))/(K))`
Same impulse is imparted to `m_(1)` by wall.
time requried
`=(T)/(4)=(1)/(4)2pisqrt((m_(2))/(K))=(pi)/(2)sqrt((m_(2))/(k))`
`ltNgt=(J)/((T)/(4))=(F_(0)sqrt((m_(2))/(K)))/((pi)/(2)sqrt((m_(2))/(k)))=(2F_(0))/(pi)`