The correct option is D π√mK+π√m2K
The mass m completes first half of oscillation (upward) under the effect of upper spring alone.
Since, T=2π√mK, time taken for one half of oscillation t1=T2=π√mK
When the mass goes downwards, it is under the combined action of both the springs.
Both springs are in parallel combination.
So, Keq=K1+K2=2K
Time taken to complete second half of oscillation
t2=T2=2π√m2K2=π√m2K
Therefore, total time for 1 complete oscillation
=t1+t2=π√mK+π√m2K