Time period of Fig. 1 can be given as Ta=2π√‌
M
k
where, M is mass of the suspended object and k is the force constant. In Fig. 2, both the springs are in series combination, therefore its time period can be given as
Tb=2π√‌
M
keq
=2π√‌
M
k/2
‌‌(∵keq=‌
k×k
k+k
)
Now, ‌
Tb
Ta
=‌
2π√
M
k/2
2π√‌
M
k
⇒‌‌‌
Tb
Ta
=√2...(i) According to question, the ratio of time period of oscillation of two SHM is Tb/Ta=√x, so on comparing it with Eq. (i) we can say, x=2