In the device shown in Fig, the block `m` is displaced down slightly and released. It starts oscillating. Pulleys are smooth and massless string and springs are also massless. Find time period of oscillations.

Explanation

The given device is equivalent to a device shown in figure. In this equivelent device, the spring `k_(2)` is replaced by a spring of force constant `k_(2)` which is equal to `(k_(2))/(4)`. The equivalent force constant
`4 = (k_(1)k'_(2))/(k_(1)+k'_(2)) = (k_(1)xx(k_(2))/(4))/(k_(1)+(k_(2))/(4)) =(k_(1)k_(2))/(4k_(1)+k_(2))`
The time period of the oscillation of the block is given by
`T = 2pi sqrt((m)/(k)) = 2pi sqrt(((m)/((k_(1)k_(2))/(4k_(1)+k_(2))))) =2pi sqrt((m(4k_(1)+k_(2)))/(k_(1)k_(2)))`