The system shown is lying on a smooth horizontal surface. Mass `m` is constrained to move along the line `3`. Springs `1` and `2` at right angles with respect to one another are symmetrically arranges with respect to `3` which is an elastic cord. The force constants of `1,2` and `3`are `k` each and they are all just taut in the condition shown. For small oscillations of `m` along `3`, time period is `pisqrt((m)/(yk)) (1+sqrt(x))`. The valye of `xy` is___________

Answer

4

Explanation

(1) Let `m` be displaced to left by `x`.

Elongation of spring is `S = x cos 45`
Spring force is `2kS cos 45^(@)` along left `2kx cos^(2) 45^(@) =ikx`
Force due to cord `= kx`
Restoring force for leftward deflection is `2kx`

`2kx = ma`
For left `T_(1) = 2pi sqrt((m)/(2k))`
(2) For right cord becomes slack and only springs provoded restoring force.
Restoring force `= kx`
`kx =ma T_(2) = 2pi sqrt((m)/(k))`
For complete cycle, time period `= (1)/(2)` of `T_(1) +(1)/(2)` of `T_(2)`
`= pi sqrt((m)/(2K)) + pi sqrt((m)/(k)) = pi sqrt((m)/(2k)) (1+sqrt(2))`