Answer
(a) greater than
(b) $\sqrt 2 T$
Work Step by Step
(a) We know that the time period is greater as compared to the time period of a single spring as the effective spring force constant is now smaller.
(b) We can calculate $T^{\prime}$ as follows:
$T^{\prime}=2\pi \sqrt{\frac{m}{K^{\prime}}}$
As $K^{\prime}=\frac{1}{2}K$
$\implies T^{\prime}=2\pi \sqrt{\frac{m}{(K/2)}}$
This can be rearranged as:
$T^{\prime}=2\pi \sqrt{2(\frac{m}{K})}$
$\implies T^{\prime}=\sqrt 2(2\pi\sqrt{\frac{m}{K}})=\sqrt 2 T$
