Two identical springs of spring constant '2k' are attached to a block of mass m and to fixed support (see figure). When the mass is displaced from the equilibrium position on either side, it executes a simple harmonic motion. The time period of oscillations of this system is: 

Calculations:

Given:

It shows two springs of mass "m" and spring constant '2k' attached to a fixed support.

As we know, Springs connections are the same as resistance circuits. Which gets added simply when they are connected in series.

Total spring constant of this circuit , K = 2k + 2k = 4k 

The time period of oscillation is = T = 2π / ω 

Where ω = √(K/m) then time period of oscillation = 2π √(m/K)     -----(1)

The time period of oscillation is = 2π √(m/K) = \( 2π√{m/4k } \) 

Which can be written as T = π \(√{m/k }\)       -----(2) 

Hence, option (3) is correct.